Representations of Catalan's Constant
David M. Bradley
Department of Mathematics & Statistics, University of Maine
5752 Neville Hall, Orono, Maine 04469 U.S.A.
January 28, 2001
Abstract
Here, we assemble a fairly exhaustive list of formulae involving
Catalan's constant. There are single integral, double integral, innite
series, continued fraction, and asymptotic representations, as well as
some additional miscellaneous results of independent interest. A separate section in which the formulae are proved is supplied at the end.
1 Introduction
Catalan's constant may be dened by means of
1 (?1)n
X
G :=
(2n + 1) = L(2; );
n
2
=0
(1)
where is the non-principal character modulo 4.
Formulae for Catalan's constant abound. The purpose of this note is to
collect and catalog those formulae and results that are most interesting, and
to give simple, elementary proofs for as many as possible. I was inspired
by [2] to supply human proofs, as some of the derivations there are simply
recitations of formulae hard-coded into Mathematica, while others are arrived
at through somewhat arcane dilogarithm identities.
[email protected], [email protected]
1
2 Single Integral Formulae
Z tan?
Z x
1
1
G
=
G
= 2
G
=
1
2
G
=
1
2
G
= ?
G
= ?2
G
= 2
G
=
G
=
G
=
G
=
G
=
G
=
G
=
G
=
G
=
G
=
x
0
1
Z =
? tan?1 x
1
4
0
dx:
x
2
(2)
dx
1 ? x2 :
(3)
(4)
dx:
Z 1 sinx x
0
0
1
4
2
Zcosh x
(x ? ) sec(x) dx:
(6)
log(2 sin x) dx:
(7)
1
Z =
0
4
Z =
0
4
Z =
0
(5)
dx:
1
2
log(2 cos x) dx:
(8)
Z =
log(cot x) dx = ?
log(tan x) dx:
Z = 1 + cos x dx:
log
1
?
cos
x
Z = 1 + sin x log 1 ? sin x dx:
Z
Z 1 log x
p dx:
? (x +log1)xpx dx =
(1
+
x) x
Z = 0 1 + p sin x 1 dx
A
p
log @ p
:
1 ? sin x 1 + cos x
1 + sin x Z =
dx
p
log 1 ? sin x :
cos
x cos 2x
Z pp22+1
?1 (x + 1) log x dx
p
:
4
x 6x ? x ? 1
Z log x
? 1 + x dx:
Z 1 log x
dx:
1
+
x
Z
? log p (1 ? x) 1 +dxx :
4
4
0
0
1
4
2
(10)
2
(11)
0
1
4
0
1
1
4
1
4
0
1
1
2
1
2
2
1
2
0
2
4
1
2
0
2
1
1
2
0
2
1
1
0
(9)
1
2
2
2
(12)
(13)
(14)
(15)
(16)
(17)
(18)
G
G
1
G
=
G
=
G
=
G
=
=
G
=
G
=
G
=
G
1
0
2
1
2
1
2
Z1
(19)
2
(20)
2
log (x ? 1) dx :
Z 1 log(1 1++x)x
? log 2 +
1 + x dx:
Z log(1 + x )
dx:
log 2 ?
1+x
p
Z
p2 log x dx:
log 2 ?
Z = x x ? 1
log(cos x + sin x) dx
log 2 +
= ?
G
dx
:
log (1 ? x ) 1 +
x
Z1 log p (x + 1) 1 +dxx :
=
= ?
G
G
Z
1
2
1
2
2
1
4
2
0
2
1
1
2
2
0
2
1
2
2
1
2
1
4
Z =
sinh?1(sin x) dx =
2
Z =
0
2
1
0
2 + 41 log 2 ?
Z
Z =
0
csch? (csc x) dx =
2
1
16
Z =
0
2
0
2
Z =
2
(28)
dx:
=
3
2
G
=
1
8
G
=
1
8
(30)
2
0
3
2
G
1
1
1
2
(29)
2
0
3
2
0
p3 1 + x2 dx:
2+
log(2 +
log(2 +
(31)
2
Z 1 log x
p
3) +
3
2
3) +
3
4
p
Z
?p3 tan?1 x
2
Z =
0
6
x
x
(32)
dx:
dx:
sin
x
Z = x
Z 2G ? (3) =
dx = 4
tan? x dx :
sin x Z
x
=
x csc x
G = ? log 2 +
cos x + sin x dx
7
2
2
2
0
0
1
4
0
1
2
0
3
1
(25)
1
2
4
1
4
(24)
csch? (sec x) dx: (27)
x
? log 2 +
dx:
sin
x
?
p Z
= log(1 + 2) ? psin x dx:
1+x
Z ?p log x
= ?
1 + x dx:
1
16
(23)
1
1
0
(22)
sinh? (cos x) dx: (26)
(tan? x)
1
(21)
2
(33)
(34)
(35)
(36)
G
=
G
=
G
=
Z =
x cos x
dx
cos
x + sin x
0
Z =2
x
? 81 2 + 14 log 2 + 2 0 cosxxsin
dx
+
sin
x
Z =2 x csc x
1
log 2 ? P:V:
4
cos x ? sin x dx:
0
1
8
2
+
1
4
log 2 ? 2
2
(37)
(38)
(39)
Remarks. Formula (33) can be found on page 266 of [3]. We give an
alternative proof, based on an elementary trigonometric identity. The closely
related (31) and (34), of which the latter appears as an exercise on page 199
of [6], both follow from the same trigonometric identity. Note that (16)
and (17) imply the identity
Z 1 log x
1 + x dx = 0;
a result which is typically proved via contour integration. However, this
result can also be proved simply and directly by the making the change of
variable y = 1=x.
2
0
3 Double Integral Formulae
G
G
G
G+
1
2
G
=
Z Z
1
1
dx dy
1 + x2 y 2 :
Z = Z =
2
tan? (sin x sin y) dx dy :
=
sin x
Z Z =
d dx
p
=
:
1
?
x sin Z Z = q
=
1 ? x sin d dx:
0
0
2
1
2
0
0
1
2
Z Z
0
1
2
2
2
0
1
4
1
0
0
1
=
2
2
2
qdx dy
1
(x + y) (1 ? x)(1 ? y)
Z Z tan? (xy)
2G ? (3) = 8
1 + x y dx dy:
Z Z tan? x
2G ? (3) = 4
dx dy:
1+x y
7
2
7
2
0
0
1
0
2
0
1
0
1
1
2
1
1
2
0
4
2
:
(40)
(41)
(42)
(43)
(44)
(45)
(46)
2G ? (3) = ?
7
2
Z Z
0
0
qlog(1 ? x y )
2
1
1
xy
2
(1 ? x )(1 ? y )
2
2
dx dy:
(47)
Remarks. The innermost integral in (42) is the complete elliptic integral of
the rst kind. The innermost integral in (43) is the complete elliptic integral
of the second kind.
4 Line Integral Formulae
G=
?
1
15
2
Z c i1
3
?(s)?(2 ? s) (2s) (4 ? 2s) ds; 1 < c < 3=2: (48)
+ i
+
?1
3
c i
5 Asymptotic Formulae
3
F2
1; 1; ? "
; 1?"
3
2
1
2
!
? 1 =
1
4
+ (2G ? 43 log 2)" + O("2);
" ! 0:
(49)
6 Innite Series Representations
G
!
1 1 X
n
X
n (?1)k
=
2n k k (2k + 1)
n
1 n (2n + 1)
X
+1
=0
G
= 1?
G
=
1
6
+
G
=
1
2
+
G
=
1
16
G
=
1
2
G
=
1
8
n=1
1
X
1
4
n=1
1
X
n=1
1
X
n=1
1
X
n=2
16n
2
:
:
n16?n (32n ? 1) (2n + 1):
(50)
(51)
(52)
n4?n (1 ? 4?n) (2n + 1):
(53)
4?n(3n ? 1)(n + 1) (n + 2):
(54)
n=1
1
X
=0
n2?n (1 ? 2?n ) (n + 1):
(55)
n2?n (n + 1; 43 ):
(56)
5
G
= 1?
1
8
1
X
n2?n (n + 1; 45 ):
1 1 n + !?
X
=
2G ? (3) =
n
n+1
n
?
1 (?1)n nX
X
1
n=2
1; 1; 1; 1 1
2; ;
2
1
2
7
2
4
F3
3
2
=0
2G ? (3) = 2
7
2
2G
2G
G
1
+1
3
2
k =0
2
=0
=
1
8
log(2 +
(60)
(61)
p
3) +
1
X
3
8
1
X
n=0
1 :
(2n + 1) nn
2
2
1 :
(4n + 3)
n
1
X 1
:
G = ? +2
(4n + 1)
n
1 (?1)n X
n
X
1:
G = log 2 +
2n + 1 k k
n
?
1 (?1)n nX
X
1 :
G = log 2 ? 2
2n + 1 k 2k + 1
n
1 (2n + 1) 2n!
X
G = log 2 ? (n + 1) 16n n :
n
p
!
1
X
2
2
n
G = ? log 2 +
(2n + 1) 8n n :
n
1 sin(n=4)
X
G = log 2 +
:
n 2n=
n
1
X
2n (n!)
2G ? (3) = ? log 2 +
:
(2n + 1)!(n + 1)
n
Let n be a non-negative integer. Then
! nX
? (k!)
; ; n + 1
?
G = F
:
1; n + ( )k
k
G
=
1
8
2 ? 2
1
8
2
=0
2
=0
=0
1
4
1
4
3
2
2
2
=1
1
4
2
3
=0
=0
1
8
2
+1
2
2
2
=0
1
2
1
2
3
2
6
(63)
(64)
(65)
(66)
=0
2
1
32
1
2
(62)
=1
1
1
4
35
8
2
=0
1
2
(58)
=0
2
2
=0
:
(59)
:
2k + 1
1
n
n
X
X
2
1 :
=
n
(2n + 1) n k 2k + 1
n
1
X
4n :
=
(2n + 1) nn
n
n2
n=1
!
(57)
1
2
1
2
1
2
=0
3 2
2
(67)
(68)
(69)
(70)
(71)
Let
z
:=
2
F1
1
2
;
1
2
; 1 x ;
y
:= 2
F1
2
=
1
8
1
X
n=0
(?1)n
(2n + 1) (e n
2
;
1
2
1
2
1
2
1; 1; 1 1 ? x
;
y ? 21 z ?1 (1 ? x)1=2 3 F2
+2
; 1 1 ? x
:
; ; 1 x
1
2
F1
Then
G
3
2
(2 +1)y
!
3
2
(72)
+ 1) :
Note that (72) generalizes the denition (1) when y = 0, x = 1 and z = 1.
From [11], we have
1
1 sech(n )
X
X
(?1)n
G =
?
: (73)
?2
(2n + 1) (e n ? 1) n
n
n
!
1
X
4
n
(?1) (2n + 1) log 1 ? (6n + 3) ;
(74)
G = x + n
5
48
2
=0
3
4
1
4
G
=
3
4
1
X
n=0
2
1
4
(2 +1)
!
=0
(?1)n(2n + 1) log 1 +
p
x2
(2n + 1)
2
2
=1
2
(75)
;
where x = log(2 + 3):
The following representation is due to the author. Let L(1) = 1, L(2) = 3,
and L(n) = L(n ? 1) + L(n ? 2), n > 2 be the Lucas numbers (M0155 in
Sloan and Ploue's Encyclopaedia). Note the Lucas numbers satisfy the
same recurrence as the Fibonacci numbers, but with dierent initial values.
Then
q
0
p1 X
1 L(2n + 1)
10 + q50 ? 22 5 A
@
(76)
G = log
p + n (2n + 1) n :
10 ? 50 ? 22 5
n
Remarks. The formula (51) is apparently due to Glaisher. The formulae (51){(52) all have similar proofs. Steven Finch [8] attributes (58) to Bill
Gosper. It is closely connected with the double integral (47). Several of the
most interesting innite series of this section are due to Ramanujan. In [3],
we nd (60), (61), (62), (68), (70). See pages 264{269 there. Additionally, the
elegant hypergeometric representation (71) can be found on page 45 of [4],
and the stunning evaluation (72) appears on page 155 of [5]. The beautiful
formulae (73){(75) were taken directly from Ramanujan's paper [11].
1
2
5
8
1
8
7
=0
2 2
7 Innite/Matrix Product Representations
"
#
0 G
0 1
1 "
Y
=
3
9(6n+7)2 (6n+11)2
32(n+1) (2n+1)
0
n=0
2
580n +976n+411
450
1
#
:
(77)
Remarks. The matrix product (77), due to Bill Gosper, can be rewritten in terms of a F at 4=729, which Macsyma uses as its internal series
representation.
4
3
8 Continued Fraction Representations
1 2 2 4 4 6 6 :
2G = 2 ? 3+
(78)
1+ 3+ 1+ 3+ 1+ 3+
(79)
2G = 1 + 1 1 1 2 2 2 3 3 3 4 :
+ + + + + + +
1
1
1
1
G =
: (80)
?
?
1+ 3 ? 1+ 1 3 5 ? 1 + 3 5 7 ? 3 +
Remarks. Both (78) and (79) can be found in [4]. See pages 151 and 153
there.
2
2
2
2
1
2
1
2
2
2
2
2
1
2
2
1
2
1
2
1
2
1
2
?
2
2
2
2
2
2
2
2
2
9 Additional Results
Let f (n) denote the number of perfect matchings which cover the 2n 2n
square planar lattice. Then [9, 12]
= n2
lim
f
(
n
)
= e G= :
n!1
1 2
(81)
2
10 Proofs
Proof of (2): Use the Maclaurin series expansion for the arctangent, and
integrate term by term.
Z tan? x
Z X
1 (?1)n
1 (?1)n
X
dx =
x n dx =
:
x
2n + 1
(2n + 1)
n
n
1
0
1
1
0
2
=0
=0
8
2
Proof of (3): Use the addition formula for the arctangent
tan? u ? tan? v = tan?
1
1
1
with u = 1, v = x. Then
2
Z 1
0
1
4
dx
? tan?1 x
1?x
2
=
=
1 + uv
1 ? x
dx
2
Z01 tan?11+y xdy 1 ? x
4
2
2
0 (1 + y ) ? (1 ? y )
Z 1 tan?1 y
dy
y
0
G; by (2);
= 2
=
Z
u?v 1
tan?1
where we have made the substitution y = (1 ? x)=(1 + x) in the antepenultimate line.
Proof of (4): Let y = tan(x=2) in (2) and apply the double-angle formula
for sine. Then
Z tan? y
Z =
Z =
x dx
x dx
dy =
=
y
tan(x=2) cos (x=2)
sin(x=2) cos(x=2)
Z = x
=
dx:
sin x
Proof of (5): Write the hyperbolic cosine in terms of exponentials, then
expand in powers of exp(?x) using the geometric series formula. Integrate
term by term. Then
Z1
Z 1 x dx Z 1 xe?x dx X
1 (?1)n
1
X
?
n
x
n
xe
dx =
=
= (?1)
:
cosh x
1 + e? x n
(2n + 1)
n
1
1
1
4
0
1
2
1
2
2
1
4
2
0
2
0
2
0
(2 +1)
2
0
0
0
=0
=0
2
Proof of (6): Make a change of variable so that the range of integration
is from ?=2 to =2. Write the secant in terms of the reciprocal of sine, and
then use the fact that the integrand is even.
?
1
4
2
Z
1
0
(x ? ) sec(x) dx = ?
1
4
= ?
1
4
1
2
9
Z
(u ? =2) sec(u) du
Z =
0
2
?=2
v dv
cos(v + =2)
=
=
=
1
4
1
2
Z =
v
2
Z =
G;
dv
?=2 sin v
v
2
sin v dv
by (4):
0
Proof of (7): Use the boundedly convergent Fourier series
? log j2 sin xj =
1 cos 2nx
X
n
n=1
x 2 R;
;
and integrate term by term. Then
Z =
?2
4
0
Z = X
1 cos 2nx
1 sin n=2
X
= 2
dx =
n
n
n
n
1
n
X (?1)
4
log(2 sin x) dx
0
=
n=0
=1
(2n + 1)
=1
2
2
:
Proof of (8): Use
(7), write sine in terms of cosine, and apply the wellR
=
known evaluation
log 2 cos d = 0.
2
?2
Z =
0
4
0
log(2 sin x) dx = ?2
= ?2
= ?2
Z =
4
Z =
0
2
Z =
=4
2
0
log(2 cos(=2 ? x)) dx
log(2 cos ) d
log(2 cos ) d + 2
Z =
4
0
log(2 cos ) d:
Proof of (9): Add (7) and (8).
Proof of (10): Start with (9), rescale the argument by one half, and then
apply the half-angle formulae for sine and cosine.
Z =
4
0
log cot x dx =
1
2
Z =
2
0
log cot(x=2) dx =
1
2
Z =
2
0
s
log 1 + cos x dx
1 ? cos x
Proof of (11): Replace x by =2 ? x in (10).
Proof of (12): Make the change of variable v = (1 + sin x)=(1 ? sin x)
in (11).
10
Proof of (13): We start with (60), which is proved from rst principles
later in the sequel. Thus,
1
n Z
n
X
X
2
n
t k dt
G =
(2n + 1) n k
n
1
X
2n Z 1 ? t n dt:
=
1?t
(2n + 1) nn
n
1
1
2
2
=0
=0
2 +2
1
1
2
2
=0
2
0
2
0
Employing the reduction formula
Z =
n)
4n (sin x) n dx = 3 25 476 (2 (2
=
n + 1) (2n + 1) n
n
1
2
(82)
2 +1
2
0
for integrating odd powers of sine on [0; =2], we obtain
G
=
1
2
=
1
2
=
1
2
= p
Z
1
0
Z
1
0
1?t
1
2
2
dt
1?t
Z Z =
2?n(1 ? t n )
2 +2
Z =
n=0
2
2
0
sin x
Z =
2
0
1
1 ? sin
1
2
2
(sin x) n
2 +1
? 1?
x
sin x dx dt
(1 ? sin x)(1 ? t sin x)
1
0 p
sin
x
1
+
A dx :
log @ p
1 ? sin x 1 + cos x
dx
t2
1 2
t sin2 x
2
!
dx
2
1
0
0
1
X
dt
Z =
2
0
1
2
2
2
1 2
2
1
2
1
2
2
p
Proof of (14): Make the change of variable sin x = 2 sin v in (13).
Proof of (15): Make the change of variable v = (1 + sin x)=(1 ? sin x)
in (14).
Proof of (16): Integrate (2) by parts.
Z log x
Z tan? x
?
dx = (log x)(tan x) ?
x
1 + x dx:
Alternative Proof of (16): Replace x by x in (12).
Proof of (17): Make the change of variable y = 1=x in (16).
Proof of (18): We begin by proving the following
1
1
1
0
0
1
1
0
2
2
Integrating by parts once denes a recursion for the integral which can be easily solved
to give the closed form product.
1
11
Lemma 1
Z
!
log 1p+ x dx = 0:
2 1+x
Proof. Expand the integrand into a double series, then split into two double sums according to whether the outer summation index is even or odd.
Interchange the summation order in one of the double sums so that the two
double sums can be re-joined. Thus
1
2
0
Z log(1 + x)
1 (?1)n X
1 (?1)k
X
dx =
1+x
n
2k + n + 1
n
k
1 (?1)k
1
1 1 X
1
X
X
(?1)k
1 X
?
=
2n ? 1 k 2k + 2n n 2n k 2k + 2n + 1
n
1
1 1 X
1 (?1)k n X
1 (?1)k n
X
1 X
+1
1
2
0
=1
=0
=1
=
=1
=0
+
?
=0
+
2n ? 1 k n 2k
2n k n 2k + 1
n
1 (?1)k X
k
1 (?1)k
1 (?1)n X
X
(?1)n + X
=
:
2k n 2n ? 1 n 2n k n 2k ? 1
k
n=1
=
=1
=1
=1
=1
=
= +1
Now switch k and n in the second double series and paste the two double
series together. We get
1 (?1)n
1 (?1)k X
X
2k
2n ? 1 =
n=1
k =1
It follows that
Z
1
8
log 2:
!
Z dx
1
+
x
dx
log p 1 + x = log 2 ? log 2 1 + x = 0:
2
To prove (18), add zero in the form of Lemma 1 and combine the two logarithms. Simplify the argument of the logarithm by making a suitable change
of variable.
!
Z
1
?
x
? log p2 1 +dxx
!
!
Z
Z
1
+
x
dx
dx
1
?
x
+ log p
= ? log p
2 1+x
2 1+x
1
2
0
1
8
1
2
1
0
2
1
2
0
1
1
2
0
12
0
2
Z
1 + x
log 1 ? x 1 +dxx
Z1
log v
2dv
=
v ? 1 (v + 1)
1+
Z 1 log v v + 1
=
dv
v +1
= G; by (17):
=
1
2
0
2
1
2
;
v
= 1+x
1?x
2
1
For (19), factor the argument of the logarithm as a dierence of squares, and
apply (18).
Z
!
? log 1 ?2 x 1 +dxx
!
!
Z
Z
dx
1
?
x
1
+
x
= ? log p 1 + x ? log p 1 +dxx
2
2
= G; by (18) and Lemma 1:
2
1
2
0
1
1
2
0
0
2
Proof of (20): Make the change of variable y = 1=x in Lemma 1 and
use (17). Then
0 =
=
=
Z1
1
Z1
1
Z1
1
log
log
log
!
y+1
p y2dy+ 1
y 2
!
y+1
p y2dy+ 1
2!
y+1
p y2dy+ 1
2
?
Z 1 log y
1
y2 + 1
dy
? G; by (17):
Proof of (21): Replace x by 1=x in (19). Then
G
= ?
= ?
= ?
Z1
1
Z1
1
Z1
1
log
log
log
!
dx
x2 ? 1
2
2x ! 1 + x2
dx
x2 ? 1
2 ! 1 + x2
x2 ? 1
dx
2
1 + x2
13
+2
Z 1 log x
1
1+x
2
dx
+ 2G; by (17):
Proof of (22): From Lemma 1, we have
Z log(1 + x)
dx = log 2:
(83)
1+x
Replacing x by 1=x in the latter integral, we obtain
Z 1 1 + x dx
log 2 =
log x 1 + x
Z 1 log(1 + x)
Z 1 log x
=
dx ?
dx
1
+
x
1
+
x
Z 1 log(1 + x)
(84)
=
1 + x dx ? G; by (17):
Adding the two equations (83) and (84) yields (22) as required.
Proof of (23): Make the obvious substitution x = tan in the integral.
Then
Z
+ t ) dt = ? Z = log(sec ) d; t = tan ? log(1
1+t
Z =
= 2
log cos d
1
1
8
2
0
1
8
2
1
2
1
2
1
2
1
4
2
0
2
1
2
0
4
Z =
0
= 2
log(2 cos ) d ? log 2
= G ? log 2; by (8):
Proof of (24): Make the change of variable x = 1 + t .
Z p 2 log x
Z log(1 + t )
p
dx =
1 + t dt = ?G + log 2; by (23):
x x ?1
Proof of (25): By (22),
Z 1 log(1 + x)
dx ? log 2
G =
1
+
x
Z =
=
log(1 + tan ) d ? log 2
4
1
2
0
1
2
2
2
2
2
Z =
0
=
=
2
Z =
0
2
Z =
0
2
0
1
2
2
0
2
0
=
2
1
1
2
1
4
1
4
log(cos + sin ) d ?
1
4
log 2 ?
log(cos + sin ) d ?
1
4
log 2 ?
log(cos + sin ) d +
1
4
log 2:
14
Z =
2
Z =
0
2
0
log cos d
log(2 cos ) d +
1
2
log 2
Proof of (26): Write the inverse hyperbolic sine as an integral whose integrand can be expanded into powers of sine. The reduction formula (82)
for integrating powers of sine on [0; =2] is then used to arrive at the denition (1).
Z =
2
0
Z =
sinh? (sin x) dx =
2
1
Z =
0
2
=
0
q
log sin x + 1 + sin
Z x cos y dy
q
dx
1 + sin y
2
x dx
2
0
1 ? !
X
(sin y) n dy dx
=
cos y
n
n
1 ? ! 1 Z =
X
=
(sin x) n dx
n
2
n
+
1
n
1 ? ! 1
X
2 4 6 (2n)
Z = Z x
1
2
2
0
0
2
=0
1
2
2
2 +1
0
=0
1
2
2n + 1 3 5 7 (2n + 1)
1 (?1)n
X
=
(2n + 1) :
n
=
n=0
n
2
=0
Replacing x by =2 ? x immediately gives
G=
Z =
2
0
sinh? (cos x) dx:
1
Proof of (27): Use (26) and the fact that csch? (y) = sinh? (1=y).
Proof of (28): Integrate by parts twice, then use the appropriate trigonometric substitution to transform the resulting integral into the form of (8).
Z ? x)
Z
? tan? x dx = ?x tan? x + 2x 1(tan
+ x dx
Z log(1 + x )
?
= ? + log(1 + x )tan x ?
1 + x dx
= ? ? log 2 + G; by (23):
1
1
2
1
1
2
1
16
2
2
0
0
1
16
1
1
2 1
0
1
2
1
1
0
2
1
0
2
1
4
Proof of (29): As in the proof of (28), integrate by parts twice. In this
case, however, no trignometric substitution is needed.
Z =
4
0
x2
sin
2
x
dx
= Z =
= ?x cot x +
2x cot x dx
4
2
4
0
0
15
= ?
= ?
= ?
=
Z =
log sin x dx
+ 2x log sin x ? 2
Z =
4
1
16
2
1
16
2 ? 14 log 2 ? 2
1
16
2 + 41 log 2 + G;
4
0
0
4
log 2 sin x dx +
by (7):
0
1
2
log 2
Proof of (30): Start with (26). Integrating by parts, we have
Z =
2
Z =
0
=
2
sinh? (sin x) dx
1
q
log sin x + 1 + sin
2
x dx
= Z = x cos x
q
p
= x log sin x + 1 + sin x ?
1 + sin
p Z sin? u
0
2
2
2
0
=
2) ?
log(1 +
1
p
2
0
x
dx
1
du; u = sin x:
1+u
Proof of (31): We shall apply the following elegant, yet elementary trigonometric identity.
Lemma 2
1
2
2
Z =
12
0
log tan(3x) dx =
0
2
Z =
12
0
log tan x dx:
(85)
Deferring the proof of Lemma 2 for the moment, we have, letting y = 3x,
?
3
2
Z =
12
0
log tan x dx = ?
p
or, since tan(=12) = 2 ? 3,
Z log x
G=?
Z =
4
0
Z
log tan y dy;
p
log x dx;
1+x
1+x
by (16). To prove Lemma 2, note (following Chris Hill) that by the addition
formula for the tangent,
x) + tan x
tan(3x) = 1tan(2
x tan 2x
?2tan
tan
x
=
+ tan x
1 ? tan x 2 tan x
1 ? tan x
1 ? tan x
1
2
0
dx = ?
2
3
2
0
?
2
3
2
2
16
!
x
= tan x 13??3tan
p tan x ! p
!
3
+
tan
x
3
?
tan
x
p
p
= tan x
1 ? 3 tan x 1 + 3 tan x
= tan(x) tan(=3 + x) tan(=3 ? x):
2
2
It follows that
Z =
12
Z =
0
=
12
0
+
=
=
=
=
12
0
12
Z =
0
12
Z =
0
12
Z =
0
12
Z =
0
= 2
log tan x dx +
Z =
Z =
= 2
log tan(3x) dx
log tan x dx +
log tan x dx ?
log tan x dx +
Z =
0
12
0
12
0
log tan(=3 + x) dx
log tan(=3 ? x) dx
log tan x dx +
12
Z =
Z
5=12
Z
=3
5=12
Z =
=4
3
=4
12
log tan x dx
Z =
=4
4
log tan y dy
Z =
0
log tan x dx
log tan x dx
log tan(=2 ? y) dy
Z =
log tan x dx ? 3
Z =
12
=4
log tan x dx ?
log tan x dx +
12
0
log tan(3x) dx:
Rearranging the latter gives Lemma 2 as required.
Proof of (32): Make the change of variable y = 1=x in (31).
Proof of (33): Simply take (31) and integrate by parts.
Proof of (34): Let x = tan(=2) in (33) and apply the double-angle
formula for sine. Then
Z ?p tan? x
p
G = log(2 + 3) +
dx:
1
8
3
2
3
2
1
Z = (=2)xsec (=2)d=2
= log(2 + 3) +
tan(=2)
Z
=
p
p
1
8
=
1
8
log(2 +
3) +
3
2
3
4
17
0
2
6
0
6
0
sin d:
Proof of (35): One the one hand, by the double-angle formula for sine,
we have
Z =
x2
dx
sin x
2
0
Z =
x2
2
=
dx
2 sin(x=2) cos(x=2)
Z = x sec (x=2)
=
dx
2 tan(x=2)
Z = 4
tan? d ; = tan(x=2):
0
2
2
2
0
1
2
1
0
Thus, it suces to show that
2G ? (3) = 4
7
2
Z tan?
1
0
1
Integrating by parts, we have
4
Z 1
0
2
tan?1 x dx
=
x
= ?8
= ?8
4
Z =
0
4
0
dx
:
x
2
Z (log )(tan? )
d
?8
1+
1
4(log )(tan?1 )2 Z =
x
1
1
0
2
0
x log tan x dx;
= tan x
x log(2 sin x) dx + 8
Z =
4
0
x log(2 cos x) dx:
Next, we want to express the cosine in terms of sine, and then apply the
Fourier series for ? log j2 sin xj. Continuing,
4
Z = ?8
= ?8
1
Z =
0
= ?8
= 8
tan?
1
4
Z =
0
4
Z =
0
2
x
2
dx
x
x log(2 sin x) dx + 8
x log(2 sin x) dx + 8
2
0
4
Z =
0
2
x
n=1
n
dx + 4
18
x log
=4
2
0
2 sin( ? x ) dx
1
2
1
2
2
Z =
( ? x) log(2 sin x) dx
Z =
=4
x log(2 sin x) dx + 4
Z = X
1 cos 2nx
0
Z =
log(2 sin x) dx
log(2 sin x) dx
?4
Z =
4
log(2 sin x) dx
1
n
X
= 2G + 2 (?1)n ? 1 ; by (7)
0
3
n=1
= 2G ? 4
1
X
1
(2n ? 1)
1
1 1 !
X1 X
= 2G ? 4
? (2n)
n
n
n
= 2G ? (3):
Proof of (36): By (22), it suces to show that
Z 1 log(1 + x)
Z = x csc x
dx =
1+x
cos x + sin x dx:
Integrating by parts, we have
1 Z 1 tan? (1=x)
Z 1 log(1 + x)
dx = tan? x ? =2 log(1 + x) +
dx
1+x
1+x
Z 1 tan? y
=
dy; y = 1=x
y (1 + 1=y )
3
n=1
3
=1
3
=1
7
2
2
2
0
0
1
1
2
0
0
0
1
Z =
0
=
=
2
Z =
0
2
0
2
sec2 tan2 (1 + cot ) d;
csc cos + sin d:
y
= tan Proof of (37): Let
Z = (=2 ? x) sin x
Z = x cos x
dx = 2
dx; x $ =2 ? x:
I := 2
cos x + sin x
cos x + sin x
Averaging the two integrals for I gives
Z =
sin x dx + Z = x cos x ? sin x dx:
I= cos x + sin x
cos x + sin x
But
Z =
sin x dx = Z =
cos x dx; x $ =2 ? x
cos x + sin x
Z =cossinx x++sincosx x
=
cos x + sin x dx
= :
2
2
0
0
1
2
2
2
0
0
2
2
0
0
1
2
2
0
1
4
19
Therefore,
I
Z = cos x ? sin x +
x
cos x + sinx dx
= Z =
log(cos x + sin x) dx
+ x log(cos x + sin x) ?
2
=
1
8
=
1
8
2
=
1
8
2 + 41 log 2 ? G;
2
0
2
2
0
0
by (25):
Alternative Proof of (37): Write the denominator of the integrand in
terms of a shifted sine wave.
Z = x cos x
?2
dx
cos
x + sin x
Z =
p x cos x dx
= ?2
2 sin(=4 + x)
Z
p = (x ? =4) sin(x ? =4)
= ? 2
dx
sin x
=
p Z = x ? =4 cos x sin x !
p + p dx
= ? 2
sin x
=
2
2
Z =
Z =
= ?
(x ? =4) cot x dx ?
(x ? =4) dx
2
0
2
0
3
4
4
3
4
4
3
4
3
4
= Z ==
Z =
= ?(x ? =4) log sin x +
log sin x dx ?
=
=
Z =
=4
4
3
4
3
4
= ? log p + 2
1
2
1
2
2
Z =
=4
4
4
0
log sin x dx ?
log(2 cos x) dx ?
= log 2 + 2
= ? log 2 + G ? ; by (8):
4
1
4
0
1
8
1
4
2
1
8
2
1
2
log 2 ? 81 2
x dx
2
Proof of (38): Let
J
:= 2
Z =
2
0
Z =2 (=2 ? x) cos x
x sin x
dx = 2
cos x + sin x
cos x + sin x dx;
0
x $ =2 ? x:
Averaging the two integrals for J gives
Z = sin x ? cos x Z =
cos
x
dx:
dx +
x
J= cos x + sin x
sin x + cos x
1
2
2
2
0
0
20
But
Z =
2
0
cos x dx = Z =
sin x dx;
cos x + sin x
Z =coscosx x++sinsinx x
dx
=
cos x + sin x
= :
2
0
1
2
x $ =2 ? x
2
0
1
4
Therefore,
J
=
1
8
=
1
8
2
=
1
8
2
Z = cos x ? sin x x
?
cos x + sinx dx
= Z =
? x log(cos x + sin x) +
log(cos x + sin x) dx
2
0
2
2
2
0
0
? log 2 + G; by (25):
1
4
Alternative Proof of (38): Write the denominator of the integrand in
terms of a shifted sine wave.
Z = x sin x
dx
2
cos
x + sin x
Z =
p x sin x
= 2
dx
2 sin(x + =4)
p Z = (x ? =4) sin(x ? =4)
dx
= 2
sin x
=
p Z = x ? =4 sin x cos x !
p ? p dx
= 2
sin x
=
2
2
Z =
Z =
=
(x ? =4) dx ?
(x ? =4) cot x dx
2
0
2
0
3
4
4
3
4
4
3
4
=4
=
=
=
=
Z =
2
0
3
= Z =
log sin x dx
+
x dx ? (x ? =4) log sin x =
=
Z =
=4
3
4
3
4
1
8
2 ? 12 log p12
1
8
2 + 14 log 2 + 2
1
8
4
+2
Z
2 ? 14 log 2 + G;
0
2
=4
=4
4
4
log sin x dx
log(2 cos x) dx ?
by (8):
21
1
2
log 2
Proof of (39): We have
Z =
x csc x
dx
cos x ? sin x
0
Z 1 tan?1 u
du
P:V:
u(1 ? u)
0
P:V:
=
2
Z 1
1 tan? u du + Z 1 + 1 tan? u du :
+
u 1?u
" u 1 ? u Z
log
u
?
?
= P:V: (log u)(tan u) ? 1 + u du ? log(1 ? u)tan u Z log(1 ? u)
1 Z 1 log u
?
+
1 + u du + (log u)(tan u) ? # 1 + u du
1 Z 1 log(u ? 1)
?
? log(u ? 1)tan u +
1 + u du
Z 1 log u
Z log(1 ? u)
Z 1 log(u ? 1)
=
1
P:V:
1
1
1
0
0
1
1
1
0
1
2
0
1
0
1
1
2
0
1
2
1
1
=
1
du +
du
1
+
u
1
+
u
1
+
u
1 ? u du
Z log(1 ? u)
Z
=
1 + u du! + log u 1 + u
Z log u
Z log p2
Z
du
1
?
u
? 1 + u du + 2 1 + u du
= 2 log p
2 1+u
= ?G + log 2; by (16) and (18):
2
0
du +
2
1
1
2
0
1
2
1
1
2
0
2
0
1
1
2
0
1
2
0
2
0
1
4
Proof of (40): Start with (2) and write the arctangent in terms of the
integral which denes it. A double integral results. The result follows on
scaling one variable by the other, a device which is often quite useful.
Z tan?
1
1
0
y
y
dy =
Z
1
0
dy Z y
y 0
du
1 + u2
=
Z
1
0
dy Z 1
y 0
y dx
1 + x2 y 2 ;
u = yx:
Proof of (41): Expand the arctangent in powers of sines, and integrate
term by term, using the reduction formula (82) for integrating odd powers
of sine on [0; =2].
2 Z = Z = tan? (sin x sin y) dx dy
x
Z sin
1 (?1)n Z =
=
X
2
= 2n + 1
(sin x) n dx dy
(sin y) n
n
2
0
2
1
0
2
=0
2
2 +1
0
0
22
2
1 (?1)n 1 3 5 (2n ? 1) Z =
X
(sin y) n dy
=
(2
n
+
1)
2
4
6
(2
n
)
n
1 (?1)n 1 3 5 (2n ? 1) 2 4 6 (2n)
X
2
=0
=
2 +1
0
(2n + 1) 2 4 6 (2n) 3 5 7 (2n + 1)
1 (?1)n
X
=
(2n + 1) :
n
n=0
2
=0
Proof of (42):
1
2
Z Z =
2
1
0
0
p d dx
1 ? x2 sin2 =
1
2
=
1
2
=
1
2
=
=
=
Z Z
p dv p dx ;
1 ? vZ 1 ? x v
Z
v
p dv
p du ;
v 1?v
1?u
Z sin? v
p
dv
v 1?v
Z
1
1
2
0
0
2
1
2
0
2
0
v
= sin u = xv
1
1
2
0
cos d
;
sin
cos 0
Z =2 1
2
sin d
0
G; by (4):
1
2
2
=2
v
= sin Proof of (43): Let x sin = sin . Then
Z Z = q
2
1
0
0
1 ? x sin
2
2
d dx
=
=
=
=
=
=
Z = Z q
2
Z = Z 0
0
d
1 ? sin cos d sin
2
cos d d
Z = Z 1 + cos 2sin d d
2
sin
Z = Z = sin 2
d +
d
sin
sin
Z = sin cos d; by (4)
G+
sin G+ :
2
0
2
0
2
0
0
1
2
2
1
4
0
1
2
2
0
2
0
1
2
Proof of (44): We make the double change of variable x = u ? v, y = u + v.
Then (u; v) runs over the square S dened by the corners (0; 0), (1=2; ?1=2),
23
(1; 0) and (1=2; 1=2). The Jacobian of the transformation is 2, and so, by
symmetry of the square, we have
Z Z
=
0
ZZ
1
4
ZZ
qdx dy
1
1
1
4
(x + y) (1 ? x)(1 ? y)
q 2 du dv
S 2u (1 ? u) ? v
0
2
2
du dv
p
S (1 ? u) u2 ? v 2
Z 1=2 du Z u dv
Z 1 du Z 1?u
1
1
p
2
1 ? u 0 u2 ? v2 + 2 1=2 1 ? u 0
0
1 ? u
Z 1 du
Z 1=2 du
?
1
1
1
+
sin
4
1 ? u 2 1=2 1 ? u
u
0
=
1
4
=
=
=
1
4
log 2 +
1
2
Z 1
1 sin?
?
t 1+t
1
1
0
p
dv
u
2
?v
2
t dt:
Now let t = sin , cos respectively. Then
1
4
=
=
1
4
1
4
Z Z
1
(x + y) (1 ? x)(1 ? y)
Z = cos Z =
d ?
log 2 +
(=2 ? ) sin d
sin 1 + cos Z =
=
log sin d
log 2 + log sin ?
0
1
4
0
1
2
?
=
qdx dy
1
1
2
Z =
2
0
log 2 ?
2
2
1
2
0
2
1
2
2
1
2
0
0
0
(=2 ? ) tan(=2) d
1
2
Z =
2
log 2 sin d +
1
4
log 2 ? 2
Z =
4
(=4 ? ) tan d
Z =
=
=
= log 2 + log cos ? 2 log cos + 2
log cos d
Z =
0
4
1
2
1
2
4
0
= log 2 + 2
log 2 cos d ?
= G; by (8):
Proof of (45): By (35),
0
2G ? (3) = 4
7
2
Z 1
0
4
0
4
1
2
0
tan?
1
x
2
24
1
2
0
log 2
dx
x
=4
Z tan? x Z x
1
1
0
x
0
du
1 + u2 dx
Z tan? x Z x dy
= 4
x
1+x y
Z Z tan? x
1
1
1
= 4
1+x y
2
0
0
2G ? (3) = 4
1
1
1
Proof of (46): By (35),
2
0
0
2
2
dx dy:
Z 1
u = xy;
dx;
Z
2
tan?1 x dx
1
dx Z x tan?1 u
du
x 0 1 + u2
=8
Z dx Z tan? (xy)
= 8 x
1 + x y x dy;
Z Z tan? (xy)
dx dy:
= 8
1+x y
7
2
x
0
2
0
0
1
0
0
1
1
1
2
u = xy;
1
1
2
0
2
Proof of (47) due to Joakim Pettersson, Lund:
?
Z Z
1
0
0
1
Z = Z = log(1 ? sin u sin t)
log(1
?
xy )
q
du dt:
dy dy = ?
sin u sin t
xy (1 ? x )(1 ? y )
2
2
2
2
2
2
2
0
0
Let f (t) denote the previous integral times sin t. Then
Z = 2 sin u sin t cos t
f 0 (t) =
du
1
?
sin
u sin t
Z =
Z =
cos
t
cos t
=
du ?
1 ? sin t sin u
1 + sin t sin u du:
But
tan(u=2) sin t cos t :
d
?
=
2
tan
du
cos t
1 sin t sin u
2
2
0
2
2
2
0
0
1
Therefore,
f 0 (t)
=
2 tan?1
? 2 tan?
1
But
1 + sin t = cot (
cos t
1
2
1 ? sin t ? 2tan? (? sin t= cos t)
cos t
+ 2tan? (sin t= cos t):
1 +cossint t 1
2
1
1
1 ? sin t = tan ( ? t):
cos t
? t);
1
2
25
1
2
2
Therefore,
f 0 (t) = 21 ? t + 2t ? 2
1
2
?
1
2
( ? t) + 2t = 2t:
1
2
Since f (0) = 0, it follows that f (t) = t . Thus,
Z Z
Z = t
log(1
?
xy )
q
?
dx dy =
dt = 2G ? (3);
sin t
xy (1 ? x )(1 ? y )
2
1
0
2
1
2
2
0
2
2
1; 1; ? "
; 1?"
F (") := 3 F2
F (0) =
1
X
n=0
1
2
3
2
(?1)n( )n=( )n =
1
2
3
2
Next, we need to show that F 0(0) = 2G ?
Let f (") := ( ? ")n=(1 ? ")n, so that
1
2
F (") =
F 0 (0)
1
X
n=0
7
2
0
by (35).
Proof of (49): Let
Then
2
3
4
!
? 1 :
1 (?1)n
X
=
2n + 1
n=0
1
4
:
log 2.
(?1)n f (")(1)n=( )n;
3
2
1
X
0
(?1)nf (0) ((1))n ff (0)
n
n
1
X
( ) 0
(?1)n n f (0)
=
( )n f
n
1
n
X (?1) f 0
=
2n + 1 f (0):
n
=
3
2
=0
1
2
3
2
=0
=0
But,
f (") =
Thus,
f0
(0)
f
=
?(1=2 + n ? ")?(1 ? ") :
?(1 + n ? ")?(1=2 ? ")
(1 + n) ? (1) + (1=2) ? (1=2 + n)
26
n
X
n
X
1
1
?
2
k
2k ? 1
Zk 1 ? tn k
Z 1?u n
=
dt ? 2
du
1?t
1?u
Z
Z
n
n
= 2 u 1??u u du ? 2 11??uu du;
=
=1
=1
1
0
F 0 (0)
2
1
2
0
= 2
2
0
2 +1
1
and so
2
1
2
0
1 (?1)n Z
X
du
(
u ? u n ? 1 + u n)
2n + 1
1?u
n
Z
Z
1 (?1)n
X
tan? u
1?u
1
2 +1
2
0
=0
2
1
2n + 1 1 ? u
1 ? u (1=u ? 1) du
Z tan? u
1
n
X
= ?2 2(n?1)
log
2
+
2
du
+1
u(1 + u)
n
Z 1
1
? u du
= ? log 2 + 2
tan
?
u 1+u
Z
?
= ? log 2 + 2G ? 2 tan u du
1+u
Z log(1 + u)
?
= ? log 2 + 2G ? 2 log(1 + u)tan u + 2
1 + u du
= log 2 + 2G ? log 2 + log 2
= 2G ? log 2:
Proof of (50): Apply Euler's acceleration method for alternating series to
the denition (1). Alternatively, from (2), we have
Z tan? x
Z dx Z x
du
G =
dx =
x
x
2 ? (1 ? u )
Z dx Z x
du
=
x
1 ? (1 ? u )
Z
1 1
X
dx Z x
=
(1 ? u )n du
n
2
x
n
!
n
1 1 Z dx Z x X
X
n
(?1)k u k du
=
n
k
2
x
n
k
!
1
n
k
X 1 X n (?1)
=
2n k k (2k + 1) :
n
= ?2
1
du + 2
2
0
n=0
1
2
0
1
1
0
=0
1
1
2
1
0
1
1
1
2
0
1
1
2
1
1
0
1
2
1
2
2
0
1
4
3
4
1
1
1
0
1
2
2
0
0
1
1
2
0
0
2
1
=0
+1
2
0
0
1
=0
=0
+1
+1
0
2
0
=0
2
=0
27
Proof of (51): As customary, we denote the logarithmic derivative of the
Euler gamma function by . The idea is to bisect the Maclaurin series for
the deriviative of . We have [1]
X
1 1
1
X
1
? ? (1 ? z) :=
= (n)zn? ; jzj < 1:
?
z?n n
(86)
1
n=1
Thus,
4
Also,
1
X
n=1
n=2
n (2n + 1)z 2n
= z 0 (1 ? z) ? z 0 (1 + z):
0 (1 ? z ) =
Putting z = 1=4, we have
1 n (2n + 1)
X
16n
n=1
=
1
16
n
1
X
n=1
o
1 (
X
0( 3 ) ? 0( 5 )
4
4
1
(n ? z)
=
1
16
2
1
?
(4n ? 1) (4n + 1)
n
1 (?1)n
X
= 1 ? (2n + 1)
n
= 1 ? G:
=
1
2
=1
(88)
:
1 (
X
n=1
(87)
)
1
?
(n ? ) (n + )
1
1 2
4
)
1 2
4
2
2
=0
Proof of (52): As in the proof of (51), we have [1] (86), and hence (87).
Thus, in view of (88) and (1), we have
1
4
1
X
n=1
=
3
64
=
3
64
n4?2n (32n ? 1) (2n + 1)
0 (1 ? 3 ) ? 3 0 (1 + 3 ) ? 1 0 (1 ? 1 ) + 1 0 (1 + 1 )
4
64
4
64
4
64
4
1
1
1
1
X
X
X
X
1
1
1
1
?
?
+
(n ? )
(n + )
(n ? )
(n + )
n
n
n
n
1
1
1
1
X 1
X 1
X 1
X 1
=
?
+
?
(4n ? 3)
(4n + 3)
(4n + 1)
(4n ? 1)
n
n
n
n
= (G + ) + (G ? 1)
= G? :
3
4
3 2
4
=1
2
=1
1
9
3
4
3
64
3
4
=1
=1
3 2
4
2
1
4
1
6
28
1
64
1
4
=1
=1
1 2
4
2
1
64
1
4
=1
=1
1 2
4
2
Proof of (53): As in the proof of (51), we have [1] (86), and hence (87).
Thus, in view of (88) and (1), we have
1
X
=
n
n=1
1
4
n4?n (1 ? 4?n ) (2n + 1)
1
2
1
X
0 (1 ?
)?
1
2
0 (1 + 1 )o ? 1 n 1 0 (1 ? 1 ) ? 1 0 (1 + 1 )o
2
4
4
4
4
4
1
1
1
X 1
X 1
X
1
?
?
+
(n ? )
(n + )
(n ? )
(n + )
n
n
n
n
1
1
1
1
X 1
X 1
X 1
X 1
=
?
?
+
(2n ? 1)
(2n + 1) n (4n ? 1) n (4n + 1)
n
n
= ? + G; by (1):
=
1
8
1
2
1
1
2
=1
1
8
1 2
2
2
=1
=1
1
2
1
16
1 2
2
=1
2
=1
1 2
4
2
=1
1
16
1 2
4
=1
2
=1
1
2
Proof of (54): As in the proof of (51), we have [1] (86), so that
? 0 (1 ? z) =
1
X
(n + 1) (n + 2)zn ; jzj < 1:
In view of (88), it follows that
1 3n ? 1
X
(n + 1) (n + 2) =
4n
n
1
16
(89)
n=0
=1
1
16
n
0 (1 ?
1 (
X
3
4
o
) ? 0 (1 ? )
1
4
1
1
)
(n ? ) ? (n ? )
)
1 (
X
1
1
=
(4n ? 3) ? (4n ? 1)
n
1 (?1)n
X
=
(2n + 1) :
n
=
1
16
n=1
3 2
4
2
=1
1 2
4
2
2
=0
Proof of (55): As in the proof of (51), we have [1] (86), so that (89) holds.
In view of (88), it follows that
1
2
=
1
8
=
1
8
1
X
n=1
n2?n (1 ? 2?n ) (n + 1)
0 (1 ? 1 ) ? 1 0 (1 ? 1 )
4
4
2
1
1
X
X
1
n=1
?
(n ? )
1 2
4
1
4
n=1
29
1
(n ? )
1 2
2
1
X
1
X
1
?
(4n ? 1) n (2n ? 1)
n
1
1 X 1
X
1 + 1 = 2 (4n +
?
3) n (4n + 3) (4n + 1)
n
1
X
1
1
=
(4n + 3) ? (4n + 1)
n
1 (?1)n
X
=
= 2
1
2
=1
2
=0
(2n + 1)
= G; by (1):
Proof of (56): Since
n=0
1
X
1
X
n=2
n2?n
n=2
1
X
k =0
2
=0
2
=0
it follows that
2
=1
nxn
2
2
=
1
x
(1 ? x)
=
(k + )n
3
4
2
+1
=
2
? x; jxj < 1;
1 1 X
1
X
k =0
1
X
k+
3
4
n=2
n
(2k + )n
1
3
2
(k + )(2k + ) 1 ? 2k 1+
1
X
1
? (k + )(2
k+ )
k
1
1
X 2
X 2
?
=
(2k + ) k (2k + )
k
1
1
X 1
X
1
= 8
?
8
(4k + 1)
(4k + 3)
k
k
= 8G; by (1):
k =0
3
2
3
4
1 2
2
=0
3 2
2
=0
2
=0
=0
Proof of (57): Since
1
X
it follows that
1
X
n=2
n2?n
n=2
1
X
k =0
1
nxn
(k + )
5
4
n+1
=
x
(1 ? x)
=
2
? x; jxj < 1;
1 1 X
1
X
k =0
k+
30
5
4
n=2
3
2
3
2
3
4
=0
!
n
(2k + )n
5
2
2
2
1
X
=
1
!
(k + )(2k + ) 1 ? 1
2k +
1
X
1
? (k + )(2
k+ )
k
1
1
X 2
X 2
?
=
(2k + ) k (2k + )
k
1
1
X
X 1
1
?
8
= 8 (4k +
3)
(4k + 5)
k
k
= 8(1 ? G); by (1):
Proof of (58): Rewrite the binomial coecient in terms of the beta integral. Thus
!?
Z
?( )?(n + 1)
n+
=
=
xn (1 ? x)? = dx;
n
?(n + 1 + )
and (58) becomes
1 1 Z Z
1 1 n + !?
X
X
xn y n(1 ? x)? = (1 ? y )? = dx dy
=
n
+
1
n
+
1
n
n
n
Z Z
qlog(1 ? xy) dx dy
= ?
xy (1 ? x)(1 ? y )
Z Z
qlog(1 ? x y ) dx dy
= ?
xy (1 ? x )(1 ? y )
= 2G ? (3);
by (47).
Proof of (59): Integrate the series expansion for (tan? x)=(1 + x ) to
obtain
n
?
1
n
? X
X
tan x = (?1)n x n 2k 1+ 1 :
k =0
5
4
5
2
3 2
2
=0
1
1
2
1
2
1
2
1
2
=0
0
=0
1 2
0
1 2
1 2
0
1
1
1
4
2
=0
1
1
1
4
5 2
2
=0
1
1
2
2
5
2
2
=0
1
2
5
2
5
4
=0
2
0
0
2
1
1
2
0
0
2
2
7
2
1
+1
2
1
n=1
1
2
k =0
Substitute the latter into (35) and integrate term by term.
Proof of (60): From (2), we have
Z tan? x
Z dx Z x 2 du
2G = 2
dx =
x
x
2 ? (1 ? u )
1
1
1
0
0
31
0
2
2
du
dx Z x
1
x 0 1 ? 2 (1 ? u2 )
0
1 1 Z 1 dx Z x
X
(1 ? u2)n du
n 0 x 0
2
n=0
Z
=
=
1
1
X
=
n=0
1
X
=
n=0
where
2?n
2?n
Z =
2
0
Z =
2
0
cot Z
0
(cos ) n
2 +1
I2n+1 () cot d;
Z
(cos ) n
Integrating by parts, we have, for n > 0,
I2n+1 () :=
I2n+1 ()
=
Z
d d
2 +1
0
(90)
d:
(cos ) n cos d
2
Z
n
= (cos ) sin + 2n (cos ) n? sin
Z
0
2
2
1
0
0
= (cos ) n sin + 2n
2
d
(cos ) n? (1 ? cos ) d:
2
2
0
1
2
Thus for n > 0,
(cos ) n sin + 2n I ();
2n + 1
2n + 1 n?
a recurrence which gives, when iterated,
n
sin X
(2k + 2)(2k + 4) (2n) (cos ) k
I n () =
2n + 1 k (2k + 1)(2k + 3) (2n ? 1)
n n
X
k
( cos ) k :
= 4 sin n
k
(2n + 1) n k
2
I2n+1 () =
2
1
2
2 +1
=0
2
2
=0
2
1
2
Substituting for I n () in (90), we obtain
1
n X
2n X
k
4?k I k (=2)
2G =
k
n
(2n + 1) n k
n
n
1
n
X
X
1 ;
2
=
(2n + 1) nn k 2k + 1
n
2 +1
2
=0
=0
2
2
32
=0
=0
2 +1
as required.
Alternative Proof of (60): Set x = ?1=2 in [3], Chapter 9, Entry 34,
p. 293, Formula (34.1):
1 (?1)n 4n xn X
1
n
X
1 x n = X
1 ; ? < x < 1:
n
(2n + 1) 1 + x
(2n + 1) n k 2k + 1
n
n
+1
+1
2
=0
2
=0
1
2
=0
Remark. Fee [7] proved (60) by rst deriving (50) and then simplifying the
inner sum.
Proof of (61): Use (4) and the fact that
Z =
2
sin 0
d
Z 2x sin? x dx Z dx X
1 (2x) n
p
=
2x n n nn
1 ? x 2x
1
X
4n =
n
1
1
=
2
2
0
2
1
2
0
=1
2
2n(2n ? 1) n
1
X
4n :
=
(2n + 1) nn
n
2
n=1
2
=0
2
Alternative proof of (61): Start with (11). Expand the integrand into
odd powers of sine, interchange the order of summation and integration,
and apply the reduction formula (82) for integrating odd powers of sine on
[0; =2]. Thus,
Z = 1 + sin x 2G =
log 1 ? sin x dx
Z = X
1
1 (sin x) n dx
=
2n + 1
n
1
X 1
2 4 6 (2n)
=
2n + 1 3 5 7 (2n + 1)
n
1
X
4n :
=
(2n + 1) nn
n
2
1
2
0
2
0
2 +1
=0
=0
2
=0
2
Proof of (62): Use (34) and the fact that
Z =
6
0
sin d =
Z
1=2
0
2px sin? x dx = Z
1 ? x 2x
1
2
33
2
0
1=2
1 (2x)2n
dx X
2x2 n=1 n 2nn
1
X
1 2n(2n ? 1) nn
n
1
X
1 :
=
(2n + 1) nn
n
Proof of (63): Start from the denition (1), splitting the sum according
to whether n is even or odd. Then
1
1
X
1
1 ?X
G =
(4n + 1) n (4n + 3)
n
(
) X
1
1
X
1
1
1
+
?
2
=
(4n + 3) (4n + 1)
(4n + 3)
n
n
1
1
X 1
X 1
?
2
=
(2n + 1)
(4n + 3)
n
n
) X
(
1 1
1
1
X
X
1
1
1
?
+
?
2
=
(2n ? 1) (2n)
(2n)
(4n + 3)
n
n
n
1
1
X1
X 1
=
?
2
n
(4n + 3)
n
n
1
X
1 :
= ? 2 (4n +
3)
n
Proof of (64): As in the proof of (63), start from the denition (1),
splitting the sum according to whether n is even or odd. Then
1
1
X
1
1 ?X
G =
(4n + 1) n (4n + 3)
n
)
1 (
1
X
X
1
1
1
= 2 (4n + 1) ?
(4n + 3) + (4n + 1)
n
n
1
1
X
X
1
1
= 2
?
(4n + 1) n (2n + 1)
n
) X
1
1 (
1 1
X 1
X
1
1
= 2
+
?
+
(4n + 1) n (2n ? 1) (2n)
(2n)
n
n
1
1 1
X
1 ? X
= 2 (4n +
1)
n
n
n
1
X 1
? :
= 2
(4n + 1)
n
= 2
1
2
2
=0
1
8
=1
2
2
2
=0
2
=0
2
=1
3
4
2
=0
2
=0
2
2
=0
2
=0
2
=1
2
2
=1
2
=0
2
=0
2
2
=0
=0
=0
=0
=0
=0
=0
2
2
=0
2
2
2
2
=0
2
=0
2
=1
2
3
4
2
1
8
2
=1
2
2
34
2
=1
2
Proof of (65): It is convenient to start with the double sum. First, interchange summation order, then write the inner sum as the integral of a
geometric series. The outer sum gives a logarithm, which can be massaged
into (23).
Z
1 (?1)n X
1 1X
1 1X
n
1 (?1)n X
1
X
1 = X
n
=
(?1)
2n + 1 k k
k n k 2n + 1 k k n k
n
k
1 1 Z (?t )k
X
=0
=1
=1
=1
=
2
1
=
Z
k =1
= ?
k
1+t
0
1
0
=
t2n dt
dt
2
1 (?1)k
X
dt
1
+1
2k
t
1+t k
k
Z log(1 + t )
= ?
dt
1+t
= G ? log 2; by (23):
2
0
=1
2
1
2
0
1
2
Alternative Proof of (65): Again, start with the double sum, but instead of
interchanging summation order, write the inner harmonic sum as an integral
of a nite geometric series. The outer sum gives an arctangent, which can
be massaged into (3).
1 (?1)n X
1 (?1)n Z 1 ? tn
n
X
1 = X
dt
2n + 1 k k
2n + 1 1 ? t
n
n
1 (?1)n Z u ? u n
X
1
=0
0
=0
=1
= 2
=
1
2
1
Z
n=0
0
1?u
2
du
Z tan? u
du ? 2
du
1
?
u
1
?
u
Z Z 1?u
u
1
1
2
0
du + 2
2
1?u
log 2; by (3):
0
1
2
1
1
2
0
= ? = G?
1
2
2n + 1
2 +1
1
0
1
4
? tan?1 u
du
1 ? u2
Proof of (66): Start with the double sum, and write the inner sum as
an integral of a nite geometric series. The outer sum gives an arctangent,
which can be massaged into Lemma 1. Thus,
?2
?
1 (?1)n nX
X
1
1
n=0
2n + 1 k 2k + 1
=0
35
1 (?1)n Z 1 ? t n
X
= ?2 2n + 1 1 ? t dt
n
Z tan? t ! dt
2
1
= ?2
2
0
=0
1
4?
1
0
= ?G + 2
= ?G + 2
= ?G + 2
Z 1
0
1?t
t
1
4
2
dt
? tan?1 t
1?t ?2
Z tan? t
(1=t ? 1) dt
1?t
Z tan? t
1
1
1
1
t(1 + t)
0
4
0
?
? tan
1
t
t
!
dt
1?t
2
by (3)
dt
1
1
0
=
=
=
1
2
0
Z 1 1 tan?
= ?G + 2
?
t 1+t
Z
=
2
Z
t dt
? t
tan
G?2
1 + t dt Z log(1 + t)
dt
G ? 2 log(1 + t)tan? t + 2
1+t
G ? log 2 + log 2; by Lemma 1
G ? log 2:
1
1
0
1
1
1
0
1
2
1
4
2
0
1
4
Proof of (69): Put z =
1+i
2
=p
1
2
ei=4
in Landen's transformation ([10])
?z Li (z) + Li 1 ? z = ? log (1 ? z)
2
2
1
2
2
for the dilogarithm, and take imaginary parts.
Proof of (71) when n = 0: Start with (42) and expand the integrand by
the binomial theorem. Integrate term by term, applying reduction formulae
as needed to integrate even powers of sines.
G
=
1
2
=
1
2
=
1
2
Z Z = X
1
!
(?1) ?n x n sin n d dx
n
1
X (?1)n ? ! Z = n
sin d
2
1
0
0
1
2
n
2
2
=0
1
2
2
2
2n + 1 n
n
1
X (?1)n ? ! 1 3 (2n ? 1)
2n + 1 n 4 2 4 (2n)
n
0
=0
1
2
=0
36
!
1
X
1
1
3
(2
n ? 1)
2 4 (2n)
= 2n + 1
n
1 ( )n ( )n !
X
2
1
4
=
=
1
4
1
4
=0
n=0
( )n
1
2
3 F2
2
1
2
1
2
3
2
n!
; ;
1; 23
1
2
1
2
!
1 :
Proof of (76): As in the proof of (31), we begin with an elementary
trigonometric identity, which should be compared with Lemma 1.
Lemma 3
Z =
20
0
log tan x dx + 2
Z =
20
0
Z =
20
log tan(5x) dx = 3
0
log tan(3x) dx:
Proof. By the addition formula for the tangent,
tan(5x) = tan(x) tan(=5 + x) tan(=5 ? x) tan(2=5 + x) tan(2=5 ? x):
Thus,
Z =
20
Z =
0
=
20
0
+
=
=
=
20
0
20
Z =
0
20
Z =
0
20
Z =
0
= 2
= 2
log tan x dx +
Z =
Z =
= 2
log tan(5x) dx
log tan x dx +
log tan x dx +
Z =
0
20
Z =
0
20
0
20
0
flog tan(=5 + x) + log tan(=5 ? x)g dx
flog tan(2=5 + x) + log tan(2=5 ? x)g dx
log tan x dx +
20
Z =
Z =
4
Z =
3=20
4
Z =
log tan x dx +
log tan x dx +
3=20
4
Z =
3=20
log tan x dx +
log tan x dx ?
log tan x dx ?
4
Z =
3=20
4
log tan x dx + 5
Z =
7=20
4
Z
log tan x dx ?
20
0
9=20
3=20
Z
=20
log tan x dx
log tan(=2 ? x) dx
3=20
log tan x dx ? 2
Z =
0
Z
log tan x dx
3=20
Z
0
3=20
0
log tan(5x) dx ? 6
37
log tan x dx
log tan x dx
Z =
20
0
log tan(3x) dx:
We now proceed to prove (76). Rescaling the variables in Lemma 3, it
follows (9) that
G
= ?
Z =
4
log tan x dx = ?
Z
5
2
3=20
=
Z =
= ? x log tan x +
=
=
0
3
5
2
= ?
=
1
8
5
2
But,
20
3
log tan 20
?
log tan
Z
3=10
=10
1
20
3
5
2
20
3
20
=20
1
20
20
sin = 2
d
= 2
=
2
1
log tan 20
? log tan3
20
log tan x dx
x sec x
tan x dx
+
Z
3
20
Z
sin(3=10)
5
4
5
2
Z
3=10
=10
3=10
sin d:
(91)
1
2
2
2
sin(=10)
n
n=1
2n+1
=2 d=2
sin =2 cos =2
2px sin? x dx
1 ? x 4x
1
X (2x) n?
dx
sin(3=10)
1 X
+
=10
sin(=10)
Z
2
2n
n
? ;
n
2n+1
(92)
(2n + 1) n
p
p
where := ( 5 + 1)=2 and := ( 5 ? 1)=2. Since the Lucas numbers have
the explicit formula (cf. Binet's formula for the Fibonacci numbers)
p !n
p !n
5
5 ; 0 n 2 Z;
1
+
1
?
L(n) =
+
2
2
it follows (91), (92) that
! X
1 L(2n + 1)
tan(
=20)
:
(93)
G = log
+
tan (3=20)
(2n + 1) nn
n
But
q p p
q p
5
+
2
5
?
5
5+2 5?1
tan = q
p p ; tan = q p :
5+2 5+ 5
5+2 5+1
Therefore, we may write
1
8
1
2
2 2
n=0
5
8
3
1
20
2
=0
2
3
20
0q
1 q
p
p
p
tan(=20) = @ q5 + 2 5 + 1 A q5 + 2 5 ? 5 = a + b ;
p
p p
tan (3=20)
5+2 5?1
5+2 5+ 5 a?b
3
3
38
where a and b are to be determined. Cross multiplying, we have
a(
p
p
q
p
p
5 ? 3)(5 + 2 5) + 5b( 5 + 3) 5 + 2 5 = 0;
p!
p
q p
5
?
5)
5
p = 2 a = 5b 5 + 2 5:
3+ 5
or
a(5 +
Cross multiplying again, we have
0
p 1
q
p
5
5
?
A
@
10b = q
p a = a 50 ? 22 5:
q
5+2 5
p
Therefore, if we take b = 50 ? 22 5 and a = 10, then (93) becomes
q
0
p1 X
1
50
?
22
5A
10
+
@
q
G = log
p +
1
8
10 ? 50 ? 22 5
5
8
n=0
L(2n + 1)
(2n + 1)
;
2 2n
n
as required.
References
[1] M. Abramowitz & I. Stegun, Handbook of Mathematical Functions,
Dover, New York, 1972.
[2] V. Adamchik, 32 representations for Catalan's constant, Wolfram
Research, Inc. ([email protected]), available at website
http://www.wolfram.com/victor/articles/catalan/catalan.html.
[3] B. C. Berndt, Ramanujan's Notebooks: Part I, Springer-Verlag, 1985.
[4] B. C. Berndt, Ramanujan's Notebooks: Part II, Springer-Verlag, 1989.
[5] B. C. Berndt, Ramanujan's Notebooks: Part III, Springer-Verlag, 1991.
[6] J. M. Borwein & P. B. Borwein, Pi and the AGM, Wiley-Interscience,
John Wiley & Sons, Toronto, 1987,
39
[7] G. J. Fee, Computation of Catalan's constant using Ramanujan's
formula, ISSAC '90 (Proc. Internat. Symp. Symbolic and Algebraic
Computation, Aug. 1990) ACM Press - Addison Wesley, 1990.
[8] S. Finch, Favorite Mathematical Constants, Research and Development
Team, MathSoft, Inc. ([email protected]) available at website
http://www.mathsoft.com/asolve/constant/catalan/catalan.html.
[9] P. W. Kasteleyn, The statistics of dimers on a lattice, I, The number
of dimer arrangements on a quadratic lattice, Physica 27(1961), pp.
1209{1225.
[10] L. Lewin, Polylogarithms and Associated Functions, Elsevier North
Holland Inc., New York, 1981.
Z x tan? t
dt, Journal of the Indian
[11] S. Ramanujan, On the integral
1
0
t
Mathematical Society, VII (1915), pp. 93{96.
[12] H. N. V. Temperley & M. E. Fisher, Dimer problem in statistical
mechanics|an exact result, Philos. Magazine 6(1961), pp. 1061-1063.
40