H
23
Chapter 9
of Ramanujan's
Second Notebook
Infinite Series Identities,
Transformations, and Evaluations
Bruce C. Berndt
Padmini T. Joshi
American Mathematical Society

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VOLUME 23 Chapter 9 of Ramanujan's second notebook-
Infinite series identities, transformations,
and evaluations
Bruce C. Berndt and Padmini T. Joshi

Contemporary
Mathematics
23
Chapter 9
of RamanujarTs
Second Notebook
Infinite Series Identities,
Transformations, and Evaluations
Bruce C. Berndt
Padmini T. Joshi
American Mathematical Society
Providence, Rhode Island

1980 Mathematics Subject Classification. Primary 33A15, 33A30, 30A10, 10A40, 40-00.
Library of Congress Cataloging in Publication Data
Berndt, Bruce C., 1939-
Chapter 9 of Ramanujan's second notebook.
(Contemporary mathematics, ISSN 0271-4132; v. 23)
Bibliography: p.
1. Series, Infinite. 2. Power series. 3. Transformations (Mathematics) I. Joshi,
Padmini T., 1927— . II. Title. III. Title: Ramanujan's second notebook.
IV. Series: Contemporary mathematics (American Mathematical Society); v. 23.
QA295.B48 1983 515'.243 83-11803
ISBN 0-8218-5024-5
Copyright © 1983 by the American Mathematical Society
Printed in the United States of America
All rights reserved except those granted to the United States Government
This book may not be reproduced in any form without the permission of the publishers
This volume was printed directly from author prepared copy.

TABLE OF CONTENTS
Preface vii
Introduction 1
The entries in Chapter 9 of Ramamijan's Second Notebook 3
v

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PREFACE
Chapter 9 represents one of the more unified chapters in the
second notebook. As in other chapters, it contains a mixture of both new
and old results. This chapter is characteristic of Ramanujan's greatest
talent and love - dazzlingly beautiful formulas for infinite series.
Apery's proof of the irrationality of £(3) is but one example of many
demonstrating that elegant formulas are often very useful as well. Chapter 9
contains many formulas bearing the same features as Apery's formula for C(3).
We trust that some of Ramanujan's formulas herein will be valuable, indeed, to
present day researchers. But we also hope that these marvelous formulas will
foster the same kind of majestic thrill and sublime upliftment that listening,
for example, to a Beethoven symphony engenders.
Urbana, Illinois
June, 1983
vii

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INTRODUCTION
Chapter 9 fully illustrates Hardy's declaration [65, p. xxxv],
"It was his insight into algebraical formulae, transformations of infinite
series, and so forth, that was most amazing." This chapter has 35 sections
containing 139 formulas of which many are, indeed, very beautiful and elegant.
In this paper, we prove (or, in some cases, disprove) each of these formulas.
In Chapter 9, Ramanujan gives several transformations of power series
leading to many striking series relations and attractive series evaluations.
Most of Ramanujan's initial efforts in this direction pertain to the
dilogarithm and related functions. As is to be expected, these results
are not new and can be traced back to Euler, Landen, Abel, and others.
However, most of Ramanujan1s remaining findings on transformations of power
series appear to be new.
The beautiful formula
(o.d to).f I (-l)k+1<;!)2
k-1 (2k)! k"3
has been made famous by Apery's proof of the irrationality of £(3) [3], [18],
[55], [67], [77]. Ramanujan evidently missed this formula, but Chapter 9
contains several intriguing formulas of the same type. Some of these
involve C(3) and Catalan's constant.
In Chapter 8, Ramanujan studies certain functions which are akin
to Log r(x + 1) [6 1. In sections 27-30 of Chapter 9, Ramanujan returns
to this topic. The generalization studied here is very closely related to
that studied by Bendersky [5] and more recently by Busing [15].
* Research partially supported by National Science Foundation grant
no. MCS-7903359.
1

2
B. C. BERNDT AND P. T. JOSHI
Except for a simple result in section 3 1, the material in sections 27-30
has no relation to the rest of Chapter 9.
In analyzing Ramanujan's work, Hardy has frequently pointed out
that "he knew no theory of functions" [^3, p. 14]. Many of the formulas
in Chapter 9 can be extended by analytic continuation to complex values of
x. However, because Ramanujan obviously intended his results to hold for
just real values of x, we have presented his theorems in this more restricted
setting. We have made exceptions to this decision in the few instances when
vacuous theorems would otherwise result.
Several of Ramanujan's formulas in Chapter 9 need minor corrections.
However, there are a few results, for example, Entry 3 and formula (11.3),
which are evidently quite wrong. In describing three beautiful formulas of
Ramanujan which he could not prove, Hardy [A3, p. 9] has written, "They must
be true because, if they were not true, no one would have had the imagination
to invent them." Clearly, Hardy's bold pronouncement is invalid here. However,
in essence, he may be correct, because very likely "corrected versions" of
Ramanujan's incorrect formulas exist. Unfortunately, we have no insights
as to what these "corrected versions" might be.

THE ENTRIES IN CHAPTER 9 OF
RAMANUJAN'S SECOND NOTEBOOK
Entry 1. For each positive integer r, define
- I( * - l 1>
k=ol (2k+l-a}r r2k+l+a)r J
S
r k-CA (2k+l-a)A (2k+l+a)
where a is real but not an odd integer. Assume that |x| < tt. Then if
r is an odd positive integer,
k 2k
fn^afJV -U 1 - ^v r*ne f9L> M 1 4- a\vl (r-l)/2 ("1) Sr_2kX
(2k)!
(i) (x) = T fcos<2k + 1 - a)x _ cos(2k + H a)x\ = y
r k=0 H2k + 1 - a)r (2k 4- 1 + a)r } k«0
while if r is an even positive integer,
.... , . = r [sin(2k + 1 - a)x sin(2k + H a)x\ _ r(?2 ~1 ("1) Sr-2k-l*
ui; s ex; - i i - r / ~ «• (2k + n»
r k=0 V (2k + 1 - a)r (2k + 1 + a)r j k=0 {• L)'
Proof. We first establish (i) for r ■ 1. Observe that
oo
/■i -i \ /\ o /\r cos(2k + l)x
(1.1) c1 (x) - 2a cos(ax) \ r—*—_-
k=0 (2k + IT - a
. 9 c,n/avx V (2k + l)sln(2k + l)x
+ 2 sin(ax) 2, o 2 •
k=0 (2k + 1) - a
Now for |x| < tt [12, p. 368],
n ?^ tt cos (ax) _ 1 r (-1) cos(nx)
{L'Z) 2a sin(aTr) 7"2 + i 2 2 »
Za n«l n — a
from which it follows that, if 0 < x < tt,
oo
/, on JL COS a(X - TT) 1_ r
U,J; 2a sin(aTr) " 0 2 " £
2 2
2a n=l n - a
3

4
B. C. BERNDT AND P. T. JOSHI
Subtracting (1.3) from (1.2), we get, for 0 < x < tt,
/i / \ o V cos(2k + l)x i\ / / \ / \\
(1.4) 2 I 2 2 * 2a sinfair) {cos<ax> " cos a(x-7r)}.
k=0 (2k + 1)Z - a a sinU7T;
Differentiating both sides of (1-2) with respect to x and proceeding as
above, we find that, for 0 < x < tt,
/i c\ o V (2k + l)sin(2k 4- l)x tt / . / \ / xi
(1.5) 2 ;> ■* '—^ j—t- = sin(a7Tx isin(ax) - sin a(x - tt)}.
k=0 (2k + 1) - a sirnair;
Substituting (1.4) and (1.5) into (1.1), we deduce that, for 0 < x < tt,
(1.6) c, (x) = :—t—r- (cos (ax) {cos (ax) - cos a(x - tt)}
l l sin (air;
+ sin (ax) (sin (ax) - sin a(x - tt)})
= y tan(aTr/2)
2a I ~2 2 = sr
k=0 (2k + 1)Z - a L
Trivially, (1.6) holds for x = 0. Replacing x by -x, we see that (1.6)
has thus been proven for |x| < tt . This completes the proof of (i) for r - 1.
If we integrate the extremal sides of (1.6) over [0,x], |x| < tt,
we readily find that
s2(x) = S1x,
which is in agreement with (ii) when r ■ 2.
We now proceed by induction. Assume that (i) is valid for some
positive odd integer r. Integrating both sides of (i) over [0,x], |x| < tt,
we find that ,
(r-l)/2 (-l)kSr 2kx2k+1
(1.7) sr+1(x) = J (2k 4 1)! '

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
5
which is precisely (ii) with r replaced by r + 1. Integrating (1.7)
over [0,x], |x| < it, we find that
(r-l)/2 (-l)kS x2k+2
-cr+2(x) + Sr+? = I
~r+2v~' r+2 ^ (2k + 2)!
c
,<*> = I
(r+l)/2 (-l)kS_0 „x2k
r+2-2k
r+2 k=0 (2k)I
which is (i) with r replaced by r+2. This completes the proofs of
both (i) and (ii).
Entry 2. For each integer r with r >^ 2, define
S
r k<
* 1 l ^ ~ r}.
=0 ^(2k + 1 - a) (2k + 1 + a) J
where a is real but not an odd integer. Assume that 0 < x < tt. Then
if r is an even positive integer,
O (x) = 1 [cos(2k + 1 - a)x | cos(2k + 1 + a)x 1
r k=0 H2k + 1 - a)r (2k + 1 + a)r /
r/2-1 (-DkSr.2kx2k (,1)r/2TOr-l
" k=0 (2k)! 2(r - 1)! 9
and if r is an odd positive integer,
(ii) s (x) = ? (sln<2k + 1 - a)x + sin(2k + 1 + a)x \
r k=0 V (2k + 1 - a)r (2k + 1 + a)r '
(r-3)/2 WV^1 ^(r-D/2^-1
" £ (2k + 1)! + 2(r - 1)!

6
B. C. BERNDT AND P. T. JOSHI
Proof. We first establish (ii) for r = 1, Using (1.4)
and (1.5), we find that, for 0 < x < it ,
(2.1) sl(x) - 2 cos(ax) I (2k * 1)s%(2k %1)x
1 k=0 (2k + 1) - a
o . / v T cos(2k + l)x
- 2a sin (ax) I * r—'—~
k=0 (2k + 1) - a
= -y—:—7—r (cos(ax)(sin(ax) - sin a(x - tt)}
- sin (ax) (cos (ax) - cos a(x - tt)})
7T_
"" 2 >
which proves (ii) when r ■ 1.
Integrating both sides of (2.1) over [0,x], 0 < x < tt, we find
that
-c2(x) + S2 = ^,
which establishes (i) for r = 2.
Proceeding by induction on r, we assume that (i) is valid for
an arbitrary even positive integer r. Integrating (i) over [0,x],
0 < x < tt, we readily achieve (ii) with r replaced by r + 1. A
second integration yields (i) with r replaced by r + 2. Since the
details are like those in the previous proof, we omit them. This completes
the induction.
Ramanujan [66, vol. 1, p. 143; vol. 2, p. 104] supplies the
following incomplete hint for his apparently invalid argument: "In both
1 & 2 expand the series in ascending powers of x and apply.11
The last term on the right sides of both (i) and (ii) in Entry 2
is absent in the notebooks.

CHAPTER 9 OF RAMANUJAN1S SECOND NOTEBOOK
In preparation for Ramanujan's next formula, we make some
definitions. Let
(3.1) H - I ±
n k-1 *
S (X) = T (-Dksin(2k + l)x ^ c (x) = y (-l)kcos(2k + l)x
n k=0 (2k + l)n n k=0 (2k + l)n
and
» (-l)V+1cos(2k + l)x
V> (x) = I -
k«0 (2k + l)n
where n is any natural number and x is real. Unfortunately, Entry 3
is false for at least n sufficiently large. We are unable to offer a
corrected version of Ramanujan's formula. It appears that if a corrected
formula exists, its shape would be significantly different.
Entry 3. Let S , C and *p be defined as above. If n is
z— n* n n
an odd integer at least equal to 3, then i
(3.2) * (x) - <P (x) = xS^ 9(x) - xS (x) + nC (x) - nC .. (x).
n-z n n-z n n-1 n+1
Disproof (for n sufficiently large). First, observe that
(3.2) is certainly false for all n if x is any odd multiple of tt/2.
If x is not an odd multiple of tt/2, then a brief calculation
shows that
(3.3) ,^M - ,uW * . L1&&L .
as n tends to ». On the other hand, a similar argument shows that
(3.4) xSn-2(x) " xSn(x) + nCn^1(x) - nCR+1(x)
% 8x sin(3x) 8n cos(3x)
3n " 3n+1
as n tends to ». For large n, (3.3) and (3.4) are incompatible.

8
B. C. BERNDT AND P. T. JOSHI
In order to state Entries 4(i) and (ii), we need to make several
definitions. For each nonnegative integer n, define
v m - V f2k\ (-D sin(2k + l)x , n r ,2k. (-l)ksin(2k + 2)x
nU; " i ^ k; ~2k n ^ ' i ( k' ~2k n
k=0 K 2 K(2k -I- l)n k=0 K 2 K(2k + 2)n
.n r .2k. (-l)ksin(2k + 2)x
k«0 K 2 K(2k + 2)n L
♦ W-I (?) (^)kcOS(2k^1)X + <-l>n I (?) ^)kcos(2k-,2)x
n k=0 R 2ZK(2k + l)n k=0 k 2ZK(2k + 2)n
( .n r ,2k (-1) cos(2k + 2)x
^" ; ^ * k ' 2k n+1
k=0 K 2 K(2k + 2)n+i
and
,2kx
*(n)=7 i (k}~2k n+T
77 k=0 k 2ZK(2k + l)n+i
Sn = (1 - 21"n)C(n),
where £ denotes the Riemann zeta-function.
Entries 4(i), (ii). Let |x| < tt/2 and let n be an integer.
If n >_ 0, then
(4.1)
[n/2] _.k n-2k k
I. !:"?^. I. 2-2jS2j^(2k-2j)
k=0
(n - 2k)!
3-0
y sin(7rn/2)F (x),
y cos(7rn/2)i|;n(x),
if n is odd,
if n is even.
If n >_ -1, then
(4.2)
k=0
[n/2] nkn-2k k
y sin(7rn/2)Fn+1(x), if n is odd,
[ j cos(7Tn/2)^n+1(x), if n is even.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
9
Before commencing a proof of (4.1) and (4.2), we offer a few
comments. Remanujan further defines
* An = ¥ + <T> + (T} +",+ (X" 2> •
where evidently x is meant to be a positive multiple of it. Ramanujan's
versions of Entries 4(i) and (ii) also contain formulas for F (x) and
\\) (x) in terms of A, , k £ n - 1. Because (4.1) and (4.2) implicitly
indicate that x is a continuous variable, with |x| < tt/2, and because
A, is defined for only values of x that are positive integral multiples
of tt, Ramanujan's formulas involving A, appear to have no sensible
interpretation, and we shall not make any further comments about these
formulas.
In Ramanujan's second published paper [61], [65, pp. 15-17), he
establishes a recursive formula for ^(n) in terms of S, , 1 £ k <_ n.
This recursion is also given by Ramanujan in Chapter 10, section 13 [7].
Proof of Entries 4(i), (ii). We proceed by induction. We first
establish (4.1) for n = 0. If L denotes the left side of (4.1) when n = 0,
then L = SQ^(0). Since C(0) = - 1/2, we find that SQ - 1/2. Now in
Proposition 4(vii) below, which actually holds for |x| <^ tt/2, set x = tt/2
to deduce that <p(0) = 1. Hence, L = 1/2. On the other hand, by Propositions
4(iil), (iv), and (vi) below,
1 ,i. t \ 1 /cos(x/2) , cos(3x/2) , ... « ^ _\ 1
-j ^qM = 2" 1 cos(x/2)/2 cos x + 1} - y ,
COS X COS X J
and so (4.1) is valid for n * 0. (Propositions 4(v), (vii), and (ix)
below can be used to provide a direct proof of (4.1) when n = 1.)
We now prove (4.2) for n = -1. In this case, the left side
of (4.2) is understood to be equal to 0. On the other hand, by Propositions

10
B. C. BERNDT AND P. T. JOSHI
4(il), (iii), and (v),
. 1 F (x) = - \ /sin(x/2) - Sin(3x/2) + sin(x/2)/T7o7T ) - 0,
W2 cos x v2 cos x '
as desired.
We also prove (4.2) for n = 0. In this case, the left side of
(4.2) is equal to
00
ty(1)-£ I (2kk)^F-j:—2 = iLog2>
0 * k=0 k 2Zk(2k + 1)Z
by Example (i) in section 31. On the other hand, by Propositions 4(vi),
(viii), and (x),
y ^,(x) = ■=■ {Log(/cos x + /2" cos(x/2)) - cos(x/2)/2 cos x + 1
+ cos(x/2)/2 cos x - Log(/cos x + f? cos(x/2)) + Log 2-1}
= \ Log 2,
and so (4.2) is valid for n = 0.
Proceeding by induction, we now assume that (4.1) and (4.2) are
valid for any fixed, positive even integer n. Integrating (4.1) over [0,x],
|x| < it/2, we readily find that
n/2 k n+l-2k k
<*-3) j0 Fli-ni ^02"2jv(2k-2j)
-y sin((n + l)ir/2)Fn+1(x).
Thus, we have established (4.1) with n replaced by n + 1.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
11
Integrating (4.3) over [0,x], |x| < ir/2, we find that
n/2 ( .k n+2-2k k
(4-4) JQ fai 2 -2k)I .\0 2 \f^ ~ ^
{00
2k> (-Dk
22k(2k + l)n+2
oo V oo If \
+ y (2k) ±±T. y (2k) tlf \
k=0 K 2Zic(2k + 2)n+Z k=0 * 2ZIC(2k + 2)n+J J
Comparing (4.4) with (4.1), with n replaced by n+2, we find that we
must show that
n/2 +1 -, r °° oi. / ,xk
"n+2
(4.5) I 2-2^S2/(n + 2 - 23) - ± \\ (2k) 2k ^
j-0 ^ <-k=0 K 2 (2k + 1
oo If oo \e \
+ y (2k) (-D _ y (2k) till I
^ V k ' 2k n+2 ^ l k } 2k n+3 / *
k=0 K 2ZK(2k + 2V k=0 R 2 R(2k + 2)n+J J
In a similar fashion, after two integrations of (4.2), we find that
it suffices to show that
n/2 +1 «. i f °° oi / nk
(4.6) I 2~2h *(n + 3-2j)=i I I (2kk) 2k (-1} n+3
j-0 J lk=0 k 2Zlc(2k + l)n+J
y (2k) (-Dk + y (2k) tnH \
k-0 k 22k(2k + 2)n+3 kto k 22k(2k + l)n+4 /*
Combining (4.5) and (4.6) together, we deduce that, in order to
prove Entries 4(i), (ii), it suffices to prove the following curious theorem.
Theorem. Let n denote a nonnegative integer. Then
n
(4,7) I 2~JS,*(n - j)
j-0 ^
2|j
\ I (2£)(-4)~k{(2k + l)"n + (-l)n(2k + 2)"n - (-l)n(2k + 2)~n~1}.
k=0
We are very grateful to R. J. Evans for providing the following
elegant proof of (4.7).

12 B. C. BERNDT AND P. T. JOSHI
Proof. Let L and R denote, respectively, the left and right
n n
sides of (4.7). Define
oo oo
L(x) = I L xn and R(x) = £ R x",
n=0 n n=0 n
where |x| is sufficiently small. It then suffices to show that
(4.8) L(x) = R(x).
Next, define, for j >_ 0,
f2~jS,, if 2 | j,
(4.9) Tj =■
J
0 , if 21]
Then, by (4.9) and the definition of L ,
and
(4.10)
n
L - I tjp(n - j),
n j-o J
L(x) = T(x)#(x),
n
n >_ 0,
where
OO 00
T(x) = I T xn and *(x) - £ *>(n)xn,
n«0 n n«0
for |x| sufficiently small. We shall compute L(x) by determining
T(x) and <f>(x).
First, since SQ = 1/2,
oo oo oo k+1
(4.11) X(x) - I S2n4-x2* - | + I 4"\2n J S^
n=0 n=l k=l k
k-1 n«l
± + I (-l)**1 X'
2 ii /J 2
k-1 4k - x

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
Next, since
(4.12) (1 - x)"1/2 - I (2kVkxk, -1 < x < 1,
k=0
we find that, for v < 1,
I (2k) i
k«0 k 4k(2k + 1 - v) JO
f1 „ 2,-1/2 -v.
I (1 - x ) x dx
(1 - u) u
1/2 -(v+D/2
ra/2>r(^> ^Tr(if)
2T<^>
2r(^f)
Differentiating n times with respect to v, we find that
,2kx
1 «T
£ V (ZK) i
T\ .Ln k ,k/01 Nn+1 r- , , n
k=0 4 (2k + 1 - v) /tt n! dv
r(if)
lr(^> J
Setting v = 0 yields
<f>(n)
1 d"
'r(^> i
/F n! dv
lr<2?> J
v=0
Hence,
(4.13) *<x) = — 2
/F r£p)
Therefore, by (4.10), (4.11), and (4.13),
(4.14) L(x) --i- r(^)r(-^) .
2/F 2 2
We next compute R(x). By (4.12),
(4.15) (1 4- X)"1/2 = J (2k)(-4)"kxk,
k=0 k
-1 < x < 1.

14
B. C. BERNDT AND P. T. JOSHI
Thus, for v < 1,
N 1 r ,2k, (-l)k 1 f ,, _,_ 2,-1/2 -v.
) J l (k)-k ' 21 (1 + X } X dx
k«0 4 (2k + 1 - v) j0
n . ,-1/2 -(v+l)/2, _ _
(1 + u) u du - F
0
say. Differentiating n times with respect to v, we find that
1 ? /21S (~Dk . F(n)(v)
2 /. * k; .k/01 . . ,n+l n!
k=0 4 (2k + 1 - v)
Recalling that R denotes the right side of (4.7), we deduce that
„ - F01"1^) , (-pV^C-l) (-l)nF(n)(-l) n>.
n " (n - 1)! + (n - 1)! nl ' n °
and, with the help of (4.15),
Ro = Si
Thus,
k ~ F(-l)•
(4.i7) R(x) i + y (i^lm + w)V^ikil xn
n—1
n«0 n'
fi + xF(x) - xF(-x-l) - F(-x-l)
1
^ + xF(x) + (-x-l)F(-x-l)
To determine xF(x), we return to (4.16), and integrating
we deduce that, for x < 1,
(4.18) xF(x) --yf (1 + u)"1/2u1/2d(u"x/2)
J0
1 ^ 1 f1 /i a. \"3/2 -(x+l)^
■ + t (1 + u) u du.
2/2 4 Jo

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
Furthermore, replacing u by 1/u, we also find that, for x < 1,
(4.19) xF(x) - - -L- + i | (1 + u)"3/2ux/2du.
2/2 * Jl
Hence, from (4.18) and (4.19), if -2 < x < 1,
xF(x) + (-x - l)F(-x - 1) « - — + y
/2 A
.00
n . ,-3/2 x/2.
(1 + u) u du
.J,^_^r(2^)r(lz2L)
•2 2/F 2 2
We therefore conclude from (4.17) that
R(x) = JL T(*±l)T(±JL)m
2/F
By (4.14) and the foregoing equality, we deduce (4.8), which completes
proof.
Proposition 4. For |x| < ir/2, we have
(il) Sx E I l-£ W>1 sin(2k + l)x
k=0 2 (k!) /2 cos x
/win r - V (-Dk(2k)! ,0. . ,* cos(x/2)
(lil) C = I —zr— ' cos(2k + l)x = -—ZZZZ »
1 k«0 2 K(k!) /2 cos x
(i2) s2 3 I (-\t+l(?)l •*«•»> -£iILi^ .
k-0 2ZK(k!) ^2 cos x
,..,, p - r (-l)k(2k)! ... , cos(x/2) ,
(ii2) C- = I -±jr— J cos(2kx) = -—ZZZ
k=0 2 (k!) /2 cos x
(ili) s, s I (-1)k(y sin(2k + 2)x - sln(3x/2) ,
k=0 2^K(k!)Z /2 cos x
(iv) C, = J (^)k(2^! cos(2k + 2)x = COS<3^2> ,
k-0 2^(k!r /2 cos x

16
B. C. BERNDT AND P. T. JOSHI
(V) S^ = I (;i)k(2^)! ^Sln?(;V22)X - sin(x/2)/rToTT .
* k=0 2ZK(kir Zk + Z
/ .n n - V (~Dk(2k)! cos(2k + 2)x , ,-.« ,
(vi) C = ), ' ' ' 2k . 2 - cos(x/2)/2 cos x -1 ,
* k=0 2ZK(k!)Z ZK Z
, ,.v c - V (-Dk(2k)t sin(2k + l)x . -1//T , , /0v*
(vii) sc = 1 ou o on u. i = sin (/2 sin(x/2)),
5 k=0 22k(k!)2 2k+1
t ---N n - V (-Dk(2k)! cos(2k + l)x T (} . «■ , ,_.
(vin) C. = ) ' ' 0 = Log(/cos x + /Z cos(x/2)),
D k=0 2ZR(k!) ZK X
(ix) s - J (-l)k(2k)i sin(2k+2)x
6 k=0 22k(k!)2 (2k +2)2
= sin(x/2)/2 cos x + sin'^/I sin(x/2)) - x,
and
, v r - V (-Dk(2k)! cos(2k + 2)x
K ' 6 = I "~~2k 2 2—
k=0 2^(k\y (2k + 2)1
= cos(x/2)/2 cos x - Log (/cos x + /2* cos(x/2)) -»- Log 2-1.
In all of the equalities of Proposition 4 and throughout all of
their proofs below, we take the principal branches of all multi-valued
relations.
Proof of (il), (iil). For |x| < ir/2,
r + *« - V (-D (2k)! r(2k+l)ix
1 lSl " '• ~!k 2
k-0 2Zk(k!)
ix,. , 2ix x"1/2
* e (1 + e )
(cos(x/2) + i sin(x/2))(2 cos x)"1/2.
Equating real and imaginary parts above yields (iil) and (il), respectively.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK 17
Proof of (i2), (ii2). For |x| < it/2,
c . is I (-l)k»k)l e2kix
2 2 k=0 22k(k!)2
2ix -1/2
(1 + e X)
= (cos(x/2) - i sin(x/2))(2 cos x)"1/2,
and the results follow as before.
Proof of (iii), (iv). For |x| < tt/2,
, + 1S, - I *£&&& e(2k+2)ix . e2ix(l + e2ix)"1/2,
J J k«0 2ZR(k!)Z
from which the desired equalities readily follow.
Proof of (v), (vi). For |x | < tt/2,
+ 1S - y (-dW e(2k+2>ix
4 ~ ^ n2k/f .,2 2k + 2
4 " k=0 2"(k!)'
ri _,. 2ixv1/2 . ix/2/= ,
(1 + e ) - 1 ■ e /2 cos x - 1.
Equating real and imaginary parts on both sides above, we complete the proof.
Proof of (vii), (viii). For lx| < tt/2,
C + 1S - V (-l)k(2k)l e(2k+1)i*
5 5\i0 22k(k!)2 2k + 1
. ,-1, iXv
smh (e )
t / ix . /i . 2ixN
= Log(e + /l + e )
. / ix , ix/2/= x
= Log(e + e /2 cos x).

18
B. C. BERNDT AND P. T. JOSHI
Hence,
(4.20)
and
i 2 2
Cc = y Log{(cos x 4- cos(x/2)/2 cos x) + (sin x 4 sin(x/2)/2 cos x) }
"5 2
= ~ Log{l 4- 2 cos x 4- 2 cos(x/2)/2 cos x;
= — Log{cos x 4 2 cos (x/2) + 2 cos(x/2)/2 cos x}
= Log(/cos x 4 il cos(x/2)),
(4.21) S. - tan
-1
sin
cos
X
X
4
4
sin(x/2)/2
cos(x/2)/T
cos
cos
X
X
-1
f/2 sin(x/2){/I cos(x/2) 4 /cos xj ]
/cos x{/cos x 4 /7 cos(x/2)}
-1
tan
f/2 sin(x/2)
[/I - 2 sin2(x/2)
sin"1(/2 sin(x/2)).
Proof of (ix), (x). For |x|<tt/2, put
c +iS = " (-Dk(2k)! e^wnx
6 6 k=0 22k(k!)2 (2k +2)
IX
2 " f(">.
where u * e . Observe that
uf'(u) = /l 4 u2 - 1.
Thus,
f(t) = [ i (A 4 u2 - l)du
h u
= {/l 4 u2 - Log(l 4/l 4 u2)}L
= /l 4 t - 1 4 Log2 4 Log t - Log(~ 4 /t"2 4 1).

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
19
Hence,
C + iS6 * elx/2/2 cos x - 1 + Log 2 - ix
- -=- Log{(cos x + cos(x/2)/2 cos x) + (sin x + sin(x/2)/2 cos x) }
+ i tan
sin
cos
X
X
+
+
sin(x/2)/2
cos(x/2)/T
cos
COS
X
x J
ix/2
/2 cos x - 1 + Log 2 - ix
Log(/cos x + vT cos(x/2)) + i sin~1(/2 sin(x/2)),
by (4.20) and (4.21). Equating real and imaginary parts above, we deduce
(x) and (ix), respectively.
Entry 5. Let a, n, and 0 be real with n >^ 0 and |6| <_ tt/2.
Then
oo
(i) S = I (£)sin(a + 2k)6 = 2nCosn6 sin(a + n)6
k=0
and
00
(ii) CE J (£)cos(a + 2k)6 = 2ncosn0 cos(a + n)0.
k«0 *
Proof. By Stirling's formula, the series in (i) and (ii), indeed,
do converge (absolutely) for n >^ 0. Now,
n j. -o V /n\ (a+2k)i6 iaS.. . 2i0.n
C + iS = J, (i,'e * e (1 + e )
k=0 k
i(a+n)0/o QXn
■ e (2 cos 0) .
Equating real and imaginary parts on both sides above, we deduce (ii) and
(i), respectively.
If |z| <_ 1 and n is a natural number with n ^ 2, the poly-
logarithm LiR(z) is defined by
(6,1) Li*(z) = J, £ •
k=l k

20
B. C. BERNDT AND P. T. JOSHI
Furthermore, set
(6.2) 2x(z) - Li (z) - Li (-z) = 2 £
2k+l
z
n n k=0 (2k + 1)°
Observe that, for |z| < 1,
(6.3) Li,(z) = -f L°6(1 - w)
2 Jo w
dw -
r2 j
dw
0 w
1 du
o1""'
where the principal branch of Log(l - w) is assumed. (The latter expression
for Li«(z) suggests the terminology "dilogarithm.") Equality (6.3) may
be used to define Li«(z) for all complex z. By employing the equality
rz Li ., (w)dw
Li (z) = -^ ,
JO
we may, by induction, analytically continue Li (z), n >^ 2, to the entire
complex z-plane. In sections 6 and 7, Ramanujan derives several properties
of the dilogarithm Li~(z) and trilogarithm Li~(z).. Since most of these
results are known, we shall not give complete proofs but refer to Lewin's
book [52] where proofs may be found.
In order to state Entry 6, we also need to recall the definition
of the Bernoulli numbers B , 0 < n < °°,
n —
B
n n
e - 1 n«0
This convention for the Bernoulli numbers is different from that of Ramanujan.
Entry 6. Let Li« and x? be defined as above. Then
(i) Li2(l - z) + Li2(l - 1/z) = - j Log2z ,
T\ 1 2
(ii) Li2(-z) + Li2(-l/z) - - --£- - j Log z,
2
(iii) Li2(z) + Li2(l - z) * -g- - Log z Log(l - z),

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK 21
1 2
(iv) Li2(z) + Li2(-z) = j Li2(z ),
2
1 - z. ir . 1 , T ,1 + zN
(v) x2(«) + x2(rTT) = T + I Log z L°8 (r^7)'
(vi) Li2(l~^) + Li2(3~^5') = Li2(z) + Li2(w)
+ Li2((l - z)(l - w)) + Log(1 " z)Log(1 " w>'
2
(vii) Li2(e~ ) = -g- + z Log z - z
„ n+1
oo B Z
+ I (n"+l)tn • I2' < 2lr'
n-i
and
« n+1
oo B Z
(viii) Li (1 - e~Z) = I I , |z| < 2tt.
1 n=0 vn ■L;-
Proof. Part (i) is proved in Lewin's text [52, p. 5, equation (1.12)]
and is due to Landen [50],
Part (ii) is found in [52, p. 4, equation (1.7)] on noting that
Li(-l) = -tt2/12. Evidently, (ii) is due to Euler [25, p. 38], [27, p. 133].
See the top of page 5 of [52] for further references.
Equality (iii) is also due to Euler [25], [27, p. 130] and can
be found in Lewin's book [52, p. 5, equation (1.11)].
Formula (iv) is rather trivial and can be found in Lewin's book
[52, p. 6, equation (1.15)].
Part (v) is again due to Landen [50] and is established in Lewinfs
treatise [52, p. 19, equation (1.67)].
Formula (vi) was first established by Abel [1, p. 193], but an
equivalent formula was proved earlier by Spence [75], The former formula is
also in Lewin's book [52, p. 8, equation (1.22)].
Formula (vii) arises from (6.4) after dividing both sides of (6.4)
by z and integrating both sides twice. See [52, p. 21, equation (1.76)].
To prove (viii), first replace z by t in (6.4) and integrate
both sides over [0,z]. Next, in the resulting integral on the left side,
set w * 1 - e . Using (6.3), we complete the proof.

22
B. C. BERNDT AND P. T. JOSHI
Example. We have
(i) Li2(l/2) = —• " \ Log22,
(ii) Li2(^^) = Yq - Log2(^f-^),
/ • . . n t • /3 - /5N it2 2 ,/5 - lx
(in) Li2( ^ ) = yj - Log ( ^ ),
X9^T- 1) =^-iLog2(/2 - 1),
(iv)
^2V " A/ 16 4
,/5" - 1, it2 3 T 2,/? - 1,
(v) X2( 2—} = H " "4 L°g ( 2 )j
and
2
(vi) Xj^" 2) =|^-|Log2(^f^).
Proof. Part (i) follows from Entry 6(iii) on setting z = 1/2.
The result is found in Lewin's book [52, p. 6, equation (1.16)]. The priority
for this evaluation seems to be clouded. According to Lewin [52, p. 6],
the result is credited to Euler in 1761, but Landen claims to have established
(i) in 1760. On the other hand, Bromwich [12, p. 520] indicates that the
result is due to Legendre.
Formula (ii) can be found in Lewin's treatise [52, p. 7, equation
(1.20)] and is apparently due to Landen [50],
Formula (iii) is found in Lewin's book [52, p. 7] and is again
due to Landen [50].
Equality (iv), which readily follows from Entry 6(v) upon setting
z = /2 - 1, is again found in Lewin's book [52, p. 19,equation (1.68)].
Part (v) is also found in Lewin's book [52, p. 19, equation
(1.69)] and is due to Landen [50].
Formula (vi) was submitted by Ramanujan as a problem in the
Journal of the Indian Mathematical Society [62], [65, p. 330], See also
Lewin's book [52, p. 19, equation (1.70)], where the result is attributed
to Landen [50].
Many other functional equations and numerical examples for the
dilogarithm can be found in [25], [27], [32], [50], [52], [53], [70] and [75].

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
Entry 7. Let Li3(z) be defined by (6.1). Then
(i) Li3(l - z) + Li3(l - 1/z) + Li3(z)
2
= C(3) + -g- Log z + -g- Log z - j Log z Log(l - z),
2
(ii) Li3(-z) - Li3(-l/z) = - -g Log z r- Log z,
and
(iii) Li3(z) + Li3(-z) - j Li3(z2).
Proof. Part (i) is due to Landen [50] and can be deduced from
[52, p. 155, equation (6.10)] by letting x = 1 - z there.
Part (ii) can be found in Levin's book [52, p. 154, equation
(6.6)].
Part (iii) is trivial. See also [52, p. 154, equation (6.4)].
Example. If X3 is defined by (6.2), then
(i) Li3(l/2) = I Log32 - L. Log 2 + X3U)
and
(ii)
• /3_-/5\ 2 T 3 VF+JU 2tt2 t ,/5 + 1, . 4
l3(—5—) = I Log (—2—) " TT g(—2—} I
Proof. The first equality follows from Entry 7(i) on setting
z » 1/2. See also [52, p. 155, equation (6.12)].
Part (ii), due to Landen [50], is again in Lewin's book [52,
p. 156, equation (6.13)]. In Ramanujan's notebooks [66, vol. II, p. 107],
the coefficient 4/5 on the right side of (ii) is inadvertently omitted.
Entry 8. For |x| < 1, define
oo U 2k"1
h. x
(8.1) f(x) - I 2k - 1 »
k=l
where
(8-2> hn " I 2k-VT
k*l

24
B. C. BERNDT AND P. T. JOSHI
Then, for |x| < 1,
*<2-r7> =iLog2(1"x) + i Li2(x)-
Proof. Taking the Cauchy product of the Maclaurin series for
1/(1- x2) and Log(* * X), we find that, for |x| < 1,
1 - X
Hence,
(2 - x)
2 f' iT^) = AxTT^O Log(1 " X)
1 T ,, N Log(l - x)
= - ^ Log(l - x) - fi 1 _ x)
Integrating the foregoing equality over [0,x], |x| < 1, and using (6.3)
d x 2
and the equality — (-r——) = =-, we complete the proof.
(2 - x)
Example. With f defined by (8.1), we have
(i) f(l/3) = -— -^-LogZ2,
2
(ii) f(l/^5) = ^ ,
and
(iii) f (/5 - 2) = ^ - | Log2(^A).
Proof. To deduce part (i), set x = 1/2 in Entry 8 and then
employ Example (i) of section 6.
To obtain (ii), set x « (/F - l)/2 in Entry 3, note that
2
1 - x = x , and use Example (ii) of section 6.
Lastly, set x = (3 - /F)/2 in Entry 8. Using Example (iii)
of section 6, we deduce the desired equality.
Both Examples (ii) and (iii) are in error in the notebooks [66,
vol. 2, p. 107], In (ii), Ramanujan has an extra term - tt Log ( -—)
on the right side. In (iii), he has written 3/4 instead of 3/8 on

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK 25
the right side. Ramanujan [63], [65, p. 330] submitted Examples (i) and
(iii) as a question in the Journal of the Indian Mathematical Society.
For other examples of this sort, see Catalan's paper [16],
Entry 9. For |z| £ 1, define
k+1
00 H.:
(9.1) g(z) = I -£
k=l (k + l)2
where H, is defined by (3.1). Then g can be analytically continued
to the entire complex plane. Furthermore,
1 2
(i) g(l - z) = j Lo8 z L°gd - z) + Li2(z)Log z
- Li3(z) + CO),
(ii) g(l - z) - g(l - 1/z) = i Log3z,
12 13
(iii) g(l - z) « -j Lo8 2 L°g(z " !) " T Lo8 z
- Li2(l/z)Log z - Li3(l/z) + C(3),
and
1 3
(iv) g(-z) + g(-l/z) = - ■£ Log z - Li2(-z)Log z
+ Li3(-z) + £(3).
Proof. Squaring the Maclaurin series for Log(l - z), we find
that
(9.2) zg»(z) = \ Log2(l - z), |z| < 1.
Thus,
(9.3) -g'(l - z) = - 2(^°g *} , 0 < z < 1.
Integrating by parts twice, we find that
g(l - z) = i Log2z Log(l - z) + Li2(z)Log z - Li3(z) + c.

26 B. C. BERNDT AND P. T. J0SH1
If we let z tend to 1~, we find that c = Li^Cl) « £(3), which completes
the proof of (i) for 0 < z < 1.
Since Li«(z) and Li«(z) can be analytically continued into
the full complex z-plane, then (i) shows that g(z) can be analytically
continued as well.
From (9.3),
2
-g
(1-z) 1 g'(i-i/z) = *^
z
2 6 v- -,-/ 2z
Integrating this equality over [l,z], we get (ii).
Consider (i) for 0 < z < 1. Replacing z by 1/z, we obtain
for z > 1,
1 2
g(l - 1/z) = J Log z(Log(z - 1) - Log z} - Li2(l/z)Log z
- Li3(l/z) + £(3).
Substituting this expression for g(l - 1/z) into (ii), we deduce (iii)
for z > 1. By analytic continuation, (iii) is valid for all complex z.
Lastly, if z > 0, we find from (9.2) that
-g' (-) + -\ g' (-i/z) - ^SL±ll . u>i2<i + iM
Z
2
Log z Log z Log(l + z)
2z z
Integrating by parts, we find that
g(-z) + g(-l/z) = - ^2*-^ - Li2(-z)Log z + J
Li (-z)
— dz
3
= - | Z - Li2(-z)Log z + Li3(-z) + c.
By analytic continuation, this holds for all complex z. Now set z = -1
and use the fact that g(l) = C(3), which can be deduced from (i). We
then find that c * 5(3), and so the proof of (iv) is complete.
Entry 9(i) is stated without proof by Lewin [52, p. 303, formula (12)].
Entry 9(iii) contains a misprint in the notebooks [66, vol. 2, p. 107];
Ramanujan has written Log(l - z) for Log(z - 1) on the right side.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
27
The formula
(9.4) I S-_« r(3)9
k=l (k + 1)
obtained from Entry 9(i) by setting z = 0, has a long history. Formula
(9.4) was evidently first discovered by Euler [24], [26, p. 228] in 1775.
This evaluation and many other results of this sort were established by
Nielsen [58], [59], [60]. In 1953, (9.4) was rediscovered by Klamkin
[49] and submitted as a problem. Briggs, Chowla, Kempner, and Mientka
[11] rediscovered the result again in 1955. Once again, in 1982, (9.4)
was rediscovered, by Bruckman [13].
« \ n-2
(9.5) 2 I -£« (n + 2)C(n + 1) - I £(n - k)C(k + 1),
k=l k k=l
where n is a positive integer at least equal to 2. This result is also
due to Euler [24], [26, p. 266]. Nielsen T58, p. 229], [59, p. 198],
[60, pp. 47-49] developed a very general method for obtaining (9.5) and
similar types of results. Formula (9.5) was rediscovered by Williams [81]
in 1953. Rao and Sarma [71] have also proved (9.5).
It might be remarked that the problem of evaluating series of the
type
oo Ir
y _L y _L
k-i kn A f '
where m and n are positive integers with n ^ 2, was first proposed
in a letter from Goldbach to Euler [28], [29] in 1742. The two
mathematicians exchanged a series of letters about this problem in 1742 and 1743,
and Euler was successful in obtaining several evaluations of series like
that depicted above. However, (9.4) and (9.5) apparently are not found in
these letters.
Matsuoka [54] has made a study of the related Dirichlet series
Ks) = I H^ k"S, Re(s) > 1.
k=l

28 B. C. BERNDT AND P. T. JOSHI
In particular, he has shown that F can be analytically continued to the
entire complex s-plane, and he has determined the poles of F and their residues.
Entry 10. For |z| <_ 1, define
k+1
00 \:
h(z) = I -i
k-1 (k + l)3
Then h(z) can be analytically continued into the entire complex z-plane.
Furthermore,
(i) h(l - z) - h(l - 1/z) = - —■ Log4z + | Log3z Log(l - z)
2
+ £(3)Log z - 2 Li4(z) + Li3(z)Log z + —
and
(ii) h(-z) - h(-l/z) = - ~- Log4z - Li3(-z)Log z
+ 2 Li4(-z) + ^(3)Log z +^gg. .
Proof. First observe that
(10.1) hf(z) = g(z)/z,
where g is defined by (9.1). It follows from Entry 9 that h can be
analytically continued into the entire complex z-plane.
By (10.1), Entry 9(ii), and Entry 9(i),
-h»(l - z) - 4 h'(l - 1/z) = - g(1 " 2) - g(1 - 1/2)
z2 l - Z z2(l - 1/z)
■ " —~- ^d - z) - i g(l - z) + £ Log3z}
3
= gC1 " z) _ Log z
z 6z(l - z)
12 1 1
= 27 Log z Log(l - z) + - Li2(z)Log z - - Li3(z)
, 1 r/«v 1 T 3 1 , 3
+ z C(3) " 6?Lo^ z - 6(1 - z) Log Z'

CHAPTER 9 OF RAMANUJANfS SECOND NOTEBOOK
29
Integrating the equality above, we find that
2
h(l - z) - h(l - 1/z) =
Log z Lofi(l - z) dz + Li3(z)Log 2
f Li3(z)
dz -
Li3(2) 1 4
--— dz + C(3)Log z - -XT Log z
1 3
+ -g- Log z Log(l - z) -
Log z Log(l - z) dz + c
2z
= Li3(z)Log z - 2 Li4(z) + C(3)Log z - ■—■ Log*z
1 3
+ -r Log z Log(l - z) + c,
where in the penultimate equality we integrated by parts twice. Letting
4
z = 1 and employing the fact that Li^d) = C(4) = tt /90, we find that
c = it /45, which completes the proof of (i).
Next, by (10.1),
-h'(-z) - 4- h1(-1/z) = \ {g(-z) + g(-l/z)}
= - ■—- Log3z - i Li2(-z)Log z + i Li3(-z) +^(3),
by Entry 9(iv). Integrating both sides above, we find that
1 4 f Li (-8)
h(-z) - h(-l/z) = - ~ Log z - Li3(-z)Log z + —^ dz
f Li (-z)
+ I — dz + C(3)Log z + c
1 4
- " 24' Log z - Li3(-z)Log z + 2 Li4(-z) + £(3)Log z + c.
4
Putting z = 1 above and using the fact that Li. (-1) = -7£(4)/8 = -7it /720,
4
we find that c = 7ir /360. This completes the proof of (ii).
In the notebooks [66, vol. 2, pp. 107,108], the right sides of
Entries 10(i) and (ii) must be multiplied by (-1).
Entry 11. For -1 <_ x £ 1, define
2k
00 h, x
F(x) = I JS_
k-1 (2k)
and
oo h X
G(x) = I k
2k
k-1 (2k)"

30
B. C. BERNDT AND P. T. JOSHI
where h, is defined by (8.2). Then for 0 <_ x <_ 1,
(i) F(fri) = iLog2x Log(fTT} + \ *2(x)Logx
and
(ii)
+ j {X3(D - X3(x))
G(x) + G(^-^) = F(x)Log x + F(fr^)Log(^^)
-^Log2x Log2(f^) + G{1).
Proof. For |x| < 1,
(11.1) xF
u 2k
°° h. x rx °° «. t
•w- x -V= J v dt
k=l ZR J0k=l
1 fX 1
2 L i 2
'0 1 - t
Log(f^T)dt
1 T 2.1 + Xv
8
vi - x7
Hence, for 0 _< x < 1,
2
(11.2)
F»(1 " x) = _ Log x
(1 + x)2 1 + X 4(1 - x2)
Integrating the equality above, we find that
1 . 2 T ,1 - x, xl r 1 f 2kT .
= ¥ Log x Log(rT7) tj J 2k-5TT X L°g X dX
Xo(x)
1 T 2 T ,1 - xx _,_ 1 , NT if A2V
= g- Log x Log(1 ^ x) + y X2(x>Log « - 2" I —^— dx
s g Log x Log(1 ~ X) + j X2(x)L°g x " J X3OO + c.
Now let x tend to 1- to find that c ■ X3(D/2, which completes the
proof of (i).
We now prove (ii). Observe that for |x| £ 1, xG'(x) ■ F(x).

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
31
Hence, for 0 <_ x < 1,
(1 + x)Z X + X x 1 - xZ X + x
An integration of this equality yields
G(x) + G(y-^) = F(x)Log x - f F1 (x)Log x dx
+ F(f^)Log(f^) + \ —±-z F'(f^)Log(f^)dx
1 (1 + x)
4 j fV°e<f^>.
J 1 - x
by (11.1) and (11.2). Integrating by parts, we get
G(x) + G(i-^) « F(x)Log x + F(±-=-|)Log(f-=-*;)
1 , 2,1 - xXT 2 1 [ . ,1 - xx Log
-16Lo* (rr7)Log x - 4 J Lo8(rrr) 7:
2
—j dx
x
♦i
1 - X
' F(x)Log x + F(^)Log(f^) - ± Log2x L0g2(^) + c.
Letting x approach 1-, we find that c ■ G(l), and this completes the
proof.
In fact, Ramanujan claims that
(11.3) C(l) - I I ^^ 3 - S- I i-5 •
* k=0 (4k + 1) 3/3 k«0 (2k + 1)
Unfortunately, this beautiful formula is incorrect. Taking the first three
terms of the series defining G(l), we find that G(l) > 0.1529320988... .
On the other hand, the right side of (11.3) is easily found to be less than
0.1442780636... . We have been unable to find any formula for G(l) which
resembles (11.3). R. Sitaramachandrarao (personal communication) has derived
several expressions for G(l) that are related to the Riemann zeta-function

32
B. C. BERNDT AND P. T. JOSHI
and similar types of series. Unfortunately, none of Sitaramachandrarao's
formulas echoes (11.3).
Entry 12. For |x| < 1, define
11 x
H(x) - I * _ ± ,
k«l ^ l
where Hk is defined by (3.1). Then for 0 < x < 1,
,1 - xN
1 + x
HC^Tj) - (Log 2 - l)Log x + x _ x
1 2 IT
+ j Log x + — + Li2(-x).
Log I
4x
Id + x)'
2„.,
Proof. First observe that, for |x| < 1, x H'(x) is the Cauchy
2 2
product of the Maclaurin series for -Log(l - x ) and 1/(1 - x ), i.e.,
H'(x)
Hence,
(1 + x)
Log(l - x )
"2 2
x*<l - xZ)
2H (r^} - ——:r Log
r
4x
2x(l - x) 1(1 + x)
2x + „ ,2
{Log 4 + Log x - 2 Log(l + x)}.
(1 - x)\
Integrating the equality above, we find that, for 0 < x < 1,
n~^) - \ Log 4 Log x + r-^ Log 4 + \ Log2x + 2 I ^SSJL
_ f Log(l + x) dx . 4 [ Log(l + x)
J X J (1 - x)2
= Log 2 Log x + r4T Log 4 + i Log2x + ^~^T " 2 j 7
(1 - x)'
dx
(1 - x)
+ Li2(-X) . 4 ^g^1 + *> + 4 ( -^.y
j 1 - x
= Log 2 Log x + 2^Log(4x) + 1^ Log2x _ 2 Log x + 2 Log(l - x)
2
+ Li2(-x) - 2-L°f(! * X) + 2 Log(l + x) - 2 Log(l - x) + cx

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
=* (Log 2 - l)Log x +
1 - x
Log
(^M
Log
+ Li2(-x) - Log
f *x 1
a + xr
+ c,
where c - c, + Log 4. Letting x tend to 1-, we fine that
2
c« -1J.A-1) = t\ 111, and this completes the proof.
Examples. We have
k-l kV 12
H. o
(ii) J 2 ■|x3(l)»
k-l (2k - 1)
(iii)
and
(iv)
I -f = 2x3(l>.
k-l k
oo h (/5 - 2)2k-1 2 -.
r k __ = T* 4. 3 T-J//5 - ly
k-l
2k - 1
60 A
+ (/5 + 2)Log 4 + (3/5 + 5 + Log 2)Log(^J l).
Proof of (i). For |x| £ 1, define
„ k
oo H X
t(x) - I -V .
k=l k
Observe that
(12.1) t(x) « Li3(x) + g(x),
where g is defined by (9.1). We wish to evaluate t(l/2). Using
Entry 9(i), and Example 6(i), we find that
t(l/2) - Li3(l/2) + g(l/2)
= Li3(l/2) - -| Log32 - Li2(l/2)Log 2
- Li3(l/2) + C(3)
j Log32 - (fj - I Log22)Log 2 + 5(3),
from which the desired result follows.

34
B. C. BERNDT AND P. T. JOSHI
Proof of (ii). We wish to evaluate (t(l) - t(-l)}/2. By (12.1),
(9.4), and Entry 9(iv),
\ {t(l) - t(-l)} = \ (g(l) - g(-l)} + \ 1Li3(l) - Li3(-l)l
- \ (CO) - |(Li3(-l) + c(3))} + x3(D
-7X3(1) + x3(D -f x3u>-
Proof of (iii). The left side of (iii) is 4F(1), where F
is defined in Entry 11. Putting x = 0 in Entry 11(i), we find that
F(l) * xAl)/2. Hence, the result follows.
Proof of (iv). The left side of (iv) is H(/5" - 2), where
H is defined in Entry 12. Putting x = (/F - l)/2 in Entry 12 and noting
that (1 - x)/(l + x) « /5" - 2, we find that
(12.2) H(/5 - 2) * (Log 2 - DLog^ ~ 1) + 3(/5 + 2)Log(^^)
2
+ 2(/5 + 2)Log 2 + J Log2(^^) + L + Li,,( - ^~^).
Since, by (6.2), Li„(-x) * Li2(x) - 2y (x), we find from Examples 6(ii)
and (v) that
t • / /F - 1N ti2x1t 2,/5" - lx
Li2( —) " ~ 15 + 2 Log (_2~>'
Using this value in (12.2) and simplifying, we deduce the sought result.
Example (ii) was first established by Nielsen [59]. Jordan [45],
[46] apparently not only first proved Example (iii) but also found a general
00 2
formula for £ h /k , where n is a positive integer. Later proofs
k=l k
of both (ii) and (iii) were found by Sita Rama Chandra Rao and Sarma [72].
For other formulas like those in Sections 9 and 12, consult the
papers of Nielsen [56], [57], [58], [59] (as well as his book [60]); Euler [24], [26];
Jordan [45], [46]; Gates, Gerst, and Kac [33]; Schaeffer [70]; Gupta [39]; Hans and

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
35
Dumir [40]; Sita Rama Chandra Roa and Sarma [71]; Sitaramachandrarao and
Subbarao [73], [74]; Buschman [14]; and Rutledge and Douglass [69].
Closely connected with the polylogarithms are the Clausen functions
CI (x) defined by [52, p. 191]
(13.1) Cl2(x) = I ^^ , n > 1,
2n k=l k2n
and
.1) Cl^Cx)-
k=l k
(13.1) Cl2n+1(x) = ISSffijL , n>0,
where x is real, with the restriction that x is not a multiple of 2tt
when n - 0. It should be noted that
00
(13.3) Cl.(x)- I cos<kx) ■ -Log[2 sin(x/2)[.
1 k=l K
Entry 13. If n is a positive integer, then
fx n
— cot(y)du
0 L
.dm/2)n!C(n + l). \ (-d3 (J^)/2 F(n | 1) ,n-J
Proof. For each positive integer k, we have upon n
integrations by parts
rx n n-1
(13.4) unsin(ku)du = - \ cos(kx) + nX, sin(kx)
Jo k
4 n(n - l)xn"2 cos(kx) _ n(n - l)(n- 2)xR~3 sin(kx)
kJ k
+•••+ fn(x) +-~^rcos(mr/2),
k
where fn(z) = (-l)m+1n!cos(kx)/kn+1, if n = 2m is even, and
fR(x) = (-l)mn!sin(kx)/kn+1, if n = 2m + 1 is odd. Now sum both
sides of (13.4) on k, 1 <_ k <_ N, and let N tend to °° to get
[37, p. 30]

36
B. C. BERNDT AND P. T. J0SH1
rx n ,
lim [ ^r- {cos(y) - cos(N + -y)u} csc(y)du
N-**> J0
fX un wu*, r . 1xj(j+D/2 r(n + 1) n-i
= Jo T cot(T)du = - ^ (-l)J TOi + 1 - j) X C1J+1(X>
+ n!£(n + l)cos(niT/2)t
where we have used the Riemann-Lebesgue lemma.
Entry 13 is equivalent to formulas in Lewin's book [52, p. 200,
equations (7.52), and (7.53)].
Next, define
00
/w i\ r» / \ V sin(2k - l)x
(14.1) D9 (x) = 2. JZ~ » "11.
2n k=l (2k - l)2n
and
(14.2) D2n+.(x) = I C05(2k ~2lll , n > 0,
2n+1 k=l (2k - l)2n+1
where x is real, with the restriction that x is not a multiple of T\
when n = 0. Observe that [37, p. 38]
(14.3) Dx(x) = - \ Log|tan(x/2)|.
Entry 14. If n is a positive integer, then
fx n
(14.4) I -y csc u du
Jo
= co.(«r/2)n,Vl(l) - I (-l,J««)/2^^ --JD (x).
3*0 J J
where x is defined by (6.2).
Proof. The proof follows along the same lines as that of Entry
13. We begin with (13.4) but with k replaced by 2k - 1. Now sum on k,
1 <^ k ^ N, aod let N tend to °°. It is easily seen that we get the right
side of (14.4). On the left side, we obtain [37. p. 30]
lim
N-*»
fx un rx un
-=- (l - cos(2Nu)} csc u du = | — csc u du,
0 l
l u
— csc u
J0
by the Riemann-Lebesgue lemma. This finishes the proof

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
37
On the right side of (14.4), Ramanujan has written Li (1)
instead of Xn+1U> t66» vol« 2» P- 1091-
Entry 15. For each nonnegative integer n, define
f (x) ■ x cot x dx.
n
Then if n > 0,
(15.1)
and
(15.2)
vk,nx n-k
2nf (2- - x) = I <-l>K(?)irn"" {f (2x) - 2Kf, (x)}.
n z k*0 K
Proof. Now, since tan x = cot x - 2 cot(2x),
2\ (f - x) - 2
n I
x) cot( y - x)d(y - x)
BJ<T
= -2n f & - x)n tan x dx
= -2n f I (£)(£)n"k(-x)k{cot x - 2 cot(2x)}dx
J k=0 k Z
r ,nx, .vk n-k
k=0 K
J (2x)kcot(2x)d(2x) - J (2x)kcot x dx]
I 0(-l)k7rn"k{fk(2x) - 2kf,(x)}.
k=0 K * K
Examples. We have
f n . A f (sin^y)" ,
x cot x dx - I - "— dy
J sin(2x)
dx
(tan z)
dz.
Proof* Equality (15.1) arises from setting y = sin x, and (15.2)
is gotten by letting z = tan x.
Recall that h is defined by (8.2).
Proposition 15. For |x| £ 1,
00
(i) \ (tan^x)2 - I
, ,vk-l, 2k
1 ,__-l„,2 r ^ hkx
k-1
2k

38
B. C. BERNDT AND P. T. JOSHI
» 02k„ iX2 2k+2
(ii) y (sln X) = JQ (2k + 2) ■
i O °° 1 1 91, 2k+1
(in) YT (sin x) = I (l+~2 +•••-♦" 2^k)
and
J* k-1 3Z (2k - 1) K 2ZK(2k + 1)
4 » , , 92k,..,2 2k+2
,. . 1 , . -1 * Y / 1 j. 1 A A 1 n 2 (kt) x
(iv) tj (sin x) = I (— + — +...+ j) .
<♦• k=1 2Z 4 (2k) U* ;*
Proof. The Maclaurin series (i) - (iii) may be found in [22,
pp. 88-90] where the methods for deriving them are clearly delineated.
Since (iv) is not given, we shall prove (iv).
Write, for |x| <_ 1,
_1 oo
,,. ON a sin x r k
(15.3) y - e = \ a x .
k=0 K
AT?,
Then y' = ay/
2w in2 2 2
(1 - x )(yf) -ay.
Differentiate both sides above with respect to x and then divide both
sides by 2y* to obtain
(1 - x2)y" - xy» - a2y = 0.
Substituting the power series (15.3) into the differential equation above
and equating coefficients of like powers of x, we find that
2 2
(k^ + a*)a.
K k > 0.
°k+2 (k + 2)(k + 1) *
Moreover, it is easily seen that a0 = 1 and a., = a. A simple inductive
argument now gives
(15 4) a = «(a2 + l2)(a2 + 32) ••• (a2 + (2n - l)2)
U5-*' a2n+l (2n + 1)! , n >_ l,
and
2 2 2 2 2 2 2
(15.5) a2n ^ , n > 1.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK 39
Expanding exp(a sin x) as a power series in a, and equating coefficients
k
of a on both sides, with the use of (15.4) and (15.5), we may deduce
-1 k
the Maclaurin series for (sin x) , k _> 1. In particular, for k = 4,
we find from (15.5) that
L i_ 2n
, . -1 ^ °° b0 x
(sin x) _ r 2n
4! ~n£2 (2n)! '
where
. Y 2242- (2n-2)2 . 22(n-l){(n . 1)|}2 "y1 1 §
j=l W) j-1 (2j)Z
which completes the proof of (iv).
Entry 16. For |x| <_ tt/2,
oo 2k+l °°
,., , * r /2kv sin x . u . i , 1 v sin(2kx)
(16.1) I ( ) -st =■ - x Log) 2 sin x| + ■=■ }. £ '- .
k=0 K 2 R(2k + 1) k-1 kZ
Proof. Making the substitution t = sin(u/2) and employing
Entry 13 with n - 1, we find, for |x| <_ tt/2, that
(16.2)
dt
rsin x °° 01 , .0v2k rsin x . -1.
V f2k\ (t/2) . _ sin t
^ ( k} 2k + 1 dt " E—
Jo k=o K Zk + L jo z
- j \ u cot(u/2)du - ± {-2x Cl1(2x) + Cl2(2x)L
If we now use (13.1) and (13.3) in (16.2), we deduce (16.1).
Let
oo k
(16.3) G = I ("1}
k«0 (2k + l)z
denote Catalan's constant.
Examples. We have
(i) I fy^ST^ 2"IL°62'
k=0 K 2 (2k + 1)
(ii) I (Zb -aJT-2 2 = — Log 2 + -^ G,
k=0 K 2J (2k + 1) 4/1 /I

40 B. C. BERNDT AND P. T. JOSHI
T ,2k, 1 3/3 " 1 /
(iii) I (k}"4k+l 2=T" i T"~»
k=0 k 2^k X(2k + 1) * k=0 (3k + 1) 6/3
and
cxj k 2 °°
,. x r .2kx _3 ir T Q 2tt r 1
(iv) £ ( . ) -7T o = Log 3 - ~27" -H £ 2 •
k=0 k 24k(2k + l)2 3/3 27 k=0 (3k + 1)Z
Proof. Part (i) follows from putting x = ir/2 in (16.1).
Put x = tt/4 in (16.1) and multiply the resulting equality by
/I. We then obtain (ii).
Next, let x « tt/6 in (16.1). The left side of (16.1) becomes
the left side of (iii), and the right side of (16.1) is found to be
l °°
2 kii k2
= 3/J r 1 /3 I v 1 , y 1
* k=0 (3k +1) * *> k=0 (6k + 1) k«0 (6k + 2)
-4
*• 2 " i 2 ,-
k=0 (6k + 4) k=0 (6k + 5) J
\
_ 3/3 V 1 _ £ / y 1 . " 1 1
4 2 2 1 i 2 i 2 /
k=0 (3k + 1) * ^k=0 (6k + 1) k=0 (6k + 5) >
= 3/3 y 1 Xfr 1 1 y 1 \
4'- 221^- 2 9 £ 21
k=0 (3k + 1) »-k=0 (2k + 1) k=0 (2k + 1) >
.3^? 1 4/Jtt2
4 k=0 (3k + I)2 ~ 9 8 '
which completes the proof of (iii).
Lastly, put x = tt/3 in (16.1) and multiply the resulting equality
by 2//T. The left side then becomes the left side of (iv), and the right
side is equal to
it . ~ , 1 r sin(2kTT/3)
—: Log 3 + — I 2
3/3 /3 k«l k
OO / 00 00 >
- — Log 3 + I 2 " 2 1 i — ' I 2 f»
3/J k=0 (3k + ir * ^k-1 k k=l (3k) i
which is readily seen to equal the right side of (iv).

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
41
Part (i) can be found in a paper of Ruscheweyh [68].
Entry 17. For |x| <ir/2,
/^^^ ^\ V (-1) tan x T ■ i , r sin(4k + 2)x
(17.1) I «— = x Log I tan x| + I 5-
k=0 (2k + 1) k=0 (2k + 1)
Proof. For |x| < tt/2, the left side of (17.1) is equal to
(17.2)
vk 2k
rtan x °° , nk Zk rtan x fc -i
Jo k*o ZK + L h c
1 f2x
= j \ u esc u du = -2xD (2x) + D„(2x),
where in the penultimate step we made the substitution u - 2 tan t, and
in the last step we utilized Entry 14 with n ■ 1. If we now employ (14.1)
and (14.3), we deduce (17.1).
Examples. We have
(i)
'1//5 tan^t
- dt = - Y^ L°8 3
5tt + 5/3
18/3 4 k«0 (3k + 1)
GO
I —^
(ii)
and
(lii)
r*-1 fn^t ^ _ w .__,.«■ „ w2 , .^ " (-l)k
I t«L_L dt . JL Log^ - i) -T6 + /2 I
JO c ° •LD k*
k«0 (4k + 1)
2-vT. -1
0 ^^dt-JLi-,(2-^+f
1 -1
tan t
dt,
where
f1 tan^t
dt = G = .915965594177...,
where G is defined by (16.3).

42
B. C. BERNDT AND P. T. JOSHI
Proof of (i). In Entry 17, put x = tt/6 and use (17.2) to obtain
fl//Jtan^t ^ tt t ^/3 n 11 1 .
dt = - — Log 3 + -y (1 - —7- + —7 J +# >
0 t X 5 7 11
= - ^ Log 3 + y i (1 - ^ + T - -^ + ^ - ^ + — f +" }
Z 2 4 5 7 8 10 11
+ ( —o o + —o o +•'•))
2Z 4Z 8 10^
IT
5/3
" " 8 22 42 52 72 82 102 ll2
77 T a ^ 5/3 " 1 5/3 / v 1 V 1 I
" 12 Log 3 "T" *• 2 " T" 1 *• ~2 " *• 2 / »
iZ * k=0 (3k + 1) * lk=l Y. k=l (3k) J
from which (i) follows.
Proof of (ii). Set x = tt/8 in Entry 17. Using (17.2) and the
fact that tan (tt/8) « /I - 1, we find that
f^"1 tfiC^t dt . £ Log(/J . 1} + J «in{(2k + l)w/4)
J0 t k=0 (2k + 1)
- 5 Log(^ - 1) +-J I -fcll^-A I -^ j—.
* /2 k=0 (4k + 1) yfl k=0 (2k + 1)
The latter sum in the equality above is L(2,x), where x ^s the real,
even, primitive character of modulus 8. By a standard formula [10, p. 48],
9
L(2,X) = tt /2"/16. This completes the proof of (ii).
Proof of (iii). Let x = tt/12 in Entry 17. Noting that
tan(ir/12) = 2 - i/J and using (17.2), we find that
f2-^ta£^tdt w ^(2 _ ^ + \ sln{(2k + l),r/6}
Jo c iZ k-o (2k + iy
«^Log(2 -•3)+i(l+-L_l_l+...)+l.l+l_l+.
b r ir r r is2 2iz

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
43
/ oo y^ oo k ^
^k=0 (2k + l)z k=0 (6k + 3)^ y
= ~Log( 2 - /I) +| G +| G.
The given integral representation for G follows easily upon integrating
the Maclaurin series for (tan t)/t termwise, and so the proof of (iii)
is completed. The decimal expansion of G is correct to the number of
places given [23, p. 246],
Entry 17 as well as Examples (ii) and (iii) may be found in
Ramanujan's paper [64], [65, pp. 40-43].
Entry 18. For 0 £ x £ tt/4,
0, 2k+l . 2k+l
(18.1) I (2k> cos x - sin x
k-0 K 2ZKL(2k + 1)
oo 2k+l
ff y tn \ 1 V /2k» sin (2x)
= y Log(2 COS X) - y \ ( ,_) ~Jk ^—^ •
Z k=0 K 2ZK(2k + 1)
Proof. Replacing x by tt/2 - x in Entry 16, we find that, for
0 £ x £ it,
oo 2k+l °° k+1
(i8.2) i (2,k) £g—*. = (i. x)LOg|2 cos x| + i y (-1} !;in(2kx)
k=0 K 2ZK(2k +1) * k-1 k
Subtracting (16.1) from (18.2), we deduce that, for 0 £x £ tt/2,
01 2k+l , 2k+l
V /2k. cos x - sin x it t /0 n t /n . /0 ..
2, ( . ) jr o = y Log (2 cos x) - x Log(2 sm(2x))
k=0 K 2 K(2k + 1) L
oo
V sin(4kx)
" L 2— '
k=l (2k)
Replacing x by 2x in Entry 16, we get, for |x| £ ir/4,
oo 2k+l °°
y (2k, sin J2j0 . 2x LQg|2 8ln(2x)| + 1 v sinJAkxi
k=0 K 2ZK(2k + 1)' Z k=l k

44 B. C. BERNDT AND P. T. JOSHI
Combining the latter two equalities, we deduce (18.1) for 0 <_ x <_ tt/4 .
Example. For |x| <_ 1, define
*(x) = r-i^dt.
Then
*(3/5) - \ * (24/25) = | Log 2 + 2*(l//5) - 2\\j(2/rf).
Proof. From (16.1) and (16.2), for |x| <_tt/2,
9, . 2k+l
(18.3) *(sin x) = I (Zk) Sin X
k=0 k 22k(2k + l)2
Let sin x « 3/5; so cos x = 4/5. Then sin(2x) » 24/25. Thus, by
Entry 18 and (18.3),
(18.4) *(4/5) -*(3/5) «yLog(8/5) -y*(24/25).
Secondly, let sin x = 1//5; so cos x = 2//5 and sin(2x) = 4/5. Again,
from Entry 18 and (18.3),
(18.5) *(2/vT) - *(l//5) = \ Log(4//5) - j*(4/5)
= j Log(8/5) + ~ Log 2 - y i|;(4/5).
Combining (18.4) and (18.5) together, we deduce the proposed equality.
Entry 19. For 0 <_ x < ir/2,
0, 2k+l . 2k+l m » , 1Nk 2k+l
V z21^ cos x + sin x ir r (-1) tan x
Z ( k) ok 2 * J Log(2 cos x) + '• 9~
k=0 K 2ZK(2k + l)z z k=0 (2k + 1)Z

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
45
Proof. Adding (16.1) and (18.2), we find that, for 0£x < tt/2,
00 ou 2k+1 j. • 2k+1
r ,2k. cos x + sin x 7T
i ( . ) ~r « - ■=■ Log (2 cos x) + x Log (tan x)
k=0 K 2ZK(2k + 1) Z
+ V sin(4k + 2)x
k=0 (2k + l)2
°° , 1Nk 2k+l
= - Log (2 cos x) + I -—- 5—,
k«0 (2k + 1)
by Entry 17.
Example. We have
00 2k+l °° k
V ,2kN 1 + 2ZK X 7T y=v 1 c (-1)
i ( k } 2k k-H T-— Log(4//5) + — I 2k4.l 2
k Zk k+1 Z ^ / «+l Z
Proof. In Entry 19, put sin x ■ 1//F; so cos x ■ 2//F and
tan x = 1/2. The proposed formula now follows.
Entry 20. Let |x| £ tt/2. Then
00 o2kn ,n2 • 2k+2 2
(20.1) I 2 (k!) Sln ^ - \ Log| 2 sin x|
k«0 (2k + l)!(2k +2)
00 00
, x r sin(2kx) 1 r cos(2kx) i r ao\
2 /■ .2 + 4 .'•. ,3 " 4 ^U;#
k=l k k=l k
Proof. By Proposition 15(ii), the left side of (20.1), for
|x| £ tt/2, is equal to
rsin x , . -1 ,2
fsin x . . -1 v^ «
Jo ~^~ «
2x 2
-=- COt(y)du
0
= ~ (-2^(3) - 4x2Cl1(2x) + 4xCl2(2x) + 2Cl3(2x)}.
In the first equality, we made the substitution u ■ 2 sin t, and to get
the last equality we used Entry 13 with n = 2. If we now employ (13.1) -
(13.3) in the equality above, we deduce (20.1).
In the notebooks [66, vol. 2, p. Ill], the term -£(3)/4 in
(20.1) has been omitted.

46
B. C. BERNDT AND P. T. JOSHI
Examples. We have
oo 2k 2 2
(i) I ^-^ 2=T L°g2-iX3(l)
k=0 (2k + l)!(2k + 2) °
and
k-1 2 2
(ii) I ^ j = ^Log 2 +£G -ilX,(l).
k=0 (2k + i)!(2k + 2)2 6* 8 16 3
where Xo is defined by (6.2) and G is defined by (16.3).
Proof. To obtain (i), simply set x - tt/2 in Entry 20.
Putting x = tt/4 in Entry 20, we see immediately that the left
side of (20.1) yields the left side of (ii). On the right side, we get
^Log2+lG+i I ^ -IC(3>.
The last two expressions are together equal to -35£(3)/128 = -5x~(l)/16,
and so the proof is complete.
Entry 21. For |x| <_ tt/4,
oo (-1) h 2
(21.1) I y^ tan2kx = ~- Log|tan x|
k-1 (2k)L z
? sin(4k+2)x , 1 7 cos(4k + 2)x 1 ,, x
k=o (2k + 1) z k=o (2k + ir
where h, is defined in (8.2).
Proof. Using Proposition 15(i), then making the substitution
u = 2 tan t, and finally employing Entry 14 with n = 2, we find that
the left side of (21.1) is equal to, for |x| <_ tt/4,
(tan^t)2 ,„ 1 f2x u2
— 2t dt " 4 L TCSC udu
ftan x -1^2 , r2x 2
0 " - JQ
j ^-2X3(D - 4x2D1(2x) + 4xD2(2x) + 2D3<2x)}.
Using (14.1) - (14.3) above, we complete the proof.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
47
Example. We have
« (-Dk~\
(21.2) I =—£ = ttG - 2x-(D,
k=l k
where G denotes Catalan's constant.
Proof. Let x = tt/4 in (21.1) and multiply the resulting equality
by 4 to achieve (21.2).
In the notebooks [66, vol. 2, p. 112], the right side of (21.2)
is incorrectly multiplied by 1/4. Nielsen [59] evidently first established
(21.2).
Entry 22fc Let 0 _< x <_ tt/2. Then
00 02k/1ix2; 2k+2 ^ . 2k+2 T 2
V 2 (k!) icos x + sin x) ir T /0 N
L -—' 2 ="T Log(2 cos x)
^(k!) (cos
k=0 (2k + l)!(2k + 2)
2L V /'2k^ cos x . I V 2 (k!) sin (2x) 1 n n
2^^k;2k 2 4^ 2"2 *2K±)'
L k=0 * 2 *(2k + 1) k=0 (2k +l)!(2k + 2) Z J
Proof. Replacing x by ir/2 - x in Entry 20, we find that,
for 0 <_ x <_ it,
» 02k/llN2 2k+2 . 2 n
(22.1) I 2 (k!) C0S 24 * (L_ - „ + x2)Log|2 cos x|
k=0 (2k + l)!(2k +2) ^
00 1f4-1 00 If
j.1/* n V (-D sin(2kx) , 1 r (-1) cos(2kx) ±r,^
+ 2" C2" " x) / 72 + 4 i. .3 "4 ^;'
k=l k k«l k
Adding (20.1) and (22.1), we deduce that, for 0 £ x _< tt/2,
« 02k/ll>k2r 2k+2 . . 2k+2 «, 2
on 9^ V 2 (k!) Icos x + sin xi tt t ,n x
(22.2) 2, ; = " -o" Log(2 cos x)
k=0 (2k + l)!(2k + 2)2 °
+ f {<£-*>Log(2co.*>+± I (-Dmsin(2kx) }
k=l k
+ i{2x2Log|2 sin(2x)i +x J sln(fx> + | J ^.i?(3))
k-1 k * k=l k *

48
B. C. BERNDT AND P. T. J0SH1
Observe that the former expression in curly brackets on the right side of
(22.2) is equal to the right side of (18.2). Secondly, note that the latter
expression in curly brackets on the right side of (22.2) is equal to the
right side of (20.1), but with x replaced by 2x. Lastly, note that
7C(3)/8 = Xo(D- Employing all of these observations, we see that (22.2)
reduces to the desired equality.
Entry 23. For |x| <_ tt/4,
°° (-D^V „, °° 92k„ ,,2 . 2k+2
(23.!, I ktan2kx_2 I 2 <k!> *™ *-
k=l (2k) k=0 (2k + l)!(2k + 2)
oo 2k 2 2k+2
_ I V 2ZK(k!) s±nK (2x)
4 k=0 (2k + l)!(2k + 2)2
where h is defined in (8.2).
Proof. By Entry 20, the right side of (23.1) is equal to, for
|x| £ 7T/4,
2 oo oo
0 i-x i0 . I x r sin(2kx) , 1 r cos(2kx) 1 ^/0*-i
2 1^- Log|2 sin x| + Y Z ^— 4* ^ 3"""" " J ^(3)}
k=l k k=l k
-i{2x2Log|2 sin(2x)| +x \ sin(*kx) + ± J cos(4kx) 1^,
* k=l k * k=l kJ *
2
x
Z k=0 (2k + IT k=0 (2k + 1)J J
t I- I j. V sin(4k + 2)x , 1 r cos(4k + 2)x 1 /n v
Log|tanx|+x 2. ? +7 1 ^ oX,(D.
Entry 21 now implies (23.1).
Entry 24. Let x,y, 8, and ^ be real numbers with xe + ye - 1,
0 <_ x,y <_ 1, and -tt < 6, <£ <_ ir. Then
°° k °° k 2
/.\ V x cos(k6) , r y cos (lop) it Q
(l) I 5 + 2, 9 = y - Log x Log y + 8v>
k=l k k=l k °
and
oo \r oo lr
/..x v x sin(k6) , r y sin(kp) „ _ Q T
(n) l 9^—- + 2, 2 — = -<£ Log x - 6 Log y.
k=l k k=l k

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
49
Proof. Using Entry 6(iii) below, we find that
oo k ike °° k ±\ap . Q
1 —2— / —2— = Li2^xe ) + Li2(ye >
k=l k k=l k
2
m \- - (Log x + i6)(Log y + i«,p).
Equating real and imaginary parts on both sides above, we deduce (i)
and (ii), respectively.
Entry 25. Let x, y, 8, and y be real numbers such that
xe1 + ye1<lP - xye1^ , 0 <_ x,y <_ 1, and -it < 6,^ £ it. Then
(i) y x cos(k9) + y y cos(top)
k-1 k2 k=l k2
12 2
= jr Log(l - 2x cos 6 + x )Log(l - 2y cos *p + y )
1 fc -1 , x sin 6 N fc -1 , y sin y N
tan (■= 5-) tan (—l r—-)
v-i .. — ^/ 1 - y cos ^
2tau v- wj
1 - x cos 6
and
r xksin(k6) y yksin(kl)
/ 2 '■ 2
k-1 k k=l k
- - t Log(l - 2x cos 6 + x2) tan"1 (. y sin * J
4 & 1 - y cos <p
1 t /1 1 „ , 2X fc -1 , x sin 6 N
- J Log(l - 2y cos *> + y ) tan (3. - x COB e*'
i6
Proof. We shall apply Entry 6(i) with 1 - z = xe . Then
1 - 1/z = xe /(xe - 1) = ye . Since also 1 - xe = 1/(1 - ye ),
we find that
Li2(xeiQ) + Li2(ye^) = - \ Log2(l - xei6)
= \ Log(l - xei6)Log(l - ye1*)
= \ {\ Log(l - 2x cos 6 + x2) - i tan"1 (x I ^ e)>
x {\ Log(l - 2y cos *> + y2) - i tan"1 (. y sin <p )}>
^ 1 - y cos yp

50 B. C. BERNDT AND P. T. JOSH1
Formulas (i) and (ii) now follow from equating real and imaginary parts
above, respectively.
Entry 26. Let x, y, 8, and <£ be real numbers satisfying
i0 i<£ i(04tf)
the conditions xe + ye + xye =1, 0 <_ x,y £_ 1, and
-7T < 0 ,<£ <_ 7T. Then
oo 2k+l oo 2k+l
... v x K cos (2k 4- 1)8 , r y cos (2k + l)g
u; L 2 I 2
k=0 (2k + 1) k=0 (2k + 1)
7T2 1 1
= "g- - 2" L°8 x L°8 y + 2* 6^
and
oo 2k+l oo 2W+1
.... r xz sin(2k + 1)8 , r y sin (2k + 1)?
^1L' A 9 i 2
k=0 (2k + 1) k=0 (2k + 1)
1 1 Q T
s - 2^ Lo& x ~ y e Los y-
Proof. Observe that yelV? = (1 - xel6)/(l + xel6). Thus,
Entry 6(v) yields
2
. 19* , . i<p. it 1 T , i8v. , i</>.
X2(xe ) -I- x2(ye ) = ~s" " 2" Lo8(xe )Log(ye ).
Equating real and imaginary parts above, we get (i) and (ii), respectively.
The topic of sections 27-30 is altogether different from that of
the remainder of this chapter, and is a continuation of Ramanujan's studies
in Chapter 8. Ramanujan considers
x
(27.1) *> (x) = I krLog k,
r k=l
where here it is assumed that r > -1; in Chapter 8, Ramanujan studies
«P (x) when r <^ -1 and when r = -1/2 [ 6 ]. In section 29, Ramanujan
examines an analytic function of x which reduces to (27.1) when x is a
positive integer. However, Ramanujan does not give any hint at all as to
how he has defined his analytic extension of V .

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
51
Entry 27(a). Let ^ (x) be defined by (27.1) for r > -1.
For each nonnegative integer k, let
k
VO-^o^T
Then there exists a constant C such that as x tends to °°,
r * > r+1
tf (x) - Log x \ I k - Z,(-r)\ ~ C - -?-
r lk=l > (r + 1)
Boir(r + DM,,. 0(r)xr
(r + 1)
r-2k+l
V 2k v* x/AA2k-2vw~
k^ (2k).T(r - 2k + 2)
where B, denotes the kth Bernoulli number and where £ denotes the
k
Riemann zeta-function.
Proof. We shall apply the Euler-Maclaurin summation formula
[12, articles 102, 107] to f(t) = trLog t. Then as x tends to °°,
we find that
rx °° B
(27.2) * (x) * I trLog tdt+y xrLog x + c' + J" j~ f(2k_1)(x),
r tl 2 r k=l C2k;'
where C is independent of x.
First, an integration by parts yields
rx r+lT r+1 ,
r, .x Log x x , 1
(27.3) t Log t dt = » 2+ 2'
h r + l (r + 1) (r + 1)
Secondly, by Leibniz's rule,
Ol £"> f(n)fi-t - f(r + 1) ,.r-nT„„ ,-
(27.4) f (t) - r(r + x _ n) t Log t
. r-n "r1 ,n. r(r + 1) (-l)"'^1 (n - k - 1)1
ki0 V r(r + 1 - k)
- r(r + 1) r-n. (-l)n+1n!tr~n "f1 f(-r + k)
' T(r + 1 - n) C L°8 C + f7=7) k^ (n - k)k!
. F(r + 1) r-nLoe t . ™°T<* ~ *K-lMt™
~ r<r + 1 - n) l L°8 C + fT^rl '

52
B. C. BERNDT AND P. T. JOSHI
by a formula from Hansen's tables [41, p. 126], Using (27.3) and (27.A)
in (27.2), we deduce that
(27.5) *(x) - \L+°S * - -^-y + *^* + Cr
r r + i (r + l)z ^ r
- B2kr(r + l)xr_2k+1 r
+ ki1 (2k)ir(r +2 - 2k) 1Log x + M2k-2(r))'
i 2
as x tends to °°, where C = C + l/(r+l) .
From Entry 1 of Chapter 7 [9], we have
X r xr+1 xr . B?kr(r + Dxr_2k+1
(27.6, ^ kr , i__ + x. + c(.r) + ^ _|^i____ ,
as x tends to ». Substituting (27.6) into (27.5), we deduce the desired
asymptotic formula.
Entry 27(b). Let C^ be as in Entry 27(a). Then if r > 0,
2r(r + l)C(r + 1) r . , /0WT ,0 v r'(r + IK
Cr = —- p~ {sin(7rr/2)(Log(27T) - y ^ x) )
(2tt)
tt / mm . 2f(r + l)sin(TTr/2) r Log k
- j cos(7rr/2)} 4- — J, -j^f •
2 (27T)r^i k=l kr+1
Proof. We shall first show that
(27.7) Cr = -C'(-r).
It is clear from (27.2) and (27.3) that
(27.8) Cr = li» {^(x) - \?K * + -5^-2 " 2L£|ti
x-*» (r 4- 1)
n Bni j2k-l
r 2k d , r_ .-»
" S T2k)T71k=T (x L°s x)}>
k«l ' dx
where n is chosen so that n > (r + l)/2. Applying the Euler-Maclaurin

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK 53
summation formula once again, we find that
(27.9) * (x) = x i°fX - -5 + 1 + 2Li£^Ji
(r + 1)^ (r + 1)Z Z
r + 1
n B01. r j2^-l
n B r d2^"l r d2k"X r I 1
+ £ T2kTr i 2k-i (x,rLo8 x) —2k^r (tr]Log t} r
k-1 UK;* ^ dx K dt: L 't-L'
rx h 2n+i
P2n+l(t>-2^(trL°8t)dt'
at
where P.(t) denotes the jth Bernoulli function. Formulas (27.8) and
(27.9) imply that
n B ,2k-l
(27.10) C - i—j - I -$- ^^ (trLog t)
r (r + 1) k-1 UK;" dt K L
t-1
r°°
.2n+l
d ,^r.
+ L P2n+l(t> T2HTT <' Lo* '>«■
'1 at
Now apply the Euler-Maclaurin formula to f(t) - t , Re s > 1, to find
that
<"•"> '<->-i4r*i-j1wjT'(2k"l><"
fl
where n > (r + l)/2. By analytic continuation, (27.11) holds for
Re s > -2n - 1. Differentiating (27.11) and then setting s * -r, we
find that
n B J2k-!
(27.12) C'(-r) = i—y+ I -|* ±__ (trLog t)
(r + 1)Z k-1 Ult;" dt: X
2n+l
t-1
f H n 1
" J1 **»i™ jpsi «*"" t)dt
A comparison of (27.10) and (27.12) yields (27.7).

54
B. C. BERNDT AND P. T. JOSHI
From the functional equation of C(s) [76, p. 13], we find that
(^(s) = 2(27T)S"1sin(TTS/2)r(l - s)Ul - s)«JLog(27T) + y cot(7Ts/2)
. r'(l - s) _ r/(l - S) 1
r(l - s) u(i - s) J"
Putting s =-r yields
^(,r) , - "fr + ^(r | Dsin(.r/2)|Log(27T) , , cQt(7Tr/2)
(27T)r+i l '
r'(l + r) 1 y Log k 1
r(l + r) Ur + 1) L r+! j
k=l kl
By (27.7), the proof is complete.
In the first notebook [66, vol. 1, p. 163], Ramanujan indicates
how he derived Entry 27(b), but his argument is not rigorous.
The following corollary is an immediate consequence of Entry 27(b).
Corollary. If r is an even positive integer, then
c cos(TTr/2)r(r + 1)5 (r + 1)
r 2(27T)r
Ramanujan next records the following particular values of C :
I,„^ r _ mi , c . _ 3£H)
4tt 4tt
C0=2 L°g(2*), C2=^, c4 - - 4
and
r _ 45^(7)
6 ' Q 6 '
In the case r = 0, we see that CQ is the constant which occurs in the
asymptotic expansion of Log T(x + 1) (Stirling's formula), and this
constant is well known to be j Log(2ir) [2, p. 329]. The values of C2,
C, , and Cfi are immediately deducible from the Corollary.
The next example is not correctly given by Ramanujan [66, vol. 2,
p. 113]. Furthermore, the additive factor of 1/4 in the denominator of
Example (i) may be deleted without affecting the limit.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
Example (i). We have
2 1, .1
2 ~~5 ~ ' 2 ^ - ok
e | I k
(27-13) lim i x(x^)rr 2 ^ nu/"31—-:rrk*
x-*» x T(x + 1) k*l
f x*(x+3)r(x2 + i)V
( x + 1/4 J
Proof. The logarithm of the left side of (27.3 3) is
(27.14) L = lim «|2 f k Log k + j (x2 - j Log x + y - y)
x-kjo ^ k-1
- ~ x(x + 3)Log x - \ Log T(x2 + 1) + -j Log(x + j)\
3
X 1,21
= lim •{ 2 I k Log k + j (x - y Log x + j - y)
X-x»
I k-1
- i x(x + 3)Log x - -j ((2x2 + l)Log x - x2 + ~ Log(2ir))
by Stirling's formula [2, p. 330].
On the other hand, by (27.8) and Entry 27(b),
(27.15) lim -j 2 J k Log k - x Log x + y x - x Log x - — Lo8 x
x-*» ^ k=l
i L^^x r'(2) 1 i " Log k
i{Log(2w)+Y-l}+i I ^
° ^ ' iT k-1 k
by [37, p. 945]. If we employ (27.15) in (27.14), we deduce that
2 L 2 *
v k=l k
which completes the proof.
Example (ii). We have
(27.16) lim ex3/9 " x/12 TT (k/x)^ - ,C<3>/<4.\
x-*» k»l

56
B. C. BERNDT AND P. T. JOSHI
Proof. The logarithm of the left side of (27.16) is
3 >
+ *_-JL I
( x o 3
(27.17) L = lim O k (Log k - Log x) + y u
x-*» ^k=l ^
r x3 x2 x x3 x ]
= lim ^2(x) - (3- + X + f)Log X+T"H f"*
On the other hand, by (27.7) and the Corollary above,
(27.18) itoLw -!&f^ + £-£fL*-*J&LJL_*-\
_ C(3)
, 2 '
Comparing (27.17) and (27.18), we readily deduce (27.16).
For each positive integer r, define
HrBr+l(x + X) v B2kr!H2k-lxr_2k+1
(28.1) f(r,x) = r rtl. . 1 2k 2k 1
r + 1 Kk<(r+l)/2 (2k)!(r + l-2k)! '
where H is defined in (3.1) and B (x) denotes the nth Bernoulli
n n
polynomial, 0 <_ n < °°.
Entry 28(a). For r >_ 2,
(28.2) f(r,x)
Br+1(x + 1) - Br+i fx
r(r + 1)
+ r f(r - l,t)dt.
Proof. Since B +1'(x) = (r + 1)B (x), a direct calculation
with the use of (28.1) yields
rx H _B (x + 1) H .B ..
r f/r . ,-W - T~1 r"t"1 1 r-1 r-H
r J^ f(r - l,t)dt j^pj rlrT-
r-2k+l
j B2kr!H2k-lX
Kk<r/2 (2k)!(r + l-2k)!
B ..(x + 1) H _B ,_
£/ N r+1 r-1 r+1
f(r,x) -
r(r + 1) r+1
H B ^.i
, r r+1
r + 1 '
from which (28.2) immediately follows.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
Ramanujan next studies an analytic extension of ^ (x) for all
real values of x and any positive integer r. He does not give us his
definition, but there exists considerable motivation for defining
(28.3) *r(x) = Cf(-r,x + 1) - S'(-r),
where £(s) denotes the Riemann zeta-function and £(s,a) denotes the
Hurwitz zeta-function. Normally, the definition of £(s,a) requires
the restriction 0 < a £ 1. We shall remove this stipulation so that in
the sequel a_ denotes any real number. In fact, we could allow a. to
be complex, but since Ramanujan evidently considered only real values of
a_, we shall do likewise.
First, note that if Re s > 1,
(28.4) Sf(s,x + 1) - £'(s,x) = x~SLog x.
By analytic continuation, (28.4) is valid for all complex nubmers s.
Putting s = -r, we find that 0 (x) - 0 (x - 1) = x Log x for any real
number x. Since V (0) = 0, we see that (28.3) is compatible with
(27.1).
Secondly, (28.3) is similar to definitions of other analogues of
Log I'Cx + 1) studied in Chapter 8 [ 6 ], and if r = 0, (28.3) reduces to
a formula of Lerch [8], [80, p. 271] for Log T(x + 1).
Thirdly, if x and r are positive integers,
/00 -* V ir r+1 r+1
(28.j) I k = -—j ,
k=l r + L
where B (x) denotes the nth Bernoulli polynomial and B denotes the
n n
nth Bernoulli number, 0 £ n < °°. Since
B,(a)
(28.6) C(l - k,a) = - -^— , k >_ 2, 0 < a £ 1,
we find that, for -1 < x £ 0 and r >^ 1,
B (x + 1) - B
(28.7) = -5(-r,x + 1) + C(-r,l).

58
B. C. BERNDT AND P. T. JOSHI
If we formally differentiate (28.5) and (28.7) with respect to r and
ignore the different restrictions on x, we formally deduce (28.3).
Entry 28(b). If |x| < 1 and r is any positive integer, then
H r+1
(28.8) „r(*) = 7iT (Br+1(x + 1) - Br+1) = jQL_
,„ „ r+1-2k
. y r!B2kH2k-l* _ *? r)c Xr-K
Kk<i+l)/2 (2HKr + l-2H! & W
+ V (-Dkr!(k - l)!g(k)xr+k
ki2 (r+k)l
where H is defined by (3.1), y denotes Euler's constant, and
Ck = -^'(-k), k >_ 0.
The theory of a certain analytic extension of (28.3) has been
extensively developed by Bendersky [5]. In fact, for |x| < 1, Bendersky
[5, p. 279] defines his analytic extension by (28.8), except that the
first sum on the right side of (28.8) does not appear in his definition.
Busing [15] has further developed Bendersky's work and has removed some
deficiencies in Bendersky's definition of the constants L, , which are
closely related to C, here.
Proof. For Re s > 1 and |x| < 1,
(28.9) C'(s,x+1) = - Y L°g(k + x)
k=l (k + x)S
88 - I -7 I C*)<$3 Log(k + x)
k=l kS j =0 3 K
k=l k j=0 J m*l
00 °° , °° n n-1 , - Nn+
j-0 J k-1 kS n-1 j=0 J n 3

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
59
"-1 -s, (-l)n+J
I (-^'(b + J)*3 + I Us + n)xn I Cp^-r
j=0 3 n-1 j-0 J n " J
= S^s) + S2(s) + S3(s) + S4(s) + S5(s) + S6(s),
where
-sN ., . , ,N r+1
S.(s) = I <"!)£'<s + j)xJ, S9(s) = ( "*)C'(8 + r + l)x ,
1 j=0 J * r^
"I1 -. (-l)n+J
S,(s) - I (~>'(s + j)xJ, S,(s) - I £(s + n)xn J ("S) i=i2—
J j=r+2 J * n-1 j-0 J n " J
Sc(s) = £(s + r
+>,,- .i «-•> $£-'
and
00 n-1 .n+j
S (s) -J Us + n)xn I C*) V , •
6 n-r+2 1=0 j n " J
By analytic continuation, the far right side of (28.9) represents
C'(s,x + 1) for all s.
We now evaluate S..(-r), S~(-r), S,(-r), S,(-r), and
lim (S2(s) + S5(s)}. First,
s->~r
(28.10) S1(-r) - I (*)C'(j - r)xj - - f (£>Ckxr~\
j=0 J k-0
since C • (-k) = -Ck, 0 £ k <_ r, by (27.7).
Secondly, since (.) - 0, j > r+2,
(28.11) S3(-r) = 0.
In the calculation of S,(-r) it will be convenient to let
(a) a r(a + k)
u;k r (a) • k - u*
Using (28.6) and the fact that £(0) = -1/2, we find that
£ B_, (1) _ n-1
SA
<-r> . [ W(1) xn f (r, (-I)***

60
B. C. BERNDT AND P. T. JOSHI
Employing the formula [41, p. 126],
m-1 (a)k (a)m m-1 1
(28'12) J0 k!(»-k) = ii- j0 TTk •
we find that
I K^ „(D(-l)r+1(-r)_xn n-1
(28.13) s4(-r) = ^ *+1-» + , _ n)nl " jo Z^
n=l k=0
r
JL
r
TT I <r>r+l-n(1)xn<Hr " Hr-n>
n=l
n=l k=l
r {B_,(x + 1) - B_, - xr+1)
r + 1 r+lv ' r+1
r«B H xr+1_2k
v r!B2kH2k-lx
l<k<(r+D/2 (2k)!(r + l-2k)! •
where we have used a familiar formula for B . (x + h) [2, formula 23.17,
p.804].
Letting n = k + r and j = r - m, we find that
S6(-r) = i (-1) cOOx J (m) ^-^ .
k=2 m=0
Now if k is a positive integer [36, formula (1.41), p. 6],
(29, 141 V tx\ <-1>"' - r!(k - 1)!
(28.14) i (m) irnr - .
m=u
Thus,
f28 151 s r r^ - V (-Dkr!(k - l)!c(k)xr^k
(28.15) S6(-r) - ^ (r + k)! '
We lastly examine S2(s) + sc(s) as s tends to -r. Letting
f
(s) " jlo ( J} -~ •

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
61
we have
(28.16) L = lim (S2(s) + S5(s)}
s-*—r
= xr+1 lim {(r+1)Cf(s + r + 1) + f(s)£(s + r + 1)}.
s-*~r
Replacing j by r - k and using (28.14), we find that
r k
(28.17) f(-r)-- J (k)^2I--7TT-
k=0
Since
r ( ..rfl J-1 j-1
f'(s) = I (r + 1- 1)1' 2 TT<b + «>.
j=l ^r ^ x J;J' k=0 m=0
ra^k
we see that
c-Dr+j ,r% J;1
(28.18)
i
f.(_r) . y (-D J (*) V _JL-
(r) jii' + l-j VA r-k
r ...r+j
= T izii _ (y)(H - H .)
j£l r + 1 " J 3 r r"3
Tl + T2«
say. We first calculate T.. We have
(-DrH r
j-1 J
(-l)'H
(28.19) Tt - r + ^ I (-1)J(, )
_I {-! - (.1)^}
^(l -<-!>*>.
Inverting the order of summation and, in the third equality below, using
well-known evaluation [37, p. 3], we find that

62
B. C. BERNDT AND P. T. JOSHI
r:x ^r+J+1 r rrJ
(28'20) T2= Z r""l - 1 fr I ¥
Z j=l r + X J J- k-1
, nr+l r-1 - r-k . _
r + x k=i K j=i J
f , xT+l r-1 - ±.
= izl> y I {(-Dr+kr r ) - i)
r + l ' k u 1} W; if
, . Nr+1 r-1 , .>k , _ Nr
(-1) v ("1) rr^ + ^"1) «
r + l /. r - k V r + 1 r-1
k=l
(-l)r+1(-r)r r-1 r
(r + l)r! £Q -r + k r + 1 r
-+TJ (1 + (-l)r H),
r+l * ' r'
where in the penultimate equality we used (28.12) again. Employing (28.19)
and (28.20) in (28.18), we deduce that
2H
(28.21) f'(-r)
r + l
We now return to (28.16) and expand each function on the right
side about z = 0, where z = s + r. The expansion for l/T(-z) can be
found in [37, p. 936]. See [37, p 945] for the coefficient of z in
the Maclaurin series of T(r - z + 1). For the first two terms in the Laurent
expansion of C (z + 1), see [76, p. 16]. Hence, using also (28.17) and
(28.21), we find that
(, l%fH - D C'(s + r + 1) + f(.)C(. + r + 1)
' (r +"i)| {_z + yz* +",}{r! + ft - Hr)r!z +"'H--^ + Cj +•••>
z
2H
+ {- _LT + _I_Z +...}{i + Y +...).
r + l r+l z
Hence,
r+l 1 Y Y 2Hr xr+1
(28.22) L - ^h-^n (Y - V - j^j - ^rj + 7TT - 7TT (Hr * Y>'

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
63
If we now utilize (28.10), (28.11), (28.13), (28.15), and (28.22)
in (28.9) with s = -r and recall the definition of <£ (x) in (28.3), we
readily deduce (28.8).
Entry 29. Let \p (x) be defined by (28.3), let C be defined
by (27.7), let -1 < x £ 0, and let n and r be natural numbers. Then
n-1 , B _,_. (x + 1) _,,
„ / x r r „ /x - kN r-H T , ,. r+lN _
V (x) - n I yp (—-—) = ———: Log n + (1 - n )C .
k=0
Proof. Let Re s > 1 and replace k by n - k below to get
n_1
C'(s,x + 1) - nr I ?'(s, iL^+ 1)
k-0 n
= C'(s,x+ 1) +nr I I L°gUjn + k + x)/n)
k=l j-0 {(jn + k + x)/n}S
- C(s,x + 1) +nr+s I Log{(m + x)/n}
m=l (m + x)
(1 - nr+SKf(s,x + 1) - nr+SLog n £(s,x + 1).
By analytic continuation, the extremal sides of the equalities above are
equal for all complex s. Now let s ■ -r and use (28.3) and (27.7)
to obtain
n-1
* (x) - nr I * (2L"=Ji) - -U-r,x + l)Log n + (1 - nr+i)C .
r k=0 r n r
Noting that -1 < x £ 0, we may employ (28.6) to complete the proof.
Corollary 1. If n and r are any positive integers, then
n;X k Br+lLog n -r
A^("")="7T^+(n"n ^
k*l (r + l)n
Proof. Putting x * 0 in Entry 29 and recalling that <p (0) * 0,
we easily deduce the desired equality.

64 B. C. BERNDT AND P. T. JOSHI
Corollary 2. If r is any positive integer, then
1 Br+lLog 2
*r(- V — r + (2 " 2 r)C .
r (r + l)2r r
Proof. Set n = 2 in Corollary 1.
Entry 30. Let 0 < x < 1. If r is positive and even, then
00
(i) *(x - 1) + V_(-x) = 2C + —^- cos(Trr/2) £ cos(27rkx) •
r (2lT)r k-1 kr+1
if r is positive and odd, then
(ii) *<* - 1) - tf(-x) - -£L- Bin(,rr/2> \ Sin(2fx) .
(27r)r k-1 kr+1
Proof. Recall Hurwitz's formula [76, p. 37]
5(s,x> = 2r(l - s)/sin<irs/2> \ COs(2fx) + cos(irs/2) J «in(2ffkx)| §
^ k=l (27Tk)1 S k=l (27Tk)i S j
where Re(s) < 1 and 0 < x < 1. Differentiating with respect to s, we
find that
(30.1)
C'(s,x) «-2r(l- s)/sin<irs/2) I cos(2irkx)+cos(7rs/2) J sin^kx) |
t k=l (2irk)i S k-1 (27Tk)i"S 1
. -,, „nj n, /0x v cos(2irkx) . , /ON v sin(2iTkx) 1
+ 7rr(l - sH cos(7Ts/2) ) -_ - sm(7rs/2) £ i r
1 k=l (2irk)i s k=l (2iTkr s >
+ 2r(l - s)(sin(TTs/2) J coS(2TTkx)Log(27Tk)^cos(7rs/2) J sin(2Trkx)Log(27rk)|>
I k=l (27Tk)i s k=l (2irk) S ^
Thus, if 0 < x < 1 and r is even,
«Pr(x - 1) + «Pr(-x) = £f(-r,x) -I- ^'(-r,l - x) + 2Cr
= 2Trr(l + r)cos(Trr/2) £ cos(2^kx) + 2C,
k=l (2irk)r -1
which completes the proof of (i). The proof of (ii) is analogous.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
65
Ramanujan remarks that "More general theorems true for all values
of r can be got..." [66, vol. 2, p. 114]. Indeed, (28.3) can be used to
define ^ (x) for any real number r, r ^ -1. Thus, (30.1) can then be
employed to obtain generalizations of (i) and (ii) for r > -1.
Entry 31(a). Suppose that </>(x) is defined by (28.3), and
define
oo 2k+l
(3i.i) *(,) = I (2kk) s|s GSL .
k=0 * 2 (2k + 1)
Then if 0 < x <_ 1/2,
♦ (x) - Tr(^1(x - 1) - ^(-x)} + ttx Log(2 sin(irx)).
Proof. By Entry 30(ii), for 0 < x < 1,
irto^x - 1) - ^(-x)} + ttx Log(2 sin(Trx))
00
1 r sin(2irkx) , T /0 . , x N
= J L 9 + ™ Log(2 sin (ttx)).
k=l k
But by Entry 16, with |x| £ 1/2,
*(x) = *x Log|2 sin(7Tx)| 4j sin(2irkx) .
Z k-1 k
The desired result now follows.
Entry 31(b). Let ^(x) be defined by (31.1) and let hR be
defined by (8.2). Then we have
k
t-\ m \ V ("1) hk-H «. 2k+l, , , , .„
(i) i|>(x) * l tan (ttx), |x| £ 1/4,
k=0 2k + 1
(ii) i|,(x) + *(? - x) -2. Log(2 cos(ttx))
00 MNk 2k+l, x
+ I ("1} tan fSl , 0< x< 1/2,
k=0 (2k + 1)

66
B. C. BERNDT AND P. T. JOSHI
(iii) *(| - x) + I *(2x) - i|>(x) = \ Log(2 cos(ttx)), 0 <_ x <_ 1/4,
and
(iv) *(y - x) + *(| + x) = tt(1 - 2x)Log|2 cos(ttx)|
+ I -^ S'n(27Tkx) , 0<x< 1.
k=l k
Proof of (i). By the Cauchy multiplication of the Maclaurin
-1 2
series of (tan t)/t and 1/(1 + t ), or by Proposition 15(i), we find
that, for |t| < 1,
^4- = I (-Dkw2k.
t(l + t ) k=0 k+i
Thus, for |x| <^ 1/4, we find that the right side of (i) is equal to
rtan(iTx) -1 f^x
(31.2) n Z ? dt = I u cot u du
J0 t(l + O J0
00
t h • /^ m . 1 V sin(2irkx)
= ttx Log I 2 sin(irx)| +t i 5 .
k=l k
In the penultimate step we made the substitution u = tan t, and in the
last step we applied Entry 13 with n = 1 and used
(13.1) and (13.3). Using Entry 16 on the far right side of (31.2) along
with the definition (31.1) of *l>(x), we complete the proof.
Proof of (ii). From the definition (31.1) of ty(x), we find that,
for 0 <_ x < 1/2,
01 . 2k+l/<tr N . 2k+l/<tr x
ii. / \ . ,i. A \ V /2k\ sin (ttx) + cos (TTx)
* (x) + * (y - x) - I ( , ) 5k 2 •
L k=0 R 2*K(2k + ir
Applying Entry 19, we finish the proof.

CHAPTER 9 OF RAMANUJAN*S SECOND NOTEBOOK
67
Proof of (iii). By Entry 18, for 0 £ x £ 1/4,
oo 2k+l
• /l x i / x n , / \n 1 v /2kN sin (2ttx)
*(=■ - x) - *(x) - x Log(2 cos(ttx)) - T J, U)"^ 9'
2 Z l k=0 k 2ZR(2k + 1)Z
Using the definition (31.1), we complete the proof.
Proof of (iv). Using the definition (31.1) and (18.2), we find
that, for 0 £ x £ 1,
2k+l,
i/l \ . i A _L \ i v /2kx cos (ttx)
lKy " X) + *(y + X) = 2 ^ ( ) -^ ^—^y
Z k=0 K 2 (2k + 1)
oo k+1
«/rr N<r i0 / \i . V (-1) sin(2irkx)
= 2(y - 7Tx)L0g|2 COS (TTX) I + J. 2 »
k-1 k
which completes the proof.
Part (iv) is not what is claimed by Ramanujan [66, vol. 2, p. 115].
Ramanujan asserts that
lM"| - X) + lKj + X) = 7T L0g(2 COS(TTX)).
Evidently, Ramanujan applied Entry 16 twice, with x replaced by -jw - ttx
and with x replaced by -^ + ttx. But the intersection of the domains
for which these two equalities are valid is only the origin. The infinite
series on the right side of (iv) cannot be evaluated in terms of elementary
functions. In fact, from [37, formula 1.441;4, p. 38], it is easily seen
that, for 0 £ x £ 1,
(-1) sin(27Tkx) _ 0 [X
V l-ij sinUTTkxj ~
k-1 kZ J
Examples. We have
Log I 2 cos(irt) |dt.
0
(i) t|i(1/2) = \ Log 2,
(ii) i|>(l/4) = \ G + J Log 2,

68 B. C. BERNDT AND P. T. JOSHI
(iii) *(l/3)-^f I -1 r —+ fLog3,
z k=0 (3k + 1) 9/3 D
(iv) Ml/6) = ^ J ^—5 - -L ,
k=0 (3k + 1) 6/3
and
oo k 2k+l
(v) 2*(x) -|*(2x) - ), ("1) tan fi^I , 0<x<lM.
k=0 (2k + 1)
Proof. Parts (i) - (iv) are merely restatements of Examples
(i) - (iv) in section 16.
Part (v) arises from Entry 31(b) by subtracting (iii) from (ii).
In Ramanujan's version [66, vol. 2, p. 115] of part (ii) above,
read 2\j,(l/4) for i|;(l/4).
Entry 32 below should be compared with Entry 17.
Entry 32. For |x| <_ tt/4, we have
oo „2k/1lN2 . 2k+l/0 . oo k 2k+l
(32 i) V 2 (k!) sin (2x) = 2 V C"1) tan x
k=0 (2k)!(2k+l)2 k=0 (2k + l)2
Proof. Using Proposition 15(ii) and then making the successive
substitutions t = sin(2u) and u = tan v, we find that the left side
of (32.1) is equal to, for |x| £tt/4,
rsin(2x) °° 92k,. n2 2k rsin(2x) . -1
*■ (2k + l)i dt " , ■ ■"- dt
k=0 UKL + L)' Jo /2
t/1 - t
/•x ftan x -1
csc(2u)du - 2 I tan v dv
J U
/•tan x °° , lNk 2k °° , lNk 2k+l
= 21 r -^Vt- dv -2 /v (~1} tan 2 x
Jo k=o k=o (2k + iy
Corollary (i). For |x| <_ 1, we have
oo 2k 2 t \ k °° k
(32.2) I 2 (k!) - f *X J = (1 + x) I (-x)
k=0 (2k)! (2k + l)2 ^(1 + x)2-1 k=0 (2k + l)2

CHAPTER 9 OF RAMANUJAN*S SECOND NOTEBOOK 69
Proof. Replace tan x by /u in Entry 32. Noting that
sin(2x) = 2 tan x/(l + tan x) * 2/u/(l + u), we readily deduce (32.2)
with x replaced by u.
Ramanujan [66, vol. 2, p. 115] has a slight misprint in the
third summand on the left side of (32.2).
Corollary (ii). If |x| £ tt/4, then
00 ( ^ko2k/i iA 2k+1^ ^ °° - 2k+1
V (-1) 2 (k!) tan (2x) « r _tan x
k=0 (2k)!(2k + l)2 k=0 (2k + l)2
2
Proof. In Corollary (i), replace x by -tan u. Noting that
2 2 2 2 2
tan (2u) = 4 tan u/(l - tan u) s -4x/(l + x) , we easily achieve the
proposed identity with x replaced by u.
Examples. We have
(i) I 2 (k!) 2 - 2G,
k=0 (2k)!(2k + 1)
°° k 2 2
,..x V 3 (k!)z * T o IOtt , e v
(n) L o = z L°g 3 " IT + 5 Z
k=0 (2k)!(2k + 1) 3/3 ' k=0 (3k + 1)
(iii) I (k!)2 2 - § G - 1 Log(2 + /3),
k-0 (2k)!(2k + 1) J J
00 k 2 2 °° k
Uv) I 2«!> 2 = -^Log(1 + /2)-^-+4l -^
k=0 (2k)! (2k + 1) 2/2 4/2 k=0 (4k -»- 1)
00 2k 2
(v) I 2 (k!) 2 (1 - -fe) - 7 Log(2 + /3),
k=0 (2k)!(2k +1) 4K L *
00 k 2k 2 2
(vi) j (-l)W) ,_^.lLoe2(1 + ^)t
k=o (2k)! (2k + ir a
and
(vii) J (-l)W . £ . 3 Log2(*fi>.
k=o (2k)! (2k + ir

70
B. C. BERNDT AND P. T. JOSHI
Proof. Example (i) follows from putting x = 1 in Corollary (i)
or x = tt/4 in Entry 32.
Putting x = 1/3 in Corollary (i) yields
£ 3K(kir m i £ (-i/3)K ^
k=0 (2k)!(2k + l)2 3 k=0 (2k + l)2
Employing Example 17(i), we deduce part (ii).
Thirdly, put x = (2 - /J)2 in Corollary (i). Then 4x/(l + x)2 = 1/4,
and so we get
I (k!)2 _, J (-l)k(2 - /3)2k^ m
k=0 (2k)! (2k + l)2 k=0 (2k + l)2
Applying Example 17(iii), we readily deduce (iii) above.
Next, put x = {/l - 1) in Corollary (i). Then 4x/(1 + x)2 = 1/2,
and thus
Y 2K(k!)Z m 2/- y (-l)K(/2 - 1)ZR+1 ^
k=0 (2k)!(2k + l)2 k«0 (2k + l)2
Appealing to Example 17(ii), we readily find (iv).
Fifthly, multiply the formula of part (iii) by -3/4 and add it
to the formula of part (i). This yields part (v).
Next, let tan x = tan(iT/8) = /2 - 1. Then tan(2x) = 1, and
so Corollary (ii) yields
I (-l)k22k(k!)2 , J (/2 - l)2^1 m
k=0 (2k)!(2k + l)2 k=0 (2k + l)2
Now use Example 6(iv) to deduce formula (vi).
Lastly, let tan x ■ /5 - 2 in Corollary (ii). Then tan(2x) = 1/2.
Using Example 6(vi), we readily achieve (vii) above.
Example (i) is originally due to Nielsen [59, p. 166] and is
reminiscent of the formula (0.1) for £(3) that was used by Apery [3] to
prove the irrationality of £(3). An interesting, animated account of

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
71
Apery's proof has been written by van der Poorten [77], Mendes-France [55]
has also described the lecture in which Apery announced his achievement.
In fact, formula (0.1) appears to be originally due to Hjortnaes [44] in
1953. Other proofs have been given by Hawkins [43], Ayoub [4], and
van der Poorten [78]. Cohen [19] and Leshchiner [51] have established different
formulas for £(n), 2£n<°°, for which (0.1) is a special case. Other results in
the spirit of Examples (i) - (vii) and (0.1) may be found in Comtet's book
[20, p. 89] and in the papers of Clausen [17], Ruscheweyh [68], van der Poorten
[78], [79], Zucker [82], and Gosper [35]. It is interesting that the
formula for G in Example (i) was discovered almost a half century before
the formula (0.1) for £(3).
Entry 33. If n is a positive integer, then
ffT/2
(i)
X COS X Sin(nx)dx = T^r H ,
Jn 9n+2 n
where H is defined in (3.1), and
n
rir/2 n 1 n 2k
(ii) I cos x sin(nx)dx = —tt J T~ •
Jo 2n+1 k=l k
Proof. We shall prove only (i). Formula (ii) is slightly easier
to establish, and a proof may be found in [31, p. 136],
By Entry 5(i) and an integration by parts,
rir/2 n 1 n n f7r/2
x cos x sin(nx)dx = — T (, ) | x sin(2kx)dx
JO 2n k-1 k J0
2n+2 ^ V k
" ff H ,
2n+2 n
by a well-known formula [37, p. 4].
The next two results are designated by Ramanujan as corollaries
of Entry 33. However, we prefer to begin the proofs anew. On the surface,

72
B. C. BERNDT AND P. T. JOSHI
it appears that these two corollaries as written by Ramanujan are devoid
of meaning. However, each can be assigned a mathematically precise
meaning.
Ramanujan defines
/ x V 2k - 1
k=l K
and claims [66, vol. 2, p. 116] that y?(n) "can be expanded in ascending
powers of n in a convergent series the first two terms being
2
S2x/2 + S3x /8 + &c." Here Sfe « COO, k _> 2. We shall need to extend
the definition of <£(x) to all real values of x. Upon integrating both
sides of
r k-1 tX - 1
k-1
over 1 < t < 2, we find that
(33.1) V(x) =
i ' -*
Thus, we shall define V(x) for every real number x by (33.1). By
expanding t - 1 in a power series and inverting the order of summation
and integration, we find that
ax
k-1 R'
where
f2 k
(33.2) a^ = J ±28-1 dt, k>l.
We now state a revised version of Ramanujan's first corollary.
Corollary 1. If a is defined by (33.2), then a = j £(2)
and a2 « ^ £(3).
Despite the fact that a is a rational multiple of C(k + 1)
for k * 2,3, it does not appear that this property persists for k > 3
(see [52, p. 199]).

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
Proof. Integrating by parts and using Example (i) in secti
we find that
k+l
1 t L
2 , r*
= yLog22-Log2 J, -i- - Li (1/2) + 5(2)
k-1 k2k 2
2
- - jLog22 - ^ + |Log22 + 5(2) -y 5(2),
as claimed.
Next, integrating by parts twice and employing Example (i)
both sections 6 and 7, we find that
2.2^
1 z L k=0
2 t 2.
LoS t dt
k+l
1 tK+±
= i Log32 - Log32 - 2 Log 2 Li2(l/2)
2 Li3(l/2) + 25(3) « | 5(3),
since x3(D - g" CO).
Corollary 2. For each positive integer n,
oo
k«n k2
where H is defined by (3.1).
n
Proof. From (33.1),
r2 .-n , tl n-1
*(-n) - I V^T1^ -4 1 '
Ji c " Ji k=o
n-1 n-1
--Log2+ I_-±-- J J
dt
k-1 k2k k=l k
OO
'• . «k n-1
k=n k2

74
B. C. BERNDT AND P. T. JOSHI
Ramanujan [66, vol. 2, p. 116] next seems to indicate that
Corollary 2, perhaps in conjunction with Corollary 1, can be used to
find the value of Li,(l/2), k > 2, where Li, is defined by (6.1).
The calculations in the proof of Corollary 1 make it clear that Li _ (1/2)
arises in the calculation of a, , k >_ 1. Since Corollary 2 is valid only
when n is a positive integer, it does not appear that these last two
facts can be utilized to determine Li, (1/2).
Entry 34. Let -1/2 < x < 1. Then
tuu T * r-2L_,k+1 ? (-i)k22k(k.)2hk+1xk+1
kio (2k + i)2 (l + x) \io <2k + 1>!
where h, is defined in (8.2).
Proof. Rearranging the double series below by absolute
convergence, we find that, for -1/2 < x < 1,
(34.2) I 1—^ (r5-j) = I -* I (-k-l)x3
k=0 (2k -1-1) L X k=0 (2k + 1) j-0 3
- I (-I)" I ('1)k 2 (^xn+1.
n=0 k«0 (2k +1) K
Comparing (34.1) and (34.2), we see that we must show that
n /,\k 02n, ,*2
(3A.3) I (~1} 2 <£> - 2n(:' )t hn+r n > 0.
k=0 (2k + IV Un ij' n+i
Since
I (-l)k<")t2k - (1 - t2)n,
k=0 R
we find after two integrations that
)k „ .n
(2k + lY
(34.4) I ("1)k 2 <") = I' ^ f (1 - t2)ndt
k=0 (2k + 1)Z K JO X J0
r«-12>^ r t
Jo Jt x
-i:
(1 - t2)nLog t dt.

CHAPTER 9 OF RAMANUJAN'S SECOND NOTEBOOK
75
Letting B(x,y) denote the beta function and ip(x) the logarithmic
derivative of the gamma function, we find that the far right side of (34.4)
is equal to [37, formulas 4.253,1, p. 538; 8.363,3, p. 944]
- j B(-|, n + l){if;(l/2) - *(n + 1 + 1/2)}
/iT n! r f 1 1 ]
(n + 3/2) ^n 1 . u " 1 . . , . u
k=0 [ -r- + k -2 + n + l + kJ
4r
22n(n!)2 ,
" (2n + 1)! n+1*
where we have used the Legend re duplication formula. This completes the
proof of (34.3) and hence of (34.1) as well.
If we let x tend to -1/2 in (34.1), we obtain a formula for
Catalan's constant that has been found in a different way by Fee [30].
In preparation for the last theorem in Chapter 9, define
00
K - T and A » (1 + cos(Trn)K (n),
r k-1 kr(k + l)r n
where r is a positive integer and n is any integer. If n = 1,
it is to be understood that A, = 0. Since Z, (-2k) « 0 for each positive
integer k, it follows that A =0 if n < 0. Also, since C(0) ■ -1/2,
we have A * -1.
Entry 35. For each positive integer r,
00
(35.1) K = I (-l)k(r+^1)Ar .
k=0 k r"k
By the remarks made above, the series on the right side of (35.1)
terminates. Formula (35.1), or formulas easily equivalent to it, are well
known. The first proof of (35.1) is apparently due to Glaisher [34] in
1913. Later proofs were found by Kesava Menon [47] and Djokovic [21],
One can also find (35.1) in Hansen's tables [41, p. 116].

76
B. C. BERNDT AND P. T. JOSHI
Examples. We have
2
2 ~ 3~
(i) K9 = \- - 3,
(ii) K3 = 10 - it2,
(iii) K4=^ + i^-35,
and
1C 2 4
(iv) K5 = 126 - -- .
All of these examples are consequences of (35.1). Formulas
(i) - (iii) are given explicitly by Hansen [41, pp. 40, 35, 31]. Greenstreet
[38] found (i) - (iii) in 1907, six years before Glaisher [34] found (35.1).
One can also find (i) in Bromwich's text [12, p. 226].
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83
81. G. T. Williams, A new method of evaluating £(2n), Amer. Math. Monthly
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Bruce C. Berndt
Department of Mathematics
University of Illinois
1409 West Green Street
Urbana, Illinois 61801
Padmini T. Joshi
Department of Mathematics
Ball State University
Muncde, Indiana 47306

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