Integrals evaluated in terms of Catalan’s constant
Graham Jameson and Nick Lord
Catalan’s constant, named after E. C. Catalan (1814–1894) and usually denoted by G,
is defined by
∞
X
(−1)n
1
1
G=
= 1 − 2 + 2 − ....
2
(2n + 1)
3
5
n=0
It is, of course, a close relative of
∞
X
n=0
1
π2
3
ζ(2)
=
.
=
4
(2n + 1)2
8
The numerical value is G ≈ 0.9159656. It is not known whether G is irrational: this remains
a stubbornly unsolved problem. The best hope for a solution might appear to be the method
of Beukers [1] to prove the irrationality of ζ(2) directly from the series, but it is not clear
how to adapt this method to G.
A remarkable assortment of seemingly very different definite integrals equate to G,
or are evaluated in terms of G. A compilation of no fewer than eighty integral and series
representations for G, with proofs, is given by Bradley [2]. Another compilation, based on
Mathematica, is [3]. Here we will present a selection of some of the simpler integrals and
double integrals that are evaluated in terms of G, including a few that are actually not to
be found in [2] or [3].
The most basic one is
Z
0
1
tan−1 x
dx = G,
x
(1)
obtained at once by termwise integration of the series
∞
tan−1 x X
x2n
=
(−1)n
.
x
2n
+
1
n=0
Termwise integration (for those who care) is easily justified. Write
n
X
x2r
s2n (x) =
(−1)n
.
2r + 1
r=0
Since the series is alternating, with terms decreasing in magnitude, we have tan−1 x/x =
R1
s2n (x) + r2n (x), where |r2n (x)| ≤ x2n+2 /(2n + 3), so that 0 r2n (x) dx → 0 as n → ∞.
We now embark on a round trip of integrals derived directly from (1). Most of them can
be seen in [4], but we repeat them here for completeness. First, the substitution x = tan θ
1
gives
π/4
Z
G=
0
θ
sec2 θ dθ =
tan θ
π/4
Z
0
θ
dθ.
sin θ cos θ
(2)
Since tan θ + cot θ = 1/(sin θ cos θ), (2) can be rewritten
π/4
Z
θ(tan θ + cot θ) dθ = G.
(3)
0
The substitution θ = 2φ gives one of the most important equivalent forms:
Z
0
π/2
θ
dθ =
sin θ
Z
π/4
0
4φ
dφ =
sin 2φ
π/4
Z
0
2φ
dφ = 2G.
sin φ cos φ
(4)
We mention in passing that (4) can be rewritten in terms of the gamma function. Recall
that by Euler’s reflection formula, Γ(1 + x)Γ(1 − x) = πx/(sin πx). Substituting πx = θ in
(4), we deduce:
Z
1/2
Γ(1 + x)Γ(1 − x) dx =
0
2G
.
π
The substitution x = e−t in (1) gives at once
Z ∞
G=
tan−1 (e−t ) dt.
(5)
0
Next, we integrate by parts in (1), obtaining
h
−1
G = tan
x log x
i1
0
Z
−
0
1
log x
dx.
1 + x2
The first term is zero, since tan−1 x log x ∼ x log x → 0 as x → 0+ , hence
Z 1
log x
dx = −G.
2
0 1+x
Now substituting x = 1/y, we deduce
Z
∞
1
log x
dx = G.
1 + x2
(6)
(7)
Substituting x = ey in (7), we obtain
Z ∞
Z ∞
y
y
y
G=
e dy =
dy,
2y
1+e
2 cosh y
0
0
hence
Z
0
∞
x
dx = 2G.
cosh x
2
(8)
Now substituting x = tan θ in (6), we obtain
Z π/4
Z π/4
log tan θ
2
−G =
sec θ dθ =
log tan θ dθ.
sec2 θ
0
0
(9)
Since (1 + cos θ)/(1 − cos θ) = cot2 2θ , we can deduce
Z π/2
Z π/2
Z π/4
1 + cos θ
θ
log
log tan dθ = −4
log tan φ dφ = 4G.
dθ = −2
1 − cos θ
2
0
0
0
(10)
This, in turn, leads to an interesting series representation. Given the series
∞
1 + x X x2n+1
1
=
log
2
1 − x n=0 2n + 1
and the well-known identity
Z π/2
Z
2n+1
cos
θ dθ =
0
π/2
sin2n+1 θ dθ =
0
2.4. . . . (2n)
,
1.3. . . . (2n + 1)
(11)
we deduce
2G =
∞
X
n=0
∞
∞
X
X
2.4. . . . (2n)
22n (n!)2
=
=
1.3. . . . (2n − 1)(2n + 1)2
(2n)!(2n + 1)2
n=0
n=0
2n
n
22n
.
(2n + 1)2
Using (11) again (but none of the earlier integrals), we now establish
Z π/2
sinh−1 (sin θ) dθ = G.
0
Write
an =
By the binomial series,
1.3. . . . (2n − 1)
.
2.4. . . . (2n)
∞
X
1
=
(−1)n an x2n .
(1 + x2 )1/2
n=0
Hence
−1
sinh
x
Z
x=
0
∞
X
1
an
dt
=
(−1)n
x2n+1 .
2
1/2
(1 + t )
2n
+
1
n=0
As above,
Z
π/2
sin2n+1 θ dθ =
0
1
,
an (2n + 1)
so by a neat cancellation of an ,
Z π/2
∞
X
(−1)n
−1
sinh (sin θ) dθ =
= G.
(2n + 1)2
0
n=0
We remark that the same method, applied to sin−1 (sin θ) (= θ) delivers the series
Z π/2
∞
X
1
π2
=
θ
dθ
=
.
(2n + 1)2
8
0
n=0
3
(12)
(13)
Integrals of logsine type and applications
A further application of (9) is to integrals of the logsine type, of which there are a rich
variety. We start by stating the most basic such integral, which does not involve G:
Z π/2
Z π/2
π
log cos θ dθ = − log 2.
log sin θ dθ =
2
0
0
(14)
π
2
− φ shows that the two integrals are
Rπ
equal: denote them both by I. Also, the substitution θ = π − φ gives π/2 log sin θ dθ = I.
Rπ
Hence 2I = 0 log sin θ dθ. Now substituting θ = 2φ and using sin 2φ = 2 sin φ cos φ, we have
Z π/2
log sin 2φ dφ
2I = 2
0
Z π/2
(log sin φ + log cos φ + log 2) dφ
= 2
Our proof follows [5, p. 246]). The substitution θ =
0
= 4I + π log 2,
hence (14).
Note that the equality of the two integrals in (14) (without knowing their value) implies
R π/2
that 0 log tan θ dθ = 0. Also, (14) can be restated neatly as follows:
Z π/2
Z π/2
log(2 sin θ) dθ =
log(2 cos θ) dθ = 0.
0
0
The integral (14) has numerous equivalent forms. For example, first substituting x =
sin θ and then integrating by parts, we find
Z 1
Z π/2
Z π/2
sin−1 x
π
dx =
θ cot θ dθ = −
log sin θ dθ = log 2.
x
2
0
0
0
(15)
Further equivalents, and a survey of Euler’s work in this area, are given in [6].
Catalan’s constant enters the scene when we integrate from 0 to π4 instead of π2 . Let
Z π/4
Z π/4
IS =
log sin θ dθ,
IC =
log cos θ dθ.
0
0
R π/2
Substituting θ = π2 − φ, we have IC = π/4 log sin θ dθ. So by (14), IS + IC =
R π/2
R π/4
π
log
sin
θ
dθ
=
−
log
2.
Meanwhile
by
(9),
I
−
I
=
log tan θ dθ = −G. So
S
C
2
0
0
we conclude
IS = − 21 G −
π
log 2,
4
IC = 21 G −
Again there is a neat restatement:
Z π/4
log(2 sin θ) dθ = − 21 G,
Z
0
0
4
π
log 2.
4
π/4
log(2 cos θ) dθ = 12 G.
(16)
However, (16) will be more useful in the ensuing applications.
P
1
Alternatively, (14) and (16) can be derived from the series log(2 sin θ) = − ∞
n=1 n cos 2nθ,
P∞ (−1)n
log(2 cos θ) = − n=1 n cos 2nθ (e.g. see [7]); however, justification of termwise convergence is more delicate in this case.
The substitution x = tan θ delivers the following reformulation of (16) in terms of the
log function:
Z
0
1
log(1 + x2 )
dx =
1 + x2
π/4
Z
Z
2
π/4
log sec θ dθ = −2
log cos θ dθ =
0
0
π
log 2 − G.
2
(17)
One can easily check that by substituting x = 1/y and combining with (7), one obtains
Z ∞
log(1 + x2 )
π
log 2.
(18)
dx
=
G
+
1 + x2
2
1
Now integrate by parts in (16): we find
π/4
Z
0
h
iπ/4 Z π/4 sin θ
log cos θ dθ = θ log cos θ
+
θ
dθ
cos θ
0
0
Z π/4
π
= − log 2 +
θ tan θ dθ,
8
0
so that
π/4
Z
0
θ tan θ dθ = 12 G −
an identity not given in [2] or [3].
R π/4
θ cot θ dθ = 12 G + π8 log 2.
0
π
log 2,
8
(19)
Similarly, or by (19) combined with (3), we have
Next, write
Z
I=
π/2
Z
log(1 + sin θ) dθ =
0
π/2
log(1 + cos θ) dθ.
0
By (16) and the identity 1 + cos θ = 2 cos2 12 θ, we obtain
Z
π/2
(log 2 + 2 log cos 21 θ) dθ
0
Z π/4
π
=
log 2 + 4
log cos φ dφ
2
0
π
=
log 2 + 2G − π log 2
2
π
= 2G − log 2,
2
I =
5
(20)
another integral not given in [2] or [3]. Combining (20) and (14), we have the pleasingly
simple result
Z
π/2
π/2
Z
log(1 + cosec θ) dθ =
log(1 + sin θ) − log sin θ dθ = 2G,
(21)
0
0
and of course the same applies with cosec θ replaced by sec θ.
Writing (1 + sin θ)(1 − sin θ) = cos2 θ, we deduce further
Z π/2
Z π/2
Z π/2
log(1 − sin θ) dθ = 2
log cos θ dθ −
log(1 + sin θ) dθ
0
0
π
= −2G − log 2.
2
0
(22)
Integrating by parts in (20), we have
Z π/2
h
iπ/2
π
log(1 + sin θ) dθ = θ log(1 + sin θ)
− J = log 2 − J,
2
0
0
where
π/2
Z
J=
0
θ cos θ
dθ,
1 + sin θ
hence
J = π log 2 − 2G.
(23)
Furthermore, the substitution x = sin θ gives
Z 1
sin−1 x
dx = J.
1+x
0
(24)
√
A further deduction from (16) is derived using the identity cos θ +sin θ = 2 cos(θ − π4 ):
Z π/2
Z π/2 1
π
log(cos θ + sin θ) dθ =
log
2
+
log
cos(θ
−
)
dθ
2
4
0
0
Z π/4
π
log 2 +
log cos φ dφ
=
4
−π/4
π
π
=
log 2 + G − log 2
4
2
π
= G − log 2.
(25)
4
By (25) and (14),
Z π/2
Z
π/2
log(1 + tan θ) dθ =
0
log(cos θ + sin θ) − log cos θ dθ
0
π
π
log 2 + log 2
4
2
π
= G + log 2,
4
= G−
6
(26)
and hence also, with the substitution x = tan θ,
Z ∞
log(1 + x)
π
dx = G + log 2,
2
1+x
4
0
(27)
which is proved by a rather longer method in [2].
Of course, tan θ can be replaced by cot θ in (26). With (20) and (21), this means that
R π/2
we have obtained the values of 0 log[1 + f (θ)] dθ for all six trigonometric functions.
Yet further integrals can be derived by integrating by parts in (25) and (26) (rather
better with cot θ): we leave it to the reader to explore this..
Double integrals
An obvious double-integral representation of G, not involving any trigonometric or
logarithmic functions, follows at once from the geometric series. For x, y ∈ [0, 1), we have
P
n 2n 2n
1/(1 + x2 y 2 ) = ∞
n=0 (−1) x y . Since
Z 1Z 1
1
x2n y 2n dx dy =
,
(2n + 1)2
0
0
termwise integration gives
Z 1Z
0
1
0
∞
X (−1)n
1
dx
dy
=
= G.
1 + x2 y 2
(2n + 1)2
n=0
(28)
Termwise integration is justified as in (1). (Alternatively, one stage of integration immediately equates the double integral to (1)).
Substituting x = e−u and y = e−v , we deduce
Z ∞Z ∞
Z ∞Z ∞
e−u e−v
1
G=
du dv =
du dv.
−2u
−2v
1+e e
2 cosh(u + v)
0
0
0
0
(29)
(This is really a pair of successive single-variable substitutions, not a full-blooded twovariable one.)
Next, we establish a much less transparent double-integral representation:
Z π/2 Z π/2
1
dθ dφ = 2G.
1 + cos θ cos φ
0
0
(30)
This, again, is not in [2] or [3]. It was given in [4], possibly its first appearance. Here we
present a proof based on (2). We actually show
Z
0
π/4
Z
0
π/4
1
dθ dφ = 21 G,
1 + cos 2θ cos 2φ
7
(31)
from which (30) follows by substituting θ = 2θ0 and φ = 2φ0 . Write
π/4
Z
1
dφ.
1 + cos 2θ cos 2φ
J(θ) =
0
Now
1 + cos 2θ cos 2φ = (cos2 θ + sin2 θ)(cos2 φ + sin2 φ) + (cos2 θ − sin2 θ)(cos2 φ − sin2 φ)
= 2(cos2 θ cos2 φ + sin2 θ sin2 φ).
(32)
The substitution t = tan φ gives
Z
π/4
0
a2
cos2
1
1
b
dφ =
tan−1 .
2
2
ab
a
φ + b sin φ
Applied with a = cos θ and b = sin θ, this gives
J(θ) =
By (2), it follows that
R π/4
0
θ
.
2 sin θ cos θ
J(θ) dθ = 21 G, which is (31).
Alternatively, substitute x = tan θ and y = tan φ in (28). Again, (31) is obtained after
applying (32). (This is the method of [4].)
We now establish two further double integrals:
Z 1Z 1
1
dx dy = G,
2
2
0
0 2−x −y
Z 1Z 1
1
dx dy = 4G.
1/2 (1 − y)1/2
0
0 (x + y)(1 − x)
(33)
(34)
Result (34) is identity (44) in [2], achieved with some effort; we present a somewhat simpler
proof. The substitution x = 1 − u2 , y = 1 − v 2 reduces (34) to (33). Denote the integral in
(33) by I. Since the integrand satisfies f (y, x) = f (x, y), the contributions with x ≤ y and
y ≤ x are the same, hence
Z
1
x
Z
0
1
dy dx.
2 − x2 − y 2
Z
sec θ
I=2
0
Transform to polar coordinates:
Z
π/4
I=
0
Now
Z
0
sec θ
0
2r
dr dθ.
2 − r2
h
isec θ
2r
2
2
dr
=
−
log(2
−
r
)
=
log
2 − r2
2 − sec2 θ
0
8
and
2
2
tan 2θ
=
=
,
2
2 − sec2 θ
1 − tan θ
tan θ
so
Z
π/4
(log tan 2θ − log tan θ) dθ.
I=
0
As seen in (14),
R π/4
0
log tan 2θ dθ =
R π/2
1
2
0
log tan φ dφ = 0. By (9), we deduce that I = G.
The discussion site [8] gives the following variant of (33):
Z 1Z 1
1
dx dy = 13 G.
2 − y2
4
−
x
0
0
The method is similar, but the writer resorts to Mathematica for the final stage, the integral
R π/4
log[4/(4 − sec2 θ)] dθ; with a bit of effort, this can be deduced from our results above
0
using the identity 4/(4 − sec2 θ) = 2 sin 2θ cos θ/(sin 3θ).
Applications to elliptic integrals
The “complete elliptic integral of the first kind” is defined, for 0 ≤ t < 1, by
π/2
Z
K(t) =
(1 −
0
t2
1
dθ.
sin2 θ)1/2
The value of K(t) is given explicitly by a theorem of Gauss:
K(t) =
π
,
2M (1, t∗ )
where M (a, b) is the arithmetic-geometric mean of a and b, and t∗ = (1 − t2 )1/2 . Two quite
different proofs of this theorem can be seen in [9] and [10]. However, without any reference
R1
to Gauss’s theorem, we can apply (4) to evaluate 0 K(t) dt. Indeed, reversing the implied
R1
R π/2
double integral, we have 0 K(t) dt = 0 F (θ) dθ, where
Z
F (θ) =
0
1
(1 −
t2
1
dt.
sin2 θ)1/2
Substituting t sin θ = sin φ, we have
θ
Z
F (θ) =
0
θ
1 cos φ
dφ =
.
cos φ sin θ
sin θ
Hence, by (4),
Z
1
K(t) dt = 2G.
0
Of course, this really amounts to another double-integral representation of G.
9
(35)
The “complete elliptic integral of the second kind” is
π/2
Z
(1 − t2 sin2 θ)1/2 dθ.
E(t) =
0
In the same way, we have
R1
0
R π/2
E(t) dt =
0
Z
G(θ) dθ, where
1
(1 − t2 sin2 θ)1/2 dt
G(θ) =
0
Z
θ
cos φ
cos φ
dθ
sin θ
0
Z θ
1
=
(1 + cos 2φ) dφ
2 sin θ 0
θ
+ 1 cos θ,
=
2 sin θ 2
=
hence
Z
0
1
E(t) dt = G + 21 .
(36)
A generalisation: the function Ti2 (x)
Going back to the original integral in (1), we can define a function of x by considering
the integral on [0, x]. The more-or-less standard notation is
Z x
tan−1 t
Ti2 (x) =
dt.
t
0
Clearly, G = Ti2 (1). Integrating termwise as in (1), we have for |x| ≤ 1,
∞
X
x3 x5
x2n+1
Ti2 (x) = x − 2 + 2 − · · · =
(−1)n
.
3
5
(2n + 1)2
n=0
This can be compared with “dilogarithm” function Li2 (x) =
P∞
n=1
(37)
xn /n2 . We remark that
G appears in the evaluation Li2 (i) = − 18 ζ(2) + iG.
We can use (37) to calculate values for |x| ≤ 1, for example Ti2 ( 21 ) ≈ 0.48722.
Since tan( π2 − θ) = 1/ tan θ for 0 < θ < π2 , we have
tan−1 x + tan−1
1
π
=
x
2
for x > 0. This translates into a corresponding functional equation for Ti2 (x):
1
π
Ti2 (x) − Ti2
= log x.
x
2
10
(38)
It is sufficient to prove this for x ≥ 1. Write Ti2 (x) − Ti2 (1/x) = F (x). Then
Z x
Z x
tan−1 t
1
1
F (x) =
dt =
tan−1 du
t
u
1/x
1/x u
Z x 1 π
− tan−1 u du
=
2
1/x u
= π log x − F (x).
So for x > 1, we have the following series expansion in powers of 1/x:
Ti2 (x) =
π
1
1
1
log x + − 2 3 + 2 5 − · · · ,
2
x 3x
5x
showing very clearly the nature of Ti2 (x) for large x.
Of course, the procedures that gave equivalent expressions for G can be also applied
to Ti2 (x). For example, the substitution t = tan θ leads to
Ti2 (x) =
1
2
Z
2 tan−1 x
0
θ
dθ.
sin θ
Integration by parts and then again the substitution t = tan θ gives
Z x
log t
−1
Ti2 (x) = tan x log x −
dt
2
0 1+t
Z tan−1 x
−1
= tan x log x −
log tan θ dθ.
0
As with the function Li2 (x), not many other explicit values of Ti2 (x) are known. One,
√
√
1
proved by ingenious methods in [2] (formula (33)) is: Ti2 (2 − 3) = 23 G − 12
π log(2 + 3).
References
1.
F. Beukers, A note on the irrationality of ζ(2) and ζ(3), Bull. London Math. Soc.
11 (1979), 268–272.
2.
David M. Bradley, Representations of Catalan’s constant (2001), at
http://germain.umemat.maine.edu/faculty/bradley/papers/c1.ps
3.
Victor Adamchik, 33 Representations for Catalan’s constant, at
www.cs.cmu.edu/˜adamchik/articles/catalan
4.
Nick Lord, Intriguing integrals: an Euler-inspired odyssey, Math. Gazette 91
(2007), 415–427.
5.
George Boros and Victor H. Moll, Irresistible Integrals, Cambridge Univ. Press
(2004).
11
6.
Nick Lord, Variations on a theme - Euler and the logsine integral, Math. Gazette
96 (2012), 451 – 458.
7.
J. A. Scott, In praise of Catalan’s constant, Math. Gazette 86 (2002), 102–103.
8.
http://mathoverflow.net/questions/181635/conjectured-integral-for-catalans-constant/
181701
9.
G. J. O. Jameson, Elliptic integrals, the arithmetic-geometric mean and the
Brent-Salamin algorithm for π, at www.maths.lancs.ac.uk/˜jameson
10.
Nick Lord, Evaluating integrals using polar areas, Math. Gazette 96 (2012),
289–296.
GRAHAM JAMESON
Dept. of Mathematics and Statistics, Lancaster University, Lancaster LA1 4YF, UK
e-mail: g.jameson@lancaster.ac.uk
NICK LORD
Tonbridge School, Tonbridge, Kent TN9 1JP
e-mail: njl@tonbridge-school.org
13 May 2016
12