Citation: Abu-Ghuwaleh, M.;
Saadeh, R.; Qazza, A. A Novel
Approach in Solving Improper
Integrals. Axioms 2022, 11, 572.
https://doi.org/10.3390/
axioms11100572
Academic Editors: Inna Kalchuk
and Mircea Merca
Received: 11 September 2022
Accepted: 13 October 2022
Published: 20 October 2022
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axioms
Article
A Novel Approach in Solving Improper Integrals
Mohammad Abu-Ghuwaleh , Rania Saadeh * and Ahmad Qazza
Department of Mathematics, Faculty of Science, Zarqa University, Zarqa 13110, Jordan
* Correspondence: [email protected]
Abstract:
To resolve several challenging applications in many scientific domains, general formulas
of improper integrals are provided and established for use in this article. The suggested theorems
can be considered generators for new improper integrals with precise solutions, without requiring
complex computations. New criteria for handling improper integrals are illustrated in tables to
simplify the usage and the applications of the obtained outcomes. The results of this research are
compared with those obtained by I.S. Gradshteyn and I.M. Ryzhik in the classical table of integrations.
Some well-known theorems on improper integrals are considered to be simple cases in the context of
our work. Some applications related to finding Green’s function, one-dimensional vibrating string
problems, wave motion in elastic solids, and computing Fourier transforms are presented.
Keywords:
improper integrals; power series; analytic function; Cauchy residue theorem; Ramanu-
jan’s master theorem
MSC: 30E20; 33E20; 44A99
1. Introduction
Numerous studies on the topic of improper integrals have been published in recent
years in a variety of scientific disciplines, including physics and engineering [
1
–
7
]. Due
to this, mathematicians have been particularly interested in finding new theorems and
methods to solve these integrals. Particularly in engineering, applied mathematical physics,
electrical engineering, and other fields, it is sometimes necessary to handle erroneous
integrals in computations or when describing models [
8
–
16
]. While some of these integra-
tions can be handled easily, others require complex calculations. Many of these integrals
require computer software to be solved as they cannot be calculated so manually. Addition-
ally, numerical techniques may be employed to resolve some incorrect integrals that the
aforementioned techniques are unable to resolve [17–23].
The process of evaluating improper integrals is not usually based on certain rules or
techniques that can be applied directly. Many methods and techniques were established
and introduced by mathematicians and physicists to present a closed form for indefinite
integrals, the technique of double integrals, series methods, residue theorems, calculus
under the integral sign, and other methods that are used to solve improper complex
integrals exactly or approximately [24–31].
The residue theorem was first established by A.L. Cauchy in 1826, which is consid-
ered a powerful theorem in complex analysis. However, the applications that can be
calculated using the residue theorem to compute integrals on real numbers require many
precise constraints that should be satisfied in order to solve the integrals, including finding
appropriate closed contours and also determining the poles. Another challenge in the
process of applying the residue theorem is the difficulty and efforts in finding solutions for
some integrations.
According to his published memoirs, Cauchy developed powerful formulas in math-
ematics using the residue theorem [
4
]. Researchers consider these formulas essential in
treating and solving improper integrals. However, these results are considered simple cases
Axioms 2022, 11, 572. https://doi.org/10.3390/axioms11100572 https://www.mdpi.com/journal/axioms
Axioms 2022, 11, 572 2 of 19
when compared to the results that we present in this article. In addition, we show that the
proposed theorems and results in this research are not based on the residue theorem.
One significant accomplishment in the sphere of definite and indefinite integrals is
found in the master theorem of Ramanujan, which presents new expressions concerning
the Milline transform of any continuous function in terms of the analytic Taylor series, and
others [
32
–
39
]. It was implemented by Ramanujan and other researchers as a powerful tool
in calculating definite and indefinite integrals and also in computing infinite series. The
obtained results are as applicable and effective as Ramanujan’s master theorem in handling
and generating new formulas of integrals with direct solutions.
In this study, we introduce new theorems to simplify the procedure of computing
improper integrals by presenting new theorems with proofs. Each theorem can generate
many improper integral formulas that cannot be solved by usual techniques or would need
a large amount of effort and time spent in order to be solved. The motivation of this work is
to generate as many improper integrals and their values as possible to be used in different
problems. The obtained results can be implemented to construct new tables of integrations
so that researchers can use them in calculations and to check the accuracy of their answers
while discovering new methods.
The main purpose of this work is to introduce simple new techniques to help re-
searchers, mathematicians, engineers, physicists, etc., to solve some difficult improper
integrals that cannot be treated or solved easily (and which require several theorems and a
large amount of effort to solve). This goal is achieved by introducing some master theorems
that can be implemented in order to solve difficult applications. The outcomes can be
generalized and introduced in tables to obtain and to use the results of some improper
integrals directly.
We organize this article as follows: In Section 2, we introduce some illustrative prelimi-
naries; then, facts concerning analytic functions, master theorems, and results are presented
in Section 3. Mathematical remarks and several applications are presented in Section 4.
Finally, the conclusion of our research is presented in Section 5.
2. Preliminaries
In this section, some basic definitions and theorems related to our work are presented
and illustrated for later use.
2.1. Basic Definitions and Lemmas
Definition 1
([
7
])
.
Suppose that a function
f
is analytic in a domain
Ω ⊆ C
, where
C
is the
complex plane. Consider a disc
D ⊆ Ω
centered at
z
0
; then, the function
f
can be expressed in the
following series expansion:
f (z) =
∞
∑
n=0
a
n
(
z − z
0
)
n
,
where a
n
is the coefficients of the series.
Definition 2
([
8
])
.
Assume that
f
is an analytic function; then, Taylor series expansion at any
point x
0
of f in its domain is given by
T(x) =
∞
∑
n=0
f
(n)
(
x
0
)
n!
(
x − x
0
)
n
,
which converges to f in a neighborhood of x
0
point wisely.
2.2. Basic Formulas of Series and Improper Integrals
In this section, we introduce some series and improper integrals that are needed in
our work.
Axioms 2022, 11, 572 3 of 19
Lemma 1. The following factorization formula holds for n∈ N, as follows
1
(
x
2
+1
2
)(
x
2
+3
2
)
...(x
2
+
(
2n+1
)
2
)
=
(
−1
)
n
4
n
(
2n+1
)
!
n
∑
k=0
(
−1
)
k
2n + 1
k
2n+1−2k
(
2n+1−2k
)
2
+x
2
(1)
Proof.
To prove Equation (1), we define an integral whose solution can be expressed by
two different forms: the left side of Equation (1) and the right side of the equation.
Let
I =
∞
Z
0
e
−px
(
sinx
)
2a+1
dx, (2)
where p > 0, a ∈ N.
Taking the indefinite integral:
J = p
2
Z
e
−px
(
sinx
)
2a+1
dx (3)
Applying integration by parts on Equation (3) twice, we obtain a reduction formula
as follows:
J = −pe
−px
(
sinx
)
2a+1
−
(
2a + 1
)
e
−px
sin
2a
xcosx
+
(
2a + 1
)
R
e
−px
[2a(
(
sinx
)
2a−1
−
(
sinx
)
2a+1
) −
(
sinx
)
2a+1
] dx.
(4)
Taking the limit of the integrals in Equation (4) from 0 to ∞, we obtain:
∞
Z
0
e
−px
(
sinx
)
2a+1
dx =
(
2a + 1
)(
2a
)
p
2
+
(
2a + 1
)
2
∞
Z
0
e
−px
(
sinx
)
2a−1
dx. (5)
Applying Equation (5)
(
a − 1
)
times to the integral
R
∞
0
e
−px
(
sinx
)
2a−1
dx, we obtain:
∞
R
0
e
−px
(
sinx
)
2a+1
dx
=
(
2a+1
)(
2a
)(
2a−1
)(
2a−2
)
...(3)(2)
(
(
2a+1
)
2
+p
2
) (
(
2a−1
)
2
+p
2
)...
(
3+p
2
)
∞
R
0
e
−px
sinxdx.
(6)
The integral
R
∞
0
e
−px
sinxdx
can be calculated easily using twice integration by parts
to obtain:
∞
Z
0
e
−px
sinxdx =
1
1 + p
2
. (7)
Substituting the fact in Equation (7) into Equation (6), we obtain:
∞
Z
0
e
−px
(
sinx
)
2a+1
dx =
(
2a + 1
)
!
(
(
2a + 1
)
2
+ p
2
) (
(
2a − 1
)
2
+ p
2
) . . .
(
3 + p
2
) (
1 + p
2
)
. (8)
Therefore, the left side of Equation (1) is obtained.
Now, we express the solution of Equation (2) in another form, that is, to obtain the
right side of Equation (1), as follows:
Using the power trigonometric formula deduced using De Moivre’s formula, Euler’s
formula, and the binomial theorem [10] (p. 31)
(
sin(x)
)
2a+1
=
(
−1
)
a
4
a
a
∑
k=0
(
−1
)
k
2a + 1
k
sin
[(
2a + 1 − 2k
)
x
]
. (9)
Axioms 2022, 11, 572 4 of 19
Substituting Equation (9) into Equation (2), we obtain:
∞
Z
0
e
−px
(
sinx
)
2a+1
dx =
∞
Z
0
e
−px
(
−1
)
a
4
a
a
∑
k=0
(
−1
)
k
2a + 1
k
sin
[(
2a + 1 − 2k
)
x
]
dx (10)
Therefore, by changing the order of the integral and the sum in Equation (10), we obtain:
∞
Z
0
e
−px
(
sinx
)
2a+1
dx =
(
−1
)
a
4
a
a
∑
k=0
(
−1
)
k
2a + 1
k
∞
Z
0
e
−px
sin
[(
2a + 1 − 2k
)
x
]
dx (11)
To evaluate the integral
R
∞
0
e
−px
sin
[(
2a + 1 − 2k
)
x
]
dx
, we apply twice integration by
parts to obtain:
∞
Z
0
e
−px
sin
[(
2a + 1 − 2k
)
x
]
dx =
2a + 1 − 2k
(
2a + 1 − 2k
)
2
+ p
2
. (12)
Substituting the result in Equation (12) into Equation (11), we obtain:
∞
Z
0
e
−px
(
sinx
)
2a+1
dx=
(
−1
)
a
4
a
a
∑
k=0
(
−1
)
k
2a + 1
k
2a + 1 − 2k
(
2a + 1 − 2k
)
2
+ p
2
. (13)
Therefore, the right side of Equation (1) is obtained.
Then, equating Equation (13) with Equation (8); this, thus, completes the proof of
Equation (21).
Lemma 2. The following factorization holds for n∈ N as,
1
x
(
x
2
+2
2
)(
x
2
+4
2
)
...
(
x
2
+4n
2
)
=
1
2
2n
(
2n
)
!
1
x
2n
n
+ 2
n−1
∑
k=0
(
−1
)
n+k
2n
k
x
(
2n−2k
)
2
+x
2
.
(14)
Proof.
The proof is obtained by repeating the same process in proving Lemma (1), but by
using the integral
R
∞
0
e
−px
(
sinx
)
2a
dx, where p > 0 and a ∈ N.
Lemma 3.
The following factorization formula holds for
n =
0, 1,
· · ·
, and
m =
1, 2,
· · ·
, as
follows:
1
[
(
x
2
+1
2
)(
x
2
+3
2
)
...
(
x
2
+
(
2n+1
)
2
)]
[
x
(
x
2
+2
2
)(
x
2
+4
2
)
...
(
x
2
+4m
2
)]
=
(
−1
)
n
(
2
2m+2n
)(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
2n+1−2s
x(
(
2n+1−2s
)
2
+x
2
)
+
(
−1
)
n
2
2m+2n−1
(2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
x
(
2n+1−2s
)
(
(
2m−2k
)
2
+x
2
) (
(
2n+1−2s
)
2
+x
2
)
.
(15)
Proof. This is a direct result obtained by multiplying Equation (1) by Equation (14).
Lemma 4. The following formulas of improper integrals are created using Lemmas (1–3):
∞
R
0
cos
(
θx
)
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n+1
n
∑
k=0
(
−1
)
k
2n + 1
k
e
θ(2k−2n−1)
,
for θ ≥ 0 , n = 0, 1, · · ·
(16)
Axioms 2022, 11, 572 5 of 19
Proof.
The formula is obtained by multiplying both sides of Equation (1) by
cos
(
θx
)
, then
integrating both sides from 0 to ∞, and using the well-known fact:
∞
Z
0
cos
(
θx
)
a
2
+ x
2
dx =
π
2a
e
−aθ
,
where a and θ > 0.
∞
R
0
xsin
(
θx
)
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n+1
n
∑
k=0
(
−1
)
k
2n + 1
k
(
2n − 2k + 1
)
e
θ(2k−2n−1)
,
for θ > 0 , n = 0, 1, · · · .
(17)
Proof.
The formula is obtained by differentiating both sides of Equation (16) with respect
to θ.
∞
R
0
sin
(
θx
)
x
(
x
2
+4
)(
x
2
+16
)
...(x
2
+
(
2n
)
2
)
dx
=
(
−1
)
n
(
2n
)
!
π
2
2n+1
(
−1
)
n
2n
n
+ 2
n−1
∑
k=0
(
−1
)
k
2n
k
e
2θ(k−n)
,
for θ > 0 , n = 1, 2, · · · .
(18)
Proof.
The formula is obtained by multiplying both sides of Equation (14) by
sin
(
θx
)
, then
integrating both sides from 0 to ∞, and using the well-known fact:
∞
Z
0
sin
(
θx
)
x
(
a
2
+ x
2
)
dx =
π
2a
2
1 − e
−aθ
,
where θ and a > 0
∞
R
0
cos
(
θx
)
(
x
2
+4
)(
x
2
+16
)
...(x
2
+
(
2n
)
2
)
dx =
(
−1
)
n
π 2
1−2n
(
2n
)
!
n−1
∑
k=0
(
−1
)
k
2n
k
(
k − n
)
e
2θ(k−n)
,
for θ ≥ 0 , n = 1, 2, · · · .
(19)
Proof.
The formula is obtained by differentiating both sides of Equation (18) with respect
to θ.
Lemma 5.
Let
θ >
0 and
n =
0, 1,
· · ·
,
m =
1, 2,
· · ·
. Then, we have the following improper integrals:
∞
R
0
sin
(
θx
)
(
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
))
(
x
(
x
2
+4
)(
x
2
+16
)
···
(
x
2
+4m
2
))
dx
=
(
−1
)
n
π
(
2
2m+2n+1
)
(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
1−e
−θ(2n+1−2s)
(
2n+1−2s
)
+
(
−1
)
n
π
2
2m+2n
(
2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n+1−2s
)
(
e
−θ(2n+1−2s)
−e
−θ(2m−2k)
)
(
2m−2k
)
2
−
(
2n+1−2s
)
2
.
(20)
Proof.
The formula is obtained by multiplying both sides of Equation (15) by
sin
(
θx
)
, then
integrating both sides from 0 to ∞, and using the well-known facts:
Axioms 2022, 11, 572 6 of 19
∞
Z
0
sin
(
θx
)
x
(
a
2
+ x
2
)
dx =
π
2a
2
1 − e
−aθ
,
and
∞
Z
0
xsin
(
θx
)
a
2
+ x
2
dx =
π
2
e
−aθ
,
where θ and a > 0
∞
R
0
cos
(
θx
)
(
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1
)
2
))
((
x
2
+4
)(
x
2
+16
)
...
(
x
2
+4m
2
))
dx
=
(
−1
)
n
π
(
2
2m+2n+1
)
(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
e
−θ(2n+1−2s)
+
(
−1
)
n
π
2
2m+2n
(
2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n+1−2s
)
(
(
2m−2k
)
e
−θ(2m−2k)
−
(
2n+1−2s
)
e
−θ(2n+1−2s)
)
(
2m−2k
)
2
−
(
2n+1−2s
)
2
Proof.
The formula can be obtained by differentiating both sides of Equation (20) with
respect to θ.
3. New Master Theorems
In this part, we present new theorems to help mathematicians, engineers, and physi-
cists solve complicated improper integrals. To obtain our objective, we introduce some
facts concerning analytic functions [7,9,12].
Assuming that
f
is an analytic function in a disc
D
centered at
α
, then using Taylor’s
expansion, where α , β and θ are real constants, we have
f (z) =
∞
∑
k=0
f
(k)
(α)
k!
(
z − α
)
k
, (21)
substituting
z = α + βe
iθx
into
f (z)
, where
β
is not completely arbitrary, since it must be
smaller than the radius of D, we obtain
f
α + βe
iθx
=
∞
∑
k=0
f
(k)
(α)
k!
β
k
e
iθkx
, x ∈ R. (22)
Using the formulas
e
iθx
+ e
−iθx
= 2cos
(
θx
)
, e
iθx
− e
−iθx
= 2isin
(
θx
)
,
one can obtain
1
2
f
α + βe
iθx
+ f
α + βe
−iθx
=
1
2
∞
∑
k=0
f
(k)
(α)
k!
β
k
e
iθkx
+ e
−iθkx
=
∞
∑
k=0
f
(k)
(α)
k!
β
k
cos
(
kθx
)
= f (α) + f
0
(α)βcos
(
θx
)
+
f
00
(α)
2!
β
2
cos
(
2θx
)
+ . . . .
(23)
Similarly,
1
2i
f
α + βe
iθx
− f
α + βe
−iθx
=
1
2
∞
∑
k=0
f
(k)
(α)
k!
β
k
e
iθkx
− e
−iθkx
= f
0
(α)βsin
(
θx
)
+
f
00
(α)
2!
β
2
sin
(
2θx
)
+ . . .
=
∞
∑
k=1
f
(k)
(α)
k!
β
k
sin
(
kθx
)
.
(24)
Axioms 2022, 11, 572 7 of 19
Next, the parameters in Equations (23) and (24) can be modified in the follow-
ing lemma.
Lemma 3. Assume that g
(
α + z
)
is an analytic function that has the following series expansion:
g
(
α + z
)
=
∞
∑
k=0
M
k
e
−kz
, (25)
whether z be real or imaginary, and
∑
∞
k=0
M
k
is absolutely convergent. Then
1
2
(
g
(
α − iθx
)
+ g
(
α + iθx
))
=
1
2
∞
∑
k=0
M
k
e
ikθx
+ e
−ikθx
=
∞
∑
k=0
M
k
cos
(
k θx
)
, (26)
and,
1
2i
(
g
(
α − iθx
)
− g
(
α + iθx
))
=
1
2i
∞
∑
k=1
M
k
e
ikθx
− e
−ikθx
=
∞
∑
k=1
M
k
sin
(
k θx
)
, (27)
where θ > 0, α ∈ R, and x is any real number.
The next part of this section includes the new master theorems that we establish. Moreover,
we mention here that Cauchy’s results in [
3
] are identical to our results with special choices of the
parameters, as will be discussed later.
Theorem 1.
Let
f
be an analytic function in a disc
D
centered at
α
, where
α ∈ R
. Then, we have
the following improper integral formula:
∞
R
0
f (α+βe
iθx
)+ f (α+βe
−iθx
)
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
f
α + βe
θ(2s−2n−1)
,
(28)
where θ ≥ 0 and n = 0, 1, 2, · · · .
Proof. Let
I =
∞
Z
0
f
α + βe
iθx
+ f
α + βe
−iθx
(
x
2
+ 1
)(
x
2
+ 9
)
· · · (x
2
+
(
2n + 1
)
2
)
dx. (29)
Now, since
f
is an analytic function around
α
, substituting the fact in Equation (23)
into Equation (29), we obtain
I =
∞
Z
0
2
∑
∞
k=0
f
(k)
(α)β
k
k!
cos
(
kθx
)
(
x
2
+ 1
)(
x
2
+ 9
)
· · · (x
2
+
(
2n + 1
)
2
)
dx. (30)
Fubini’s theorem implies changing the order of the summation and the improper
integral to obtain
I = 2
∞
∑
k=0
f
(k)
(α)β
k
k!
∞
Z
0
cos
(
kθx
)
(
x
2
+ 1
)(
x
2
+ 9
)
· · · (x
2
+
(
2n + 1
)
2
)
dx. (31)
The fact in Equation (1) implies that Equation (31) becomes
I = 2
∞
∑
k=0
f
(k)
(α)β
k
k!
(
−1
)
n
(
2n + 1
)
!
π
2
2n+1
n
∑
s=0
(
−1
)
s
2n + 1
s
e
kθ(2s−2n−1)
. (32)
Axioms 2022, 11, 572 8 of 19
The result comes directly, by comparing the definition of the function
g
in Equation (25)
with the definition of the function f in Equation (22), to obtain
I =
(
−1
)
n
(
2n + 1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
f
α + βe
θ(2s−2n−1)
.
Theorem 2.
Let
f
be an analytic function in a disc
D
centered at
α
, where
α ∈ R
. Then, we have
the following improper integral formula:
∞
R
0
x
(
f (α+βe
iθx
)− f (α+βe
−iθx
)
)
i
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
(
2n − 2s
+1)
f
α + βe
θ(2s−2n−1)
− f (α)
,
(33)
where θ > 0 and n = 0, 1, 2, · · · .
Proof. Let
I =
∞
Z
0
x
f
α + βe
iθx
− f
α + βe
−iθx
i
(
x
2
+ 1
)(
x
2
+ 9
)
. . . (x
2
+
(
2n + 1
)
2
)
dx. (34)
Now, since
f
is an analytic function around
α
and substituting the fact in Equation (24)
into Equation (34), we obtain
I = 2
∞
∑
k=1
f
(k)
(α)β
k
k!
∞
Z
0
x
(
sin
(
kθx
))
(
x
2
+ 1
)(
x
2
+ 9
)
. . . (x
2
+
(
2n + 1
)
2
)
dx. (35)
Substituting the fact in Equation (2) into Equation (35), we obtain
I = 2
∞
∑
k=1
f
(k)
(α)β
k
k!
(
−1
)
n
(
2n + 1
)
!
π
2
2n+1
n
∑
s=0
(
−1
)
s
2n + 1
s
(
2n − 2s + 1)e
θk(2s−2n−1)
. (36)
The fact in Equation (22) implies that Equation (36) becomes
I =
(
−1
)
n
(
2n + 1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
(
2n − 2s + 1
)
f
α + βe
θ(2s−2n−1)
− f (α)
.
Hence, this completes the proof.
We should point out that
f (α)
appears in Equation (33) because the lower index of the
infinite summation started from
k =
1 and not from
k =
0, as is the case in Equation (29).
Thus, when we want to express the answer in terms of the original function
f
, we add and
subtract f (α) to obtain our result.
Theorem 3.
Let
f
be an analytic function in a disc
D
centered at
α
, where
α ∈ R
. Then, we have the
following improper integral formula:
∞
Z
0
f
α + βe
iθx
− f
α + βe
−iθx
i x
(
x
2
+ 4
)(
x
2
+ 16
)
. . . (x
2
+
(
2n
)
2
)
dx =
(
−1
)
n
(
2n
)
!
π
2
2n
(
−1
)
n
2n
n
ψ + 2
n−1
∑
s=0
(
−1
)
s
2n
s
φ(s)
!
, (37)
where θ > 0 , n = 1, 2, · · · , ψ = f
(
α + β
)
− f (α) and φ(s) = f
α + βe
2θ(s−n)
− f (α).
Axioms 2022, 11, 572 9 of 19
Proof.
The proof of Theorem 3 can be obtained by similar arguments to Theorem 2 and
using the fact (3) in Lemma 1.
Theorem 4.
Let
f
be an analytic function in a disc
D
centered at
α ∈ R
. Then, we have the
following improper integral formula:
∞
Z
0
f
α + βe
iθx
+ f
α + βe
−iθx
(
x
2
+ 4
)(
x
2
+ 16
)
. . . (x
2
+
(
2n
)
2
)
dx =
(
−1
)
n
π 2
2−2n
(
2n
)
!
n−1
∑
s=0
(
−1
)
s
2n
s
(
s − n
)
f
α + βe
2θ(s−n)
, (38)
where θ ≥ 0 , n = 1, 2, · · · .
Proof.
The proof of Theorem 4 can be obtained by similar arguments to Theorem 1 and
using the fact (4) in Lemma 1.
Theorem 5.
Let
f
be an analytic function in a disc
D
centered at
α
, where
α ∈ R
. Then, we have
the following improper integral formula:
∞
R
0
f
(
α+βe
iθx
)
− f
(
α+βe
−iθx
)
i
(
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1
)
2
))
(
x
(
x
2
+4
)(
x
2
+16
)
···
(
x
2
+4m
2
))
dx
=
(
−1
)
n
π
(
2
2m+2n
)(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
(
ϕ−ψ(s)
)
(
2n+1−2s
)
+
(
−1
)
n
π
2
2m+2n−1
(
2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n + 1
−2s)
(
ψ(s)−φ(k)
)
(
(
2m−2k
)
2
−
(
2n+1−2s
)
2
)
,
(39)
where
θ >
0,
n =
0, 1, 2,
· · ·
,
m =
1, 2,
· · ·
,
ψ(s) = f
α + βe
−θ(2n+1−2s)
,
φ(k) =
f
α + βe
−θ(2m−2k)
, and ϕ = f
(
α + β
)
.
Proof. Let
I =
∞
Z
0
f
α + βe
iθx
− f
α + βe
−iθx
i
(
x
2
+ 1
)(
x
2
+ 9
)
. . . (x
2
+
(
2n + 1
)
2
))
(
x
(
x
2
+ 4
)(
x
2
+ 16
)
· · ·
(
x
2
+ 4m
2
))
dx. (40)
Now, since
f
is an analytic function around
α
and substituting the fact in Equation (24)
into Equation (40), we obtain
I = 2
∞
∑
k=1
f
(k)
(α)β
k
k!
∞
Z
0
sin
(
θkx
)
(
x
2
+ 1
)(
x
2
+ 9
)
. . . (x
2
+
(
2n + 1
)
2
))
(
x
(
x
2
+ 4
)(
x
2
+ 16
)
· · ·
(
x
2
+ 4m
2
))
dx. (41)
Substituting the fact in Equation (9) into Equation (41), we obtain
I = 2
∞
∑
k=1
f
(k)
(α)β
k
k!
(
A + B
)
, (42)
where
A =
(
−1
)
n
π
(
2
2m+2n+1
)(
m!
)
2
(
2n + 1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
1 − e
−θk(2n+1−2s)
(
2n + 1 − 2s
)
,
Axioms 2022, 11, 572 10 of 19
B =
(
−1
)
n
π
2
2m+2n
(
2m
)
!
(
2n + 1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n + 1 − 2s)
e
−θk(2n+1−2s)
− e
−θk(2m−2k)
(
2m − 2k
)
2
−
(
2n + 1 − 2s
)
2
!
.
The fact in Equation (22) implies that Equation (42) becomes
I
=
(
−1
)
n
π
(
2
2m+2n
)(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
ϕ−ψ(s)
2n+1−2s
+
(
−1
)
n
π
2
2m+2n−1
(
2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n
+1 − 2s)
(
ψ(s)−φ(k)
)
(
(
2m−2k
)
2
−
(
2n+1−2s
)
2
)
,
where ψ(s) = f
α + βe
−θ(2n+1−2s)
, φ( k) = f
α + βe
−θ(2m−2k)
, and ϕ = f
(
α + β
)
.
Hence, this completes the proof of Theorem 5.
Theorem 6.
Let
f
be an analytic function in a disc
D
centered at
α
, where
α ∈ R
. Then, we have
the following improper integral formula:
∞
R
0
f
(
α+βe
iθx
)
+ f
(
α+βe
−iθx
)
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1)
2
)
((
x
2
+4
)(
x
2
+16
)
...
(
x
2
+4m
2
))
dx
=
(
−1
)
n
π
(
2
2m+2n
)(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
ψ(s)
+
(
−1
)
n
π
2
2m+2n−1
(
2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n+1−2s
)((
2m−2k
)
φ(k)−
(
2n+1−2s
)
ψ(s)
)
(
(
2m−2k
)
2
−
(
2n+1−2s)
2
)
,
(43)
where
θ ≥
0 ,
n =
0, 1, 2,
· · ·
,
m =
1, 2,
· · ·
,
ψ(s) = f
α + βe
−θ(2n+1−2s)
, and
φ(k) =
f
α + βe
−θ(2m−2k)
.
Proof
The proof of Theorem 6 can be obtained by similar arguments to Theorem 5 and
using the fact (6) in Lemma 2.
The following table, Table 1 illustrates some corollaries of the theorems with special
cases and presents some values of improper integrals under certain conditions.
Axioms 2022, 11, 572 11 of 19
Table 1. Improper integral formulas with the series representation as detailed in Equation (25).
f(x)
∞
R
0
f(x)dx
Conditions No. of Theorem
1.
g
(
α−iθx
)
+g
(
α+iθx
)
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
)
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
g
(
α − θ
(
2s − 2n − 1
))
,
θ ≥ 0,
n = 1, 2, · · ·
Theorem 1
2.
x
(
g
(
α−iθx
)
−g
(
α+iθx
))
i
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1
)
2
)
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
(
2n − 2s + 1
)(
g
(
α − θ
(
2s − 2n − 1
))
− g(α)
)
,
θ > 0,
n = 1, 2, · · ·
Theorem 2
3.
g
(
α−iθx
)
+ f
(
α+iθx
)
(
x
2
+4
)(
x
2
+16
)
···(x
2
+
(
2n
)
2
)
(
−1
)
n
π 2
2−2n
(
2n
)
!
n−1
∑
s=0
(
−1
)
s
2n
s
(
s − n
)
g
(
α − 2θ
(
s − n
))
,
θ ≥ 0,
n = 1, 2, · · ·
Theorem 4
4.
g
(
α−iθx
)
+g
(
α+iθx
)
(
(
x
2
+1
)(
x
2
+9
)
...(x
2
+
(
2n+1)
2
)
)
1
((
x
2
+4
)(
x
2
+16
)
...
(
x
2
+4m
2
))
(
−1
)
n
π
(
2
2m+2n
)(
m!
)
2
(
2n+1
)
!
n
∑
s=0
(
−1
)
s
2n + 1
s
ψ(s) +
(
−1
)
n
π
2
2m+2n−1
(
2m
)
!
(
2n+1
)
!
m−1
∑
k=0
n
∑
s=0
(
−1
)
m+k+s
2m
k
2n + 1
s
(
2n+1−2s
)((
2m−2k
)
φ(k)−
(
2n+1−2s
)
ψ(s)
)
(
(
2m−2k
)
2
−
(
2n+1−2s
)
2
)
,
where ψ(s) = g
(
α + θ
(
2n + 1 − 2s
))
and φ(k) = g
(
α + θ
(
2m − 2k
))
θ ≥ 0,
n = 0, 1, 2, · · ·
m = 1, 2, · · ·
Theorem 6
Axioms 2022, 11, 572 12 of 19
4. Applications and Examples
In this section, we present the results, applications, and observations of the proposed
theorems. We also show that the simple cases of the master theorems are identical to
the results obtained by Cauchy, as detailed in his memoirs, using Residue Theorem 4.
Additionally, some examples on difficult integrals that cannot be treated directly by usual
methods are addressed. In this section, we show the applicability of our results in handling
such problems.
4.1. Some Remarks on the Theorems
Remark 1. Letting α = 0 and n = 1 in Theorem 3, we obtain
∞
Z
0
f
βe
iθx
− f
βe
−iθx
i x
(
x
2
+ 4
)
dx =
π
4
f (β) − f
βe
−2θ
, (44)
where θ > 0.
By letting
x
2
= y,
1
4
∞
Z
0
f
βe
2iθy
− f
βe
−2iθy
i y
(
y
2
+ 1
)
dy =
π
4
f (β) − f
βe
−2θ
.
Letting 2θ = ϕ
∞
Z
0
f
βe
iϕy
− f
βe
−iϕy
i y
(
y
2
+ 1
)
dy = π
f (β) − f
βe
−ϕ
.
This result appears in [
10
] (Theorem 4). Further, we show that Cauchy made a mistake in this
result (see [4]) (P. 62 formula (10)).
The following table, Table 2 presents some remarks on improper integrals.
Axioms 2022, 11, 572 13 of 19
Table 2. Remarks on improper integrals, where θ > 0.
Conditions Theorem g(x)
∞
R
0
g(x) dx
Remarks
1 α = 0, β = 1 and n = 0 1
f
(
e
iθx
)
+ f
(
e
−iθx
)
(
1+x
2
)
π f
e
−θ
Cauchy’s theorem [4] (p. 62 Formula (8)) and in [10] (3.037
Theorem 1) is identical.
2 α = 0, β = 1 and n = 0 2
x
(
f
(
e
iθx
)
− f
(
e
−iθx
))
i
(
1+x
2
)
π
f
e
−θ
− f (0)
Cauchy made a mistake in this result see [
4
] (p. 62 Formula
(8)). He corrected his result in his next memoir see [5,6].
3 n = 1 1
f
(
α+βe
iθx
)
+ f
(
α+βe
−iθx
)
(
x
2
+1
2
)(
x
2
+3
2
)
π
4!
3 f
α + βe
−θ
− f
α + βe
−3θ
This result does not appear in [4,5,10].
4 n = 1 1
x
(
f
(
α+βe
iθx
)
− f
(
α+βe
−iθx
))
i
(
x
2
+1
2
)(
x
2
+3
2
)
π
8
f
α + βe
−θ
− f
α + βe
−3θ
This result does not appear in [4,5,10].
5 n = 2 3
f
(
α+βe
iθx
)
− f
(
α+βe
−iθx
)
i x
(
x
2
+2
2
)(
x
2
+4
2
)
π
192
3
(
f
(
a + β
))
+ f
α + βe
−4θ
− 4 f
α + βe
−2θ
This result does not appear in [4,5,10].
6 n = 1 4
f
(
α+βe
iθx
)
+ f
(
α+βe
−iθx
)
(
x
2
+2
2
)(
x
2
+4
2
)
π
48
2 f
α + βe
−2θ
− f
α + βe
−4θ
This result does not appear in [4,5,10].
Axioms 2022, 11, 572 14 of 19
4.2. Generating Improper Integrals
In this section, we show the mechanism of generating an infinite number of integrals
by choosing the function
f (z)
and finding the real or imaginary part. It is worth noting that
some of these integrals with special cases appear in [
40
–
43
] when solving some applications
related to finding Green’s function, one-dimensional vibrating string problems, wave
motion in elastic solids, and when using Fourier cosine and Fourier Sine transforms.
To illustrate the idea, we show some general examples that are applied on Theorems 1,
2, and 3, as follows:
1. Setting f (z) = z
m
, m ∈ R
+
:
• Using Theorem (1) and setting α = 0 and β = 1 we have:
f
e
iθx
+ f
e
−iθx
= e
iθmx
+ e
−iθmx
= 2cos
(
θ mx
)
.
Thus,
∞
Z
0
2cos
(
θ mx
)
(
x
2
+ 1
)(
x
2
+ 9
)
· · ·
x
2
+
(
2n + 1
)
2
dx =
(
−1
)
n
(
2n + 1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
e
mθ(2s−2n−1)
.
where θ ≥ 0 and n = 0, 1, 2, · · ·
Setting m = 1, the obtained integral is a Fourier cosine transform [
40
,
41
] of the function
f (t) =
1
(
t
2
+1
)(
t
2
+9
)
···
(
t
2
+
(
2n+1
)
2
)
.
• Using Theorem (3), and setting α = 0, β = 1 we have:
1
i
f
e
iθx
− f
e
−iθx
=
1
i
e
iθmx
+ e
−iθmx
= 2sin
(
θ mx
)
.
Thus,
∞
Z
0
2sin
(
θ mx
)
x
(
x
2
+ 4
)(
x
2
+ 16
)
. . .
x
2
+
(
2n
)
2
dx =
(
−1
)
n
(
2n
)
!
π
2
2n
(
−1
)
n
2n
n
+ 2
n−1
∑
s=0
(
−1
)
s
2n
s
e
2θm(s−n)
!
.
Setting m = 1, the obtained integral is a Fourier sine transform [
40
,
41
] of the function
f (t) =
1
t
(
t
2
+4
)(
t
2
+16
)
...
(
t
2
+
(
2n
)
2
)
.
2. Setting f (z) = e
z
.
• Using Theorem (1), we have:
f
α + βe
iθx
+ f
α + βe
−iθx
= e
α+βe
iθx
+ e
α+βe
−iθx
= 2e
α+βcos(θx)
cos
(
βsin
(
θx
))
.
∞
R
0
2e
α+βcos(θx)
cos
(
βsin
(
θx
))
(
x
2
+1
)(
x
2
+9
)
···
(
x
2
+
(
2n+1
)
2
)
dx =
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
e
α+βe
θ(2s−2n−1)
,
where θ ≥ 0 and n = 0, 1, 2, · · · .
• Using Theorem (2), we have:
1
i
f
α + βe
iθx
− f
α + βe
−iθx
=
1
i
e
α+βe
iθx
− e
α+βe
−iθx
= 2e
α+βcos(θx)
sin
(
βsin
(
θx
))
.
Axioms 2022, 11, 572 15 of 19
Thus,
∞
R
0
2xe
α+βcos(θx)
sin
(
βsin
(
θx
))
i
(
x
2
+1
)(
x
2
+9
)
···
(
x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
(
2n − 2s + 1
)
e
α+βe
θ(2s−2n−1)
− e
α
.
3. Setting f (z) = sinhz.
• Using Theorem (1), we have:
f
α + βe
iθx
+ f
α + βe
−iθx
= sinh
α + βe
iθx
+ sinh
α + βe
−iθx
= 2cos
(
βsin
(
θx
))
sinh
(
α + βcos
(
θx
))
Thus,
∞
R
0
2cos
(
βsin
(
θx
))
sinh
(
α+βcos
(
θx
))
(
x
2
+1
)(
x
2
+9
)
···
(
x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
sinh
α + βe
θ(2s−2n−1)
• Using Theorem (3), we have:
1
i
f
α + βe
iθx
− f
α + βe
−iθx
=
1
i
sinh
α + βe
iθx
− sinh
α + βe
−iθx
= 2sin
(
βsin
(
θx
))
cosh
(
α + βcos
(
θx
))
.
Thus,
∞
R
0
2sin
(
βsin
(
θx
))
cosh
(
α+βcos
(
θx
))
x
(
x
2
+4
)(
x
2
+16
)
...
(
x
2
+
(
2n
)
2
)
dx
=
(
−1
)
n
(
2n
)
!
π
2
2n
(
−1
)
n
2n
n
(
sinh
(
α + β
)
− sinh(α)
)
+2
n−1
∑
s=0
(
−1
)
s
2n
s
sinh
α + βe
2θ(s−n)
− sinh(α)
,
where θ > 0, n = 1, 2, · · · .
4. Setting f (z) = cos
(
e
z
)
• Using Theorem (1), we have:
f
α + βe
iθx
+ f
α + βe
−iθx
= cos
e
α+βe
iθx
+ cos
e
α+βe
−iθx
= 2cos
e
α+βcos(θx)
cos
(
βsin
(
θx
))
cosh
sin
(
βsin
(
θx
))
e
α+βcos(θx)
.
Thus,
∞
R
0
2cos
(
e
α+βcos(θx)
cos
(
βsin
(
θx
))
)
cosh
(
sin
(
βsin
(
θx
))
e
α+βcos(θx)
)
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
cos
α + βe
θ(2s−2n−1)
.
5. Setting f (z) = ln
(
1 + z
)
,
• Using Theorem (1), we have:
f
1 + α + βe
iθx
+ f
1 + α + βe
−iθx
= ln
1 + α + βe
iθx
+ ln
1 + α + βe
−iθx
= ln(
(
α + 1
)
2
+ β
2
+ 2
(
α + 1
)
βcos
(
θx
)
).
Axioms 2022, 11, 572 16 of 19
Thus,
∞
R
0
ln(
(
α+1
)
2
+β
2
+2
(
α+1
)
βcos
(
θx
)
)
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
ln
1 + α + βe
θ(2s−2n−1)
.
• Setting α = 0 and β = 1, we have:
f
e
iθx
+ f
e
−iθx
= ln
1 + e
iθx
+ ln
1 + e
−iθx
= 2ln
2cos
θx
2
.
Thus,
∞
R
0
2ln
|
2cos
(
θx
2
)
|
(
x
2
+1
)(
x
2
+9
)
···(x
2
+
(
2n+1
)
2
)
dx
=
(
−1
)
n
(
2n+1
)
!
π
2
2n
n
∑
s=0
(
−1
)
s
2n + 1
s
ln
1 + e
θ(2s−2n−1)
.
4.3. Solving Improper Integrals
In this section, some applications on complicated problems are introduced and solved
directly depending on our new theorems. We note that the Mathematica and Maple
software cannot solve such examples.
Example 1. Evaluate the following integral:
∞
Z
0
ln
2
tan
θx
2
−
π
4
(
x
2
+ 4
)(
x
2
+ 16
)
dx,
where θ > 0.
Solution:
Using Theorem 1 and setting
α =
0,
β =
1,
and n =
1 or using Remark 6 Table 2
and setting α = 0 and β = 1, we set
f (z) =
tan
−1
z
2
=
−1
4
ln
2
1 − iz
1 + iz
.
Therefore, we have f
e
iθx
= −
1
4
ln
2
1−ie
iθx
1+ie
iθx
, and f
e
iθx
+ f
e
−iθx
= 2 Re f
e
iθx
.
Thus, we obtain
∞
R
0
f
(
e
iθx
)
+ f
(
e
−iθx
)
(
x
2
+4
)(
x
2
+16
)
dx =
−1
4
∞
R
0
ln
2
1−ie
iθx
1+ie
iθx
+ln
2
1−ie
−iθx
1+ie
−iθx
(
x
2
+4
)(
x
2
+16
)
dx
=
−1
4
∞
R
0
2 Re
ln
2
1+sin
(
θx
)
−icos
(
θx
)
1−sin
(
θx
)
+icos
(
θx
)
(
x
2
+4
)(
x
2
+16
)
dx
=
−1
2
∞
R
0
Re
(
ln
(
1+sin
(
θx
)
−icos
(
θx
))
−ln
(
1−sin
(
θx
)
+icos
(
θx
)))
2
(
x
2
+4
)(
x
2
+16
)
dx
=
−1
2
∞
R
0
Re
(
±i
π
2
−ln
|
tan
(
θx
2
−
π
4
)
|
)
2
(
x
2
+4
)(
x
2
+16
)
dx
=
−1
2
∞
R
0
Re
−
π
2
4
+ln
2
|
tan
(
θx
2
−
π
4
)
|
±iπln
|
tan
(
θx
2
−
π
4
)
|
(
x
2
+4
)(
x
2
+16
)
dx
=
−1
2
∞
R
0
−
π
2
4
+ln
2
|
tan
(
θx
2
−
π
4
)
|
(
x
2
+4
)(
x
2
+16
)
dx =
π
24
3
tan
−1
e
−θ
2
−
tan
−1
e
−3θ
2
.
∴
∞
R
0
ln
2
|
tan
(
θx
2
−
π
4
)
|
(
x
2
+4
)(
x
2
+16
)
dx =
π
3
384
−
π
48
3
tan
−1
e
−θ
2
−
tan
−1
e
−3θ
2
.
Axioms 2022, 11, 572 17 of 19
Example 2. Evaluate the following integral:
PV
∞
Z
0
x tan
(
πx
)
(
x
2
+ 2
2
)(
x
2
+ 4
2
)
. . .
x
2
+
(
2n
)
2
dx, (45)
where n = 1, 2, · · · .
Solution. Using Theorem 4, let α = 0, β = 1 and f (z) = ln
(
1 + z
)
.
Therefore, we have
f
e
iθx
+ f
e
−iθx
= ln
1 + e
iθx
+ ln
1 + e
−iθx
= ln
(
2cos
(
θx
)
+ 2
)
= 2ln
2cos
θx
2
.
Therefore, we have
I(θ) = PV
∞
Z
0
2ln
2cos
θx
2
(
x
2
+ 2
2
)(
x
2
+ 4
2
)
. . .
x
2
+
(
2n
)
2
dx =
(
−1
)
n
π 2
2−2n
(
2n
)
!
n−1
∑
s=0
(
−1
)
s
2n
s
(
s − n
)
ln
1 + e
2θ(s−n)
!
.
Now, taking the derivative of I(θ) with respect to θ, we obtain
∂I
∂ θ
= PV
∞
Z
0
−x tan
θx
2
(
x
2
+ 2
2
)(
x
2
+ 4
2
)
· · · (x
2
+
(
2n
)
2
)
dx =
(
−1
)
n
π 2
2−2n
(
2n
)
!
n−1
∑
s=0
(
−1
)
s
2n
s
(
s − n
)
2
(
s − n
)
e
2θ(s−n)
e
2θ(s−n)
+ 1
!
.
Therefore,
PV
∞
Z
0
x tan
(
πx
)
(
x
2
+ 2
2
)(
x
2
+ 4
2
)
· · · (x
2
+
(
2n
)
2
)
dx =
(
−1
)
n+1
π 2
2−2n
(
2n
)
!
n−1
∑
s=0
(
−1
)
s
2n
s
(
s − n
)
2
(
s − n
)
e
4π(s−n)
e
4π(s−n)
+ 1
!
Putting n = 1 in Equation (45), we obtain the following integral:
PV
∞
Z
0
x tan
(
πx
)
x
2
+ 4
dx = π
e
−4π
e
−4π
+ 1
=
π
(
e
4π
+ 1
)
.
Example 3. Evaluate the following integral:
∞
Z
0
1 + 2cos
(
θx
)
(
x
2
+ 1
)(
x
2
+ 4
) (
1 + 4cos
(
θx
)
+ 4
)
dx,
where θ ≥ 0.
Solution. Using Theorem 5 and taking α = 0 and β = 1, let f (z) =
1
1+2e
z
.
Thus, we have
f
e
i θx
+ f
e
−i θx
=
1
1 + 2e
iθx
+
1
1 + 2e
−iθx
=
2
(
1 + 2 cos
(
θx
))
1 + 4cos
(
θx
)
+ 4
.
Therefore, setting n = 0 and m = 1, in Theorem 5, we obtain
∞
Z
0
1 + 2cos
(
θx
)
(
x
2
+ 1
)(
x
2
+ 4
) (
1 + 4cos
(
θx
)
+ 4
)
dx =
π
12
2
1 + 2e
−θ
−
1
1 + 2e
−θ
.
Axioms 2022, 11, 572 18 of 19
5. Conclusions
In this research, we introduce new theorems that simplify calculating improper inte-
grals. These results can establish many instances of formulas of improper integrals and
solve them directly without complicated calculations or computer software. We illustrate
some remarks that analyze our work.
•
The proposed theorems are considered powerful techniques for generating improper
integrals and testing the results when using other methods to solve similar examples.
•
These theorems can be illustrated in tables of integrations, with different values of
functions and generate more results.
•
The obtained improper integrals cannot be solved manually (simply) or by computer
software such as Mathematica and Maple.
We intend to generalize the proposed theorems and make tables and algorithms to
simplify their use during the applications. Additionally, these results can be used to solve
differential equations by inverting the integrals into differential equations.
Author Contributions:
Conceptualization, M.A.-G., R.S. and A.Q.; methodology, M.A.-G., R.S. and
A.Q.; software, M.A.-G., R.S. and A.Q.; validation, M.A.-G., R.S. and A.Q.; formal analysis, M.A.-G.,
R.S. and A.Q.; investigation, M.A.-G., R.S. and A.Q.; resources, R.S. and A.Q.; data curation, M.A.-G.,
R.S. and A.Q.; writing—original draft preparation, M.A.-G., R.S. and A.Q.; writing—review and
editing, M.A.-G., R.S. and A.Q.; visualization, M.A.-G., R.S. and A.Q.; supervision, M.A.-G., R.S. and
A.Q.; project administration, R.S. and A.Q.; funding acquisition, M.A.-G., R.S. and A.Q. All authors
have read and agreed to the published version of the manuscript.
Funding: This research received no external funding.
Data Availability Statement: Not applicable.
Conflicts of Interest: The authors declare no conflict of interest.
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