SCIENTIA
Series A: Mathematical Sciences, Vol. 22 (2012), 101–107
Universidad Técnica Federico Santa Marı́a
Valparaı́so, Chile
ISSN 0716-8446
c Universidad Técnica Federico Santa Marı́a 2012
On some definite integrals connecting with sums related to
inverse tangent function
Ramesh Kumar Muthumalai
Abstract. Some definite integrals are evaluated through sums related to inverse
tangent function that found in chapter 2 of Ramanujan notebooks, Part-I. In
addition, integrals involving complicated arguments of inverse tangent function
are evaluated through infinite products.
1. Introduction
The chapter 2 of Ramanujan notebooks, Part-I [1] contains many elementary
formulas and identities on finite and infinite sums. Many of these identities involve
sums related to harmonic and inverse tangent functions. In particular, it contains
following infinite sums related to inverse tangent function
(1.1)
(1.2)
(1.3)
∞
X
(−1)k tan−1
k=−∞
∞
X
k=−∞
∞
X
tan−1
x
= tan−1 (tanh x cot a) ,
kπ + a
(−1)k tan−1
k=0
x
= tan−1 (sinh x csc a) ,
kπ + a
x
πx = tan−1 tanh
,
2k + 1
4
for a, x ∈ R [1, p. 39-40].
In the present study, some definite integrals that involving various combinations
of elementary functions are evaluated through above mentioned infinite series either
by direct formulas or infinite products. The solutions of integrals presented here are
not found in the classical table of integrals by Gradshteyn and Ryzhik [2].
2000 Mathematics Subject Classification. Primary 30E20, 33B10, 40A05, 40A20 .
Key words and phrases. Definite integrals; Infinite products; Inverse tangent function; Ramanujan notebooks.
101
102
RAMESH KUMAR MUTHUMALAI
2. Definite integrals involving elementary functions.
In this section, the direct formulas for integrals involving various combinations
of powers and elementary functions are derived through the sums related to inverse
tangent function.
2.1. Combinations of powers, trigonometric and hyperbolic function.
Consider the following definite integral for a positive integer m and an arbitrary a
Z ∞
sin2m+1 at dt
cosh t
t
0
Z ∞
∞
sin2m+1 at X
=2
(−1)k e−(2k+1)t dt
t
0
k=0
Z ∞
∞
X
sin2m+1 at
dt
=2
e−(2k+1)t
(−1)k
t
0
k=0
=
(−1)m
22m−1
R∞
2m+1
e−(2k+1)t sin t at dt [2, p. 494]
∞
m
X
X
(2m − 2i + 1)a
k
i 2m + 1
(−1)
(−1)
tan−1
.
i
2k + 1
i=0
Then, using the solution of the integral
0
k=0
Using the identity (1.3), yields
Z ∞
sin2m+1 at dt
(2.1)
cosh t
t
0
m
m X
(−1)
aπ i 2m + 1
= 2m−1
(−1)
tan−1 tanh(2m − 2i + 1)
.
2
i
4
i=0
Further, the definite integrals involving powers, hyperbolic cosine function with various
combinations of sin ax and cos ax can be evaluated by differentiating (2.1) with respect
to a.
Example 2.1. Let m = 1 and a = 2 in (2.1). Then
Z ∞
1
3π
3
π
sin3 2t dt
= − tan−1 tanh
+ tan−1 tanh
.
cosh t t
2
2
2
2
0
Similar integrals involving powers, hyperbolic and trigonometric functions are
listed in sections 4.11-4.12 of classical table of integrals [2].
2.2. Combinations of powers, exponentials, trigonometric and hyperbolic functions. Consider the following definite integral for a positive integer m and
|a| < π
Z ∞
Z ∞
at
−at
sinh at sin2m+1 tx
sin2m+1 tx
−πt e − e
2
dt
=
e
dt
t (eπt − 1)
1 − e−πt
t
0
0
ON SOME DEFINITE INTEGRALS
103
After simplification, gives
=
∞ Z
X
k=1
0
∞
sin2m+1 tx −t(kπ−a)
e
− e−t(kπ+a) dt
t
R∞
Then, using solution of the integral 0 e−px sin2m+1 ax dx
x , p > 0[2, p. 494]
"m
X
∞
(−1)m X
2m + 1
(2m − 2i + 1)x
= 2m
(−1)i
tan−1
2
kπ − a
i
i=0
k=1
!#
∞
X
−1 (2m − 2i + 1)x
−
.
tan
kπ + a
k=1
Using the identity (1.2), yields
Z ∞
m
sinh at sin2m+1 tx
(−1)m+1 X
i 2m + 1
dt = 2m+1
(−1)
(2.2)
×
t (eπt − 1)
2
i
0
i=0
(2m − 2i + 1)x
tanh(2m − 2i + 1)x
− tan−1
.
tan−1
tan a
a
Similarly, using the identity (1.1), it is find that
Z ∞
m
sinh at sin2m+1 tx
(−1)m+1 X
i 2m + 1
dt = 2m+1
(−1)
(2.3)
×
t (eπt + 1)
2
i
0
i=0
(2m − 2i + 1)x
sinh(2m − 2i + 1)x
−1
−1
− tan
.
tan
sin a
a
Example 2.2. Let m = 1 in (2.2) and (2.3). Then
Z ∞
sinh at sin3 tx
1
tanh 3x
3x
−1
−1
dt
=
tan
−
tan
πt
t (e − 1)
8
tan a
a
0
x tanh
x
3
− tan−1
− tan−1
8
tan a
a
and
Z
0
∞
1
sinh 3x
3x
sinh at sin3 tx
−1
−1
dt =
tan
− tan
t (eπt + 1)
8
sin a
a
3
sinh x
−1
−1 x
− tan
− tan
.
8
sin a
a
Remark 2.3. The solutions of the following definite integrals can be expressed in
finite terms by differentiating (2.2) and (2.3) 2n times with respect to a
Z ∞
Z ∞
2m+1
2m+1
tx
tx
2n−1 sinh at sin
2n−1 sinh at sin
dt
and
t
dt.
t
πt
πt
e −1
e +1
0
0
104
RAMESH KUMAR MUTHUMALAI
The following integrals can be expressed in finite terms by differentiating (2.2) and
(2.3) 2n + 1 times with respect to a
Z ∞
Z ∞
cosh at sin2m+1 tx
cosh at sin2m+1 tx
t2n
dt
and
t2n
dt.
πt
e −1
eπt + 1
0
0
Similarly, differentiating (2.2) and (2.3) with respect to x, the various forms of definite integrals in the combinations powers, exponentials, hyperbolic and trigonometric
functions can be expressed in finite terms.
Remark 2.4. Since sinh2r+1 x can be expressed in finite terms of hyperbolic sine
of integral multiples of x, the following integrals can be expressed in finite terms
Z ∞
Z ∞
sinh2r+1 at sin2m+1 tx
sinh2r+1 at sin2m+1 tx
dt
and
dt.
t(eπt − 1)
t(eπt + 1)
0
0
Example 2.5. Let r = 1 and m = 1 and using integrals in example 2.2. Then
Z ∞
sinh3 at sin3 tx
1
tanh 3x
−1
−1 x
dt =
tan
− tan
t(eπt − 1)
32
tan 3a
a
0
x
tanh
x
tanh
3x
3
3
−1
−1
−1
tan
−
tan
− tan
−
32
tan 3a
3a
32
tan a
3
9
3x
tanh
x
−1
−1
−1 x
+ tan
+
tan
− tan
.
32
a
32
tan a
a
Similar integrals involving powers, exponentials, hyperbolic and trigonometric
functions are listed in section 4.13 of classical table of integrals [2].
2.3. Combinations of logarithmic gamma and rational functions. Consider the following definite integral [2, p. 556] for b > 0 and c > 0
Z ∞
π
c
dx
x2
= − tan−1 .
log 1 + 2
2
2
b
c −x
c
b
0
It can be easily find that for a ∈
/N
2
x
Z ∞
1 + k+a
π
c
c
dx
−1
−1
log
=
−
tan
−
tan
.
2 2
c − x2
c
k+a
k−a
x
0
1 + k−a
Taking summation on both sides k = 1, 2, . . . ∞ and after simplification
2
x
Z ∞
∞ 1+
∞ Y
k+a
dx
πX
c
c
−1
−1
log
=
−
−
tan
.
tan
2 2
c − x2
c
k+a
k−a
x
0
k=1 1 +
k=1
k−a
Using the following identity [2, p. 886] for x, y real and x 6= 0, −1, −2, . . .
∞ Y
Γ(x) 2
y2
.
1+
=
(x + k)2
Γ(x − iy) k=0
ON SOME DEFINITE INTEGRALS
105
After simplification, yields
Z ∞
Γ(a)Γ(−a − ix) dx
log (2.4)
Γ(−a)Γ(a − ix) c2 − x2
0
c π
tanh cπ
=−
.
tan−1
− tan−1
2c
tan aπ
a
2.4. Complicated arguments of inverse tangent function. Multiply the
identity (1.3) by x/(p2 + x2 )2 and then integrating on [0, ∞), we have
Z ∞
Z ∞
∞
x
X
x tan−1 2k+1
x tan−1 tanh πx
k
4
dx =
dx.
(−1)
(p2 + x2 )2
(p2 + x2 )2
0
0
k=0
It is well known that [2, p. 599]
Z ∞
x tan−1 qx
πq 1
dx =
(2.5)
,
2
2
2
(p + x )
4p 1 + pq
0
p > 0, q > 0.
Using the identity in Ref[2, p. 897], it gives
Z ∞
∞
x tan−1 tanh πx
π X (−1)k
4
(2.6)
dx
=
(p2 + x2 )2
4p
2k + 1 + p
0
k=0
p+1
π
= β
.
8p
2
where β(x) = 12 ψ x+1
− ψ x2 [2, p. 896]. Replacing x by xπ and a by aπ in
2
identity (1.2), gives
∞
tan−1
∞
X
x X
x
x
+
tan−1
−
tan−1
= tan−1 (tanh xπ cot aπ) .
a
k+a
k−a
k=1
2
k=1
2 2
Multiply by x/(p + x ) , integrating on [0, ∞) and using (2.5), then
Z ∞
x tan−1 (tanh xπ cot aπ)
dx
(p2 + x2 )2
0
∞ π
π X
1
1
=
+
−
.
4p(a + p) 4p
k+p+a k+p−a
k=1
Using an identity of psi function ψ in Ref[2, p. 893] and after simplification
Z ∞
π
x tan−1 (tanh xπ cot aπ)
(2.7)
dx =
[ψ(p + 1 − a) − ψ(p + a)] .
2 + x2 )2
(p
4p
0
Similarly, using the identity (1.1), yields
Z ∞
x tan−1 (sinh πx csc aπ)
π
(2.8)
dx = [β(p + 1 − a) + β(p + a)] .
2 + x2 )2
(p
4p
0
A variety of integrals involving inverse tangent function is listed in section 4.5 of Ref
[2].
106
RAMESH KUMAR MUTHUMALAI
Remark 2.6. The collection of integrals given in this section cannot be evaluated
using a symbolic language. For instance, the current version of Mathematica 8.0
cannot compute the integrals in examples 2.1, 2.2 and 2.5.
3. Definite integrals connecting with infinite products.
In this section, definite integrals involving complicated arguments of inverse tangent function are evaluated by connecting with infinite products.
Divide both sides of identity (1.3) by x(p2 + x2 ) and integrate on (0, ∞], then
∞
X
(−1)
k
Z
0
k=0
∞
tan−1
x
2k+1
x(p2 + x2 )
∞
Z
dx =
0
tan−1 tanh πx
4
dx.
x(p2 + x2 )
Using the following integral [2, p. 599]
∞
Z
(3.1)
0
π
tan−1 qx
dx = 2 log(1 + pq),
2
2
x(p + x )
2p
p > 0, q > 0.
After simplification, gives
∞
Z
0
(−1)k
∞ Y
tan−1 tanh πx
π
p
4
dx = 2 log
1+
.
x(p2 + x2 )
2p
2k + 1
k=0
Similarly, using the identity (1.3) and following integrals [2, p. 599]
Z
∞
(3.2)
0
Z
(3.3)
0
1
π
q2
tan−1 qx
dx = log 1 + 2 ,
p > 0, q > 0.
x(1 − p2 x2 )
4
p
p
π
tan−1 qx
√
dx = log q + 1 + q 2 ,
q > 0.
2
x 1 − x2
The following integrals can be expressed in terms of infinite products as follows
Z
0
Z
0
(−1)k
∞ Y
tan−1 tanh πx
π
1
4
dx = log
1+
.
x(1 − p2 x2 )
4
(2k + 1)2 p2
k=0
s
!(−1)k
∞
Y
tan−1 (tanh πx
)
π
1
1
4
√
dx = log
+ 1+
.
2
2k + 1
(2k + 1)2
x 1 − x2
∞
1
k=0
ON SOME DEFINITE INTEGRALS
107
The following infinite products representations of integrals are obtained by using the
identity (1.2) and integrals (3.1), (3.2) and (3.3). For |a| ∈
/ W,
Q
∞
p
Z ∞
−1
1
+
k=0
k+a
tan (tanh πx cot aπ)
π
.
dx = 2 log Q
2 + x2 )
∞
p
x(p
2p
0
1
+
k=1
k−a
Q
∞
1
Z ∞
−1
1
+
2
2
k=0
p (k+a)
tan (tanh πx cot aπ)
π
.
dx = log Q
2 x2 )
∞
1
x(1
−
p
4
0
1
+
2
2
k=1
p (k−a)
q
Q
∞
1
1
Z 1
1 + (k+a)
2
k=0 k+a +
π
tan−1 (tanh πx cot aπ)
.
√
q
dx = log Q
∞
1
2
x 1 − x2
0
+ 1+ 1 2
k=1
k−a
(k−a)
Similarly, the following integrals are derived from the identity (1.1) and the integrals
given in (3.1), (3.2) and (3.3).
(−1)k
Q∞ p
Z ∞
−1
1
+
k=0
k+a
π
tan (sinh πx csc aπ)
dx = 2 log
(−1)k .
2
2
Q
x(p + x )
2p
∞
p
0
1
+
k=1
k−a
(−1)k
Q∞
1
Z ∞
−1
1
+
2
2
k=0
p (k+a)
tan (sinh πx csc aπ)
π
dx = log
(−1)k .
2 x2 )
Q
x(1
−
p
4
∞
0
1
k=1 1 + p2 (k−a)2
(−1)k
q
Q∞ 1
1
Z 1
+
1
+
−1
2
k=0 k+a
(k+a)
π
tan (sinh πx csc aπ)
√
dx = log
(−1)k .
q
2
Q
2
x 1−x
∞
0
1
1
+
1
+
k=1 k−a
(k−a)2
4. Conclusion.
The integrals presented in this paper are evaluated through sums related to inverse tangent function [1] either by direct formulas or through infinite products. The
solutions of these integrals are not found in the classical tables by Gradshteyn and
Ryzhik [2]. The collection presented here will give a valuable addition to the existing
list of integrals.
References
[1] B. C. Bernt, Ramanujan notebooks, Part-I, Springer-Verlag, New York, 1985.
[2] I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, 6 Ed, Academic
Press, New York, 2000.
Received 12 04 2012, revised 29 08 2012
Department of Mathematics,
D.G. Vaishnav college,
Chennai-600106, Tamil Nadu, India.
E-mail address: ramjan [email protected].