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Article

Evaluation of Certain Definite Integrals Involving Generalized Hypergeometric Functions

1
Department of Mathematics, Manipal Institute of Technology, Manipal Academy of Higher Education, Manipal 576104, Karnataka, India
2
Department of Mathematics Education, Andong National University, Andong 36729, Republic of Korea
3
Department of Mathematics, Vedant College of Engineering & Technology (Rajasthan Technical University), Bundi 323021, Rajasthan, India
4
School of Mathematical Sciences, College of Computing, Informatics and Mathematics, Universiti Teknologi MARA, Shah Alam 40450, Selangor, Malaysia
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Axioms 2024, 13(12), 887; https://doi.org/10.3390/axioms13120887
Submission received: 28 November 2024 / Revised: 20 December 2024 / Accepted: 20 December 2024 / Published: 23 December 2024

Abstract

:
In 2012, Chu investigated the generalization of classical Watson–Whipple–Dixon summation theorems in the form of analytical formulas. By employing four generalized Watson summation formulas, the objective of this paper is to evaluate a new class of several Eulerian-type integrals (single and double) and Laplace-type integrals involving a hypergeometric function. Several interesting special cases are also given. Symmetry arises spontaneously in the hypergeometric function.

1. Introduction

The generalized hypergeometric function is defined [1,2,3,4] by
F q p [ α 1 , α 2 , , α p β 1 , β 2 , , β q | ζ ] = n = 0 i = 1 p ( α i ) n j = 1 q ( β j ) n ζ n n ! ,
where α i and β j are complex parameters such that the denominators of the summands on the right-hand side are not equal to zero or negative integers.
The symbol ( α ) n is the well-known Pochhammer symbol [5] for any complex number α given by
( α ) n = α ( α + 1 ) ( α + n 1 ) , n N 1 , n = 0 .
In terms of gamma function, it can be written as
( α ) n = Γ ( α + n ) Γ ( α ) , α Z 0 .
The series given by (1) is convergent for all p q by the ratio test [1]. For p < q + 1 , it converges everywhere and converges nowhere for p > q + 1 . Further, if p = q + 1 , it converges absolutely for ζ = 1 provided δ = j = 1 q β j i = 1 p α i > 0 holds and is conditionally convergent for | ζ | = 1 and ζ 1 if 1 < δ 0 and diverges for | ζ | = 1 and ζ 1 if δ 1 .
It is interesting to mention here that the generalized hypergeometric function exhibits symmetry in both its numerator and denominator parameters. Any reordering of the numerator parameters α 1 , α 2 , , α p does not alter the function, and, similarly, any reordering of the denominator parameters β 1 , β 2 , , β q also yields the same function [6].
The result will be significant whenever a generalized hypergeometric series can be summed in terms of gamma functions because there are very few of these summation theorems in the literature. However, in this paper, we shall mention the following Watson summation theorem viz,
F 2 3 [ α 1 , α 2 , α 3 α 1 + α 2 + 1 2 , 2 α 3 | 1 ] = π Γ ( 1 + α 1 + α 2 2 ) Γ ( 1 2 + α 3 ) Γ ( 1 α 1 α 2 2 + α 3 ) Γ ( 1 + α 1 2 ) Γ ( 1 + α 2 2 ) Γ ( 1 α 1 2 + α 3 ) Γ ( 1 α 2 2 + α 3 ) = W 0 , 0 ( α 1 , α 2 , α 3 )
provided ( 2 α 3 α 1 α 2 ) > 1 .
For the two integer parameters m and n, let us denote
W m , n ( α 1 , α 2 , α 3 ) = F 2 3 [ α 1 , α 2 , α 3 1 + α 1 + α 2 + m 2 , 2 α 3 + n | 1 ] .
Clearly, for m = n = 0 , we obtain (3).
In 1992, Lavoie, Grondin, and Rathie [3] obtained the generalization of Watson series (3) with two integer parameters m and n in the following form:
F 2 3 α 1 , α 2 , α 3 1 + α 1 + α 2 + m 2 , 2 α 3 + n | 1 = G m , n 2 α 1 + α 2 + m 2 Γ 1 2 ( α 1 + α 2 + m + 1 ) Γ ( α 3 + n 2 + 1 2 ) Γ ( α 3 α 1 2 α 2 2 m + n 2 + n 2 + 1 2 ) Γ ( 1 2 ) Γ ( α 1 ) Γ ( α 2 ) × H m , n Γ α 1 2 + 1 ( 1 ) m 4 Γ ( α 2 2 ) Γ α 3 α 1 2 + 1 2 + n 2 ( 1 ) n ( 1 ( 1 ) m ) 4 Γ α 3 α 2 2 + 1 2 + n 2 + I m , n Γ α 1 2 + 1 + ( 1 ) m 4 ) Γ ( α 2 2 + 1 2 ) Γ α 3 α 1 2 + n + 1 2 + ( 1 ) n ( 1 ( 1 ) m ) 4 Γ α 3 α 2 2 + n + 1 2
for m , n = 2 , 1 , 0 , 1 , 2 . Here, as usual, x is the greatest integer less than or equal to x and its modulus is denoted by | x | . For the values of the coefficients G m , n , H m , n , and I m , n one can refer to [3]. Later, in 2012, Chu [7] established analytical formulas explicitly for Watson series (3) with two integer parameters m ∈ N 0 and n∈ N 0 where N 0 = { 0 , 1 , 2 , } in the following form:
W m , n ( α 1 , α 2 , α 3 ) = i = 0 m j = 0 n ( 1 ) i + j m i n j α 1 α 2 m + 2 i α 1 α 2 m + i ( α 1 ) i + j ( α 2 ) m i + j ( α 2 α 1 i ) m ( 2 α 3 1 ) 2 j × ( α 3 ) j ( 2 α 3 1 ) j 1 + α 1 + α 2 + m 2 j ( 2 α 3 + n ) j W 0 , 0 ( α 1 + i + j , α 2 + m i + j , α 3 + j )
W m , n ( α 1 , α 2 , α 3 ) = i = 0 m j = 0 n ( 1 ) i m i n j α 1 α 2 m + 2 i α 1 α 2 m + i ( α 1 ) i + j ( α 2 ) m i + j ( α 2 α 1 i ) m ( 2 α 3 2 n 1 ) 2 j × ( α 3 n ) j ( 2 α 3 2 n 1 ) j 1 + α 1 + α 2 + m 2 j ( 2 α 3 n ) j W 0 , 0 ( α 1 + i + j , α 2 + m i + j , α 3 + j n )
W m , n ( α 1 , α 2 , α 3 ) = 1 α 1 + α 2 m 2 m 1 + α 1 + α 2 m 2 m i = 0 m j = 0 n ( 1 ) j m i n j α 1 α 2 m + 2 i α 1 α 2 m + i ( α 1 ) i + j ( α 2 ) m i + j ( α 2 α 1 i ) m ( 2 α 3 1 ) 2 j × ( α 3 ) j ( 2 α 3 1 ) j 1 + α 1 + α 2 + m 2 j ( 2 α 3 + n ) j W 0 , 0 ( α 1 + i + j , α 2 + m i + j , α 3 + j )
W m , n ( α 1 , α 2 , α 3 ) = 1 α 1 + α 2 m 2 m 1 + α 1 + α 2 m 2 m i = 0 m j = 0 n m i n j α 1 α 2 m + 2 i α 1 α 2 m + i ( α 1 ) i + j ( α 2 ) m i + j ( α 2 α 1 i ) m ( 2 α 3 2 n 1 ) 2 j × ( α 3 n ) j ( 2 α 3 2 n 1 ) j 1 + α 1 + α 2 + m 2 j ( 2 α 3 n ) j W 0 , 0 ( α 1 + i + j , α 2 + m i + j , α 3 + j n )
On the other hand, the Eulerian integrals are an important class of finite integrals. The integrals of this type can be written in the form
I = 0 1 x c 1 ( 1 x ) d 1 f ( x ) d x
provided ( c ) > 0 and ( d ) > 0 .
Next, if the function f(x) in the integrand is such that it can be expanded in a power series such as
f ( x ) = n = 0 c n x n
provided that the radius of convergence of (10) is not greater than unity, then, from (9), we have
I = 0 1 x c 1 ( 1 x ) d 1 n = 0 c n x n d x = n = 0 c n { 0 1 x c + n 1 ( 1 x ) d 1 } d x
Evaluating the beta integral, and after a little simplification, we have
I = Γ ( c ) Γ ( d ) Γ ( c + d ) n = 0 c n ( c ) n ( c + d ) n
This is a general result.
Further, f(x) is in the form of a Gauss hypergeometric function 2F1 as follows:
f ( x ) = F 1 2 a , b e | s x .
Hence,
I = Γ ( c ) Γ ( d ) Γ ( c + d ) n = 0 ( a ) n ( b ) n ( c ) n ( e ) n ( c + d ) n ( s x ) n n ! = Γ ( c ) Γ ( d ) Γ ( c + d ) F 2 3 a , b , c e , c + d | s .
Many summations theorems may be employed to sum the 3F2 function that appeared on the right-hand side of (14). For example, if, in (14), we take s = 1 , d = c , and e = 1 2 ( a + b + 1 ) , then (14) takes the following form:
0 1 x c 1 ( 1 x ) c 1 F 1 2 a , b 1 2 ( a + b + 1 ) | x d x = Γ ( c ) Γ ( c ) Γ ( 2 c ) F 2 3 a , b , c 1 2 ( a + b + 1 ) , 2 c | 1
We now observe that the 3F2 appearing on the right-hand side of (15) can be evaluated with the help of the classical Watson summation theorem (3), and, finally, we obtain the following Eulerian integral:
0 1 x c 1 ( 1 x ) c 1 F 1 2 a , b 1 2 ( a + b + 1 ) | x d x = Γ ( c ) Γ ( c ) Γ ( 2 c ) W 0 , 0 ( a , b , c )
provided ( c ) > 0 , ( 2 c a b ) > 1 , and W 0 , 0 ( a , b , c ) is the same as given in (3).
Similar results can easily be obtained by employing the following Edwards double integral [8]:
0 1 0 1 ν α 1 ( 1 μ ) α 1 1 ( 1 ν ) α 2 1 ( 1 μ ν ) 1 α 1 α 2 d μ d ν = Γ ( α 1 ) Γ ( α 2 ) Γ ( α 1 + α 2 )
provided ( α 1 ) > 0 and ( α 2 ) > 0 and the following well-known gamma integral
0 e s t t ν 1 d t = Γ ( ν ) s ν
provided ( s ) > 0 and ( ν ) > 0 .
The details are given in the following sections.
The rest of the paper is organized as follows.
In Section 2, eight Eulerian-type single integrals involving a hypergeometric function have been evaluated in terms of a gamma function by using results (5) to (8). Section 3 deals with sixteen Eulerian-type double integrals involving a hypergeometric function and eight Laplace-type integrals using a generalized hypergeometric function are obtained in Section 4. We discuss several interesting special cases of our main results in each section.
We conclude this section by remarking that the integrals (single and multiple) involving hypergeometric functions play an important role in the area of applied mathematics, statistics, engineering, physics, and several other branches. A large number of interesting applications can be seen in the seventh chapter of the standard text of Exton [9].

2. Evaluation of Eulerian-Type Integrals Involving Hypergeometric Function

The eight new classes of Eulerian-type integrals involving generalized hypergeometric functions to be established in this section are asserted in the following theorems.
Theorem 1.
For m , n N 0 , ( α 3 ) > 0 , and ( 2 α 3 α 1 α 2 + m + 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 3 1 ( 1 μ ) α 3 + n 1 2 F 1 α 1 , α 2 α 1 + α 2 + m + 1 2 | μ d μ = Γ ( α 3 ) Γ ( α 3 + n ) Γ ( 2 α 3 + n ) W m , n ( α 1 , α 2 , α 3 ) = Ψ 1
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (5).
Proof. 
To prove the result (18), denoting the left-hand side of (18) by I, expressing 2F1 as a series, changing the order of integration and summation, which is easily seen to be justified due to the uniform convergence of the series in the interval [ 0 , 1 ] , and evaluating the beta integral, we have
I = k = 0 ( α 1 ) k ( α 2 ) k [ 1 2 ( α 1 + α 2 + m + 1 ) ] k k ! Γ ( α 3 + k ) Γ ( α 3 + n ) Γ ( 2 α 3 + k + n )
Using (2) and after some simplification, we have
I = Γ ( α 3 ) Γ ( α 3 + n ) Γ ( 2 α 3 + n ) k = 0 ( α 1 ) k ( α 2 ) k ( α 3 ) k [ 1 2 ( α 1 + α 2 + m + 1 ) ] k ( 2 α 3 + n ) k k !
Summing up the series, we finally obtain
I = Γ ( α 3 ) Γ ( α 3 + n ) Γ ( 2 α 3 + n ) 3 F 2 [ α 1 , α 2 , α 3 1 + α 1 + α 2 + m 2 , 2 α 3 + n | 1 ]
Now, we observe that the 3F2’s appearance can be evaluated with the help of the result (5) and we easily arrive at the right side of (18). This completes the proof of the result (18) asserted in Theorem 1. □
Corollary 1.
In Theorem 1, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 3 1 ( 1 μ ) α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | μ d μ = 2 2 α 3 2 Γ ( α 3 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 α 1 + α 2 1 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 Γ α 1 + 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 + 1 2 = δ 1 .
0 1 μ α 3 1 ( 1 μ ) α 3 1 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | μ d μ = 2 2 α 3 2 Γ ( α 3 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ α 1 2 + 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + 1 2 = δ 2 .
0 1 μ α 3 1 ( 1 μ ) α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | μ d μ = 2 2 α 3 2 Γ ( α 3 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 1 + 2 α 3 α 1 ) Γ ( 1 + 2 α 3 α 2 ) × ( 2 α 3 α 1 + α 2 ) Γ α 1 2 + 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 2 Γ ( α 3 α 2 2 + 1 ) ( 2 α 3 + α 1 α 2 ) Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 2 = δ 3 .
In the same manner, the results and the special cases given in the following theorems and corollaries can be easily established by applying the corresponding summation Formulas (5) to (8). Hence, they are given here without proof.
Theorem 2.
For m , n N 0 , ( α 2 ) > 0 , ( α 1 α 2 + m ) > 0 , and ( 2 α 3 α 1 α 2 + m + 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 + m 1 ) 2 F 1 α 1 , α 3 2 α 3 + n | μ d μ = Γ ( α 2 ) Γ [ 1 2 ( α 1 α 2 + m + 1 ) ] Γ [ 1 2 ( α 1 + α 2 + m + 1 ) ] W m , n ( α 1 , α 2 , α 3 ) = Ψ 2
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (5).
Corollary 2.
In Theorem 2, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 1 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | μ d μ = 2 4 α 3 2 Γ ( 1 + α 1 α 2 2 ) Γ ( α 3 + 1 2 ) Γ α 3 ( α 1 + α 2 1 2 ) π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 Γ α 1 + 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 + 1 2 = δ 4 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 | μ d μ = 2 4 α 3 3 Γ α 3 + 1 2 Γ 1 + α 1 α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 ( α 1 α 2 ) Γ ( α 1 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ α 1 + 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + 1 2 = δ 5 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | μ d μ = 2 4 α 3 2 Γ ( α 3 + 1 2 ) Γ ( 1 + α 1 α 2 2 ) Γ α 3 ( α 1 + α 2 2 ) π 3 2 ( α 1 α 2 ) Γ ( α 1 ) Γ ( 1 + 2 α 3 α 1 ) Γ ( 1 + 2 α 3 α 2 ) × ( 2 α 3 α 1 + α 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ 1 + α 3 α 2 2 ( 2 α 3 + α 1 α 2 ) Γ ( α 1 2 ) Γ 1 + α 2 2 Γ 1 + α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 6 .
Theorem 3.
For m , n N 0 , ( α 3 n ) > 0 , and ( 2 α 3 α 1 α 2 + m 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 3 1 ( 1 μ ) α 3 n 1 2 F 1 α 1 , α 2 α 1 + α 2 + m + 1 2 | μ d μ = Γ ( α 3 ) Γ ( α 3 n ) Γ ( 2 α 3 n ) W m , n ( α 1 , α 2 , α 3 ) = Ψ 3
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (6).
Corollary 3.
In Theorem 3, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 3 1 ( 1 μ ) α 3 2 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | μ d μ = 2 2 α 3 4 Γ ( α 3 1 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 1 + α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2 = δ 7 .
0 1 μ α 3 1 ( 1 μ ) α 3 1 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | μ d μ = 2 2 α 3 2 Γ ( α 3 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ α 1 + 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 2 + 1 Γ α 3 α 1 2 Γ α 3 α 2 2 + 1 2 = δ 8 .
0 1 μ α 3 1 ( 1 μ ) α 3 2 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | μ d μ = 2 2 α 3 3 Γ ( α 3 1 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 + 1 2 Γ α 2 2 Γ α 3 α 1 2 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 2 + 1 Γ α 3 α 1 2 Γ α 3 α 2 2 1 2 = δ 9 .
Theorem 4.
For m , n N 0 , ( α 2 ) > 0 , ( α 1 α 2 + m + 1 ) > 0 , and ( 2 α 3 α 1 α 2 + m 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 + m 1 ) 2 F 1 α 1 , α 3 2 α 3 n | μ d μ = Γ ( α 2 ) Γ [ 1 2 ( α 1 α 2 + m + 1 ) ] Γ [ 1 2 ( α 1 + α 2 + m + 1 ) ] W m , n ( α 1 , α 2 , α 3 ) = Ψ 4
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (6).
Corollary 4.
In Theorem 4, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 1 ) 2 F 1 α 1 , α 3 2 α 3 1 | μ d μ = 2 4 α 3 6 Γ ( α 3 1 2 ) Γ 1 + α 1 α 2 2 Γ α 3 ( 1 + α 1 + α 2 2 ) π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2 = δ 10 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 | μ d μ = 2 4 α 3 3 Γ ( α 3 + 1 2 ) Γ ( 1 + α 1 α 2 2 ) Γ α 3 α 1 + α 2 2 π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ α 1 + 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 11 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 1 | μ d μ = 2 4 α 3 5 Γ ( 1 + α 1 α 2 2 ) Γ ( α 3 1 2 ) Γ α 3 α 1 + α 2 2 π 3 2 ( α 1 α 2 ) Γ ( α 1 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 + 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2 = δ 12 .
Theorem 5.
For m , n N 0 , ( α 3 ) > 0 , and ( 2 α 3 α 1 α 2 m + 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 3 1 ( 1 μ ) α 3 + n 1 2 F 1 α 1 , α 2 α 1 + α 2 m + 1 2 | μ d μ = Γ ( α 3 ) Γ ( α 3 + n ) Γ ( 2 α 3 + n ) W m , n ( α 1 , α 2 , α 3 ) = Ψ 5
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (7).
Corollar 5.
In Theorem 5, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 3 1 ( 1 μ ) α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | μ d μ = 2 2 α 3 2 Γ ( α 3 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 + 1 α 1 α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2 Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2 = δ 13 .
0 1 μ α 3 1 ( 1 μ ) α 3 1 2 F 1 α 1 , α 2 α 1 + α 2 2 | μ d μ = 2 2 α 3 3 Γ ( α 3 ) Γ α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 14 .
0 1 μ α 3 1 ( 1 μ ) α 3 2 F 1 α 1 , α 2 α 1 + α 2 α 2 | μ d μ = 2 2 α 3 2 Γ ( α 3 ) Γ ( α 1 + α 2 2 ) Γ α 3 + 1 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2 = δ 15 .
Theorem 6.
For m , n N 0 , ( α 2 ) > 0 , ( α 1 α 2 m + 1 ) > 0 , and ( 2 α 3 α 1 α 2 m + 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 m 1 ) 2 F 1 α 1 , α 3 2 α 3 + n | μ d μ = Γ ( α 2 ) Γ [ 1 2 ( α 1 α 2 m + 1 ) ] Γ [ 1 2 ( α 1 + α 2 m + 1 ) ] W m , n ( α 1 , α 2 , α 3 ) = Ψ 6
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (7).
Corollary 6.
In Theorem 6, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 1 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | μ d μ = 2 4 α 3 2 Γ ( α 3 + 1 2 ) Γ ( 1 + α 1 α 2 2 ) Γ α 3 + 1 α 1 α 2 2 π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ 1 + α 3 α 1 2 Γ 1 + α 3 α 2 2 Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2 = δ 16 .
0 1 μ α 2 ( 1 μ ) 1 2 ( α 1 α 2 2 ) 2 F 1 α 1 , α 3 2 α 3 | μ d μ = 2 4 α 3 4 Γ ( α 3 + 1 2 ) Γ ( α 1 α 2 2 ) Γ α 3 α 1 + α 2 2 π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 17 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 2 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | μ d μ = 2 4 α 3 2 Γ ( α 3 + 1 2 ) Γ ( α 1 α 2 2 ) Γ 1 + α 3 α 1 + α 2 2 π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ 1 + α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ 1 + α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 18 .
Theorem 7.
For m , n N 0 , ( α 3 n ) > 0 , and ( 2 α 3 α 1 α 2 m 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 3 1 ( 1 μ ) α 3 n 1 2 F 1 α 1 , α 2 α 1 + α 2 m + 1 2 | μ d μ = Γ ( α 3 ) Γ ( α 3 n ) Γ ( 2 α 3 n ) W m , n ( α 1 , α 2 , α 3 ) = Ψ 7
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (8).
Corollary 7.
In Theorem 7, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 3 1 ( 1 μ ) α 3 2 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | μ d μ = 2 2 α 3 4 Γ ( α 3 1 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 1 + α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2 = δ 19 .
0 1 μ α 3 1 ( 1 μ ) α 3 1 2 F 1 α 1 , α 2 α 1 + α 2 2 | μ d μ = 2 2 α 3 3 Γ ( α 3 ) Γ α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 20 .
0 1 μ α 3 1 ( 1 μ ) α 3 2 2 F 1 α 1 , α 2 α 1 + α 2 2 | μ d μ = 2 2 α 3 5 Γ ( α 3 1 ) Γ α 1 + α 2 2 Γ α 3 1 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × ( 2 α 3 α 1 + α 2 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 + ( 2 α 3 + α 1 α 2 2 ) Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2 = δ 21 .
Theorem 8.
For m , n N 0 , ( α 2 ) > 0 , ( α 1 α 2 m + 1 ) > 0 , and ( 2 α 3 α 1 α 2 m 2 n + 1 ) > 0 , the following result holds true:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 m 1 ) 2 F 1 α 1 , α 3 2 α 3 n | μ d μ = Γ ( α 2 ) Γ [ 1 2 ( α 1 α 2 m + 1 ) ] Γ [ 1 2 ( α 1 + α 2 m + 1 ) ] W m , n ( α 1 , α 2 , α 3 ) = Ψ 8
where W m , n ( α 1 , α 2 , α 3 ) is the same as in (8).
Corollary 8.
In Theorem 8, if we take m = 0, n = 1; m = 1, n = 0; and m = n = 1, we respectively obtain the following integrals:
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 1 ) F 2 2 α 1 , α 3 2 α 3 1 | μ d μ = 2 4 α 3 6 Γ ( α 3 1 2 ) Γ 1 + α 1 α 2 2 Γ α 3 ( 1 + α 1 + α 2 2 ) π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2 = δ 22 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 2 ) 2 F 1 α 1 , α 3 2 α 3 | μ d μ = 2 4 α 3 4 Γ ( α 3 + 1 2 ) Γ ( α 1 α 2 2 ) Γ α 3 α 1 + α 2 2 π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2 = δ 23 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 2 ) 2 F 1 α 1 , α 3 2 α 3 1 | μ d μ = 2 4 α 3 7 Γ ( α 3 1 2 ) Γ α 1 α 2 2 Γ α 3 1 α 1 + α 2 2 π 3 2 Γ ( α 1 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × ( 2 α 3 α 1 + α 2 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 + ( 2 α 3 + α 1 α 2 2 ) Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2 = δ 24
Remark 1.
It is interesting to mention here that in Theorem 1, 3, 5, or 7 and in Theorem 2, 4, 6, or 8, if we set m = n = 0 , then we respectively obtain the following interesting integrals:
0 1 μ α 3 1 ( 1 μ ) α 3 1 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | μ d μ = [ Γ ( α 3 ) ] 2 Γ ( 2 α 3 ) W 0 , 0 ( α 1 , α 2 , α 3 ) = δ 25 .
0 1 μ α 2 1 ( 1 μ ) 1 2 ( α 1 α 2 1 ) 2 F 1 α 1 , α 3 2 α 3 | μ d μ = Γ ( α 2 ) Γ [ 1 2 ( α 1 α 2 + 1 ) ] Γ [ 1 2 ( α 1 + α 2 + 1 ) ] W 0 , 0 ( α 1 , α 2 , α 3 ) = δ 26 .

3. Evaluation of Double Integrals Involving Hypergeometric Function

The sixteen new classes of double integrals involving a hypergeometric function to be established in this section are asserted in the following theorems.
Theorem 9.
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 + n 1 ( 1 μ ν ) 1 2 α 3 n 2 F 1 α 1 , α 2 α 1 + α 2 + m + 1 2 | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 1
provided ( α 3 ) > 0 , ( 2 α 3 α 1 α 2 + m + 2 n + 1 ) > 0 , and Ψ 1 is the same as given in (18).
Proof. 
To prove (26), we express the hypergeometric function as a series, change the order of summation and integration, which can be easily justified due to the uniform convergence of the series in the interval [ 0 , 1 ] , and evaluate the double integral with the help of (17); we finally arrive at the right-hand side of (26). □
Corollary 9.
In Theorem 9, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 [ ν ( 1 ν ) ] α 3 ( 1 μ ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 1
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 2
0 1 0 1 [ ν ( 1 ν ) ] α 3 ( 1 μ ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 3
where δ 1 , δ 2 , and δ 3 are the same as given in Corollary 1.
Theorem 10.
0 1 0 1 ν α 3 + n ( 1 μ ) α 3 + n 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 n 2 F 1 α 1 , α 2 α 1 + α 2 + m + 1 2 | 1 ν 1 μ ν d μ d ν = Ψ 1
provided ( α 3 ) > 0 , ( 2 α 3 α 1 α 2 + m + 2 n + 1 ) > 0 , and Ψ 1 is the same as given in (18).
Corollary 10.
In Theorem 2, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 3 + 1 ( 1 μ ) α 3 ( 1 ν ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | 1 ν 1 μ ν d μ d ν = δ 1
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 μ 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | 1 ν 1 μ ν d μ d ν = δ 2
0 1 0 1 ν α 3 + 1 ( 1 μ ) α 3 ( 1 ν ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 2 2 | 1 ν 1 μ ν d μ d ν = δ 3
where δ 1 , δ 2 , and δ 3 are the same as given in Corollary 1.
Theorem 11.
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 + m 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 m ) 2 F 1 α 1 , α 3 2 α 3 + n | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 2
provided ( α 2 ) > 0 , ( α 1 α 2 + m ) > 0 , ( 2 α 3 α 1 α 2 + m + 2 n + 1 ) > 0 , and Ψ 2 is the same as given in (19).
Corollary 11.
In Theorem 11, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 4
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 ) ( 1 μ ν ) 1 2 ( α 1 + α 2 ) 2 F 1 α 1 , α 3 2 α 3 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 5
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 ) ( 1 μ ν ) 1 2 ( α 1 + α 2 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 6
where δ 4 , δ 5 , and δ 6 are the same as given in Corollary 2.
Theorem 12.
0 1 0 1 ν 1 2 ( α 1 α 2 + m + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 + m 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 m ) × 2 F 1 α 1 , α 3 2 α 3 + n | 1 ν 1 μ ν d μ d ν = Ψ 2
provided ( α 2 ) > 0 , ( α 1 α 2 + m ) > 0 , ( 2 α 3 α 1 α 2 + m + 2 n + 1 ) > 0 , and Ψ 2 is the same as given in (19).
Corollary 12.
In Theorem 12, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν 1 2 ( α 1 α 2 + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 + 1 | 1 ν 1 μ ν d μ d ν = δ 4
0 1 0 1 ν 1 2 ( α 1 α 2 + 2 ) ( 1 μ ) 1 2 ( α 1 α 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( α 1 + α 2 ) × 2 F 1 α 1 , α 3 2 α 3 | 1 ν 1 μ ν d μ d ν = δ 5
0 1 0 1 ν 1 2 ( α 1 α 2 + 2 ) ( 1 μ ) 1 2 ( α 1 α 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( α 1 + α 2 ) × 2 F 1 α 1 , α 3 2 α 3 + 1 | 1 ν 1 μ ν d μ d ν = δ 6
where δ 4 , δ 5 , and δ 6 are the same as in Corollary 2.
Theorem 13.
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 n 1 ( 1 μ ν ) 1 2 α 3 + n × 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + m + 1 ) | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 3
provided ( α 3 n ) > 0 , ( 2 α 3 α 1 α 2 + m 2 n + 1 ) > 0 , and Ψ 3 is the same as given in (20).
Corollary 13.
In Theorem 13, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 2 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) | ν ( 1 μ ) 1 μ ν d μ d ν = δ 7
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) | ν ( 1 μ ) 1 μ ν d μ d ν = δ 8
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 2 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) | ν ( 1 μ ) 1 μ ν d μ d ν = δ 9
where δ 7 , δ 8 , and δ 9 are the same as given in Corollary 3.
Theorem 14.
0 1 0 1 ν α 3 n ( 1 μ ) α 3 n 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 + n × 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + m + 1 ) | 1 ν 1 μ ν d μ d ν = Ψ 3
provided ( α 3 n ) > 0 , ( 2 α 3 α 1 α 2 + m 2 n + 1 ) > 0 , and Ψ 3 is the same as given in (20).
Corollary 14.
In Theorem 14, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 3 1 ( 1 μ ) α 3 2 ( 1 ν ) α 3 1 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) | 1 ν 1 μ ν d μ d ν = δ 7
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) | 1 ν 1 μ ν d μ d ν = δ 8
0 1 0 1 ν α 3 1 ( 1 μ ) α 3 2 ( 1 ν ) α 3 1 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) | 1 ν 1 μ ν d μ d ν = δ 9
where δ 7 , δ 8 , and δ 9 are the same as given in Corollary 3.
Theorem 15.
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 + m 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 m ) × 2 F 1 α 1 , α 3 2 α 3 n | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 4
provided ( α 2 ) > 0 , ( α 1 α 2 + m + 1 ) > 0 , ( 2 α 3 α 1 α 2 + m 2 n + 1 ) > 0 , and Ψ 4 is the same as given in (21).
Corollary 15.
In Theorem 15, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 10
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 ) ( 1 μ ν ) 1 2 ( α 1 + α 2 ) 2 F 1 α 1 , α 3 2 α 3 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 11
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 ) ( 1 μ ν ) 1 2 ( α 1 + α 2 ) 2 F 1 α 1 , α 3 2 α 3 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 12
where δ 10 , δ 11 , and δ 12 are the same as given in Corollary 4.
Theorem 16.
0 1 0 1 ν 1 2 ( α 1 α 2 + m + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 + m 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 m ) × F 1 2 α 1 , α 3 2 α 3 n | 1 ν 1 μ ν d μ d ν = Ψ 4
provided ( α 2 ) > 0 , ( α 1 α 2 + m + 1 ) > 0 , ( 2 α 3 α 1 α 2 + m 2 n + 1 ) > 0 , and Ψ 4 is the same as given in (21).
Corollary 16.
In Theorem 16, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν 1 2 ( α 1 α 2 + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 1 | 1 ν 1 μ ν d μ d ν = δ 10
0 1 0 1 ν 1 2 ( α 1 α 2 + 2 ) ( 1 μ ) 1 2 ( α 1 α 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( α 1 + α 2 ) × 2 F 1 α 1 , α 3 2 α 3 | 1 ν 1 μ ν d μ d ν = δ 11
0 1 0 1 ν 1 2 ( α 1 α 2 + 2 ) ( 1 μ ) 1 2 ( α 1 α 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( α 1 + α 2 ) × 2 F 1 α 1 , α 3 2 α 3 1 | 1 ν 1 μ ν d μ d ν = δ 12
where δ 10 , δ 11 , and δ 12 are the same as given in Corollary 4.
Theorem 17.
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 + n 1 ( 1 μ ν ) 1 2 α 3 n 2 F 1 α 1 , α 2 α 1 + α 2 m + 1 2 | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 5
provided ( α 3 ) > 0 , ( 2 α 3 α 1 α 2 m + 2 n + 1 ) > 0 , and Ψ 5 is the same as given in (22).
Corollary 17.
In Theorem 17, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 [ ν ( 1 ν ) ] α 3 ( 1 μ ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 13
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 14
0 1 0 1 [ ν ( 1 ν ) ] α 3 ( 1 μ ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 15
where δ 13 , δ 14 , and δ 15 are the same as given in Corollary 5.
Theorem 18.
0 1 0 1 ν α 3 + n ( 1 μ ) α 3 + n 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 n 2 F 1 α 1 , α 2 α 1 + α 2 m + 1 2 | 1 ν 1 μ ν d μ d ν = Ψ 5
provided ( α 3 ) > 0 , ( 2 α 3 α 1 α 2 m + 2 n + 1 ) > 0 , and Ψ 5 is the same as given in (22).
Corollary 18.
In Theorem 18, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 3 + 1 ( 1 μ ) α 3 ( 1 ν ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | 1 ν 1 μ ν d μ d ν = δ 13
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | 1 ν 1 μ ν d μ d ν = δ 14
0 1 0 1 ν α 3 + 1 ( 1 μ ) α 3 ( 1 ν ) α 3 1 ( 1 μ ν ) 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | 1 ν 1 μ ν d μ d ν = δ 15
where δ 13 , δ 14 , and δ 15 are the same as given in Corollary 5.
Theorem 19.
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 m 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 + m ) × 2 F 1 α 1 , α 3 2 α 3 + n | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 6
provided ( α 2 ) > 0 , ( α 1 α 2 m + 1 ) > 0 , ( 2 α 3 α 1 α 2 m + 2 n + 1 ) > 0 , and Ψ 6 is the same as given in (23).
Corollary 19.
In Theorem 19, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 16
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 2 ) ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 17
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 2 ) ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 + 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 18
where δ 16 , δ 17 , and δ 19 are the same as given in Corollary 6.
Theorem 20.
0 1 0 1 ν 1 2 ( α 1 α 2 m + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 m 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 + m ) × F 1 2 α 1 , α 3 2 α 3 + n | 1 ν 1 μ ν d μ d ν = Ψ 6
provided ( α 2 ) > 0 , ( α 1 α 2 m + 1 ) > 0 , ( 2 α 3 α 1 α 2 m + 2 n + 1 ) > 0 , and Ψ 6 is the same as given in (23).
Corollary 20.
In Theorem 20, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν 1 2 ( α 1 α 2 + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 + 1 | 1 ν 1 μ ν d μ d ν = δ 16
0 1 0 1 ν 1 2 ( α 1 α 2 ) ( 1 μ ) 1 2 ( α 1 α 2 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 | 1 ν 1 μ ν d μ d ν = δ 17
0 1 0 1 ν 1 2 ( α 1 α 2 ) ( 1 μ ) 1 2 ( α 1 α 2 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 + 1 | 1 ν 1 μ ν d μ d ν = δ 18
where δ 16 , δ 17 , and δ 19 are the same as given in Corollary 6.
Theorem 21.
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 n 1 ( 1 μ ν ) 1 2 α 3 + n 2 F 1 α 1 , α 2 α 1 + α 2 m + 1 2 | 1 ν 1 μ ν d μ d ν = Ψ 7
provided ( α 3 n ) > 0 , ( 2 α 3 α 1 α 2 m 2 n + 1 ) > 0 , and Ψ 7 is the same as given in (24).
Corollary 21.
In Theorem 21, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 2 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | 1 ν 1 μ ν d μ d ν = δ 19
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | 1 ν 1 μ ν d μ d ν = δ 20
0 1 0 1 ν α 3 ( 1 μ ) α 3 1 ( 1 ν ) α 3 2 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | 1 ν 1 μ ν d μ d ν = δ 21
where δ 19 , δ 20 , and δ 21 are the same as given in Corollary 7.
Theorem 22.
0 1 0 1 ν α 3 n ( 1 μ ) α 3 n 1 ( 1 ν ) α 3 1 ( 1 μ ν ) 1 2 α 3 + n × 2 F 1 α 1 , α 2 α 1 + α 2 m + 1 2 | 1 ν 1 μ ν d μ d ν = Ψ 7
provided ( α 3 n ) > 0 , ( 2 α 3 α 1 α 2 m 2 n + 1 ) > 0 , and Ψ 7 is the same as given in (24).
Corollary 22.
In Theorem 22, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 [ ν ( 1 ν ) ] α 3 1 ( 1 μ ) α 3 2 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | 1 ν 1 μ ν d μ d ν = δ 19
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | 1 ν 1 μ ν d μ d ν = δ 20
0 1 0 1 [ ν ( 1 ν ) ] α 3 1 ( 1 μ ) α 3 2 ( 1 μ ν ) 2 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 2 | 1 ν 1 μ ν d μ d ν = δ 21
where δ 19 , δ 20 , and δ 21 are the same as given in Corollary 7.
Theorem 23.
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 m 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 + m ) × 2 F 1 α 1 , α 3 2 α 3 n | ν ( 1 μ ) 1 μ ν d μ d ν = Ψ 8
provided ( α 2 ) > 0 , ( α 1 α 2 m + 1 ) > 0 , ( 2 α 3 α 1 α 2 m 2 n + 1 ) > 0 , and Ψ 8 is the same as given in (25).
Corollary 23.
In Theorem 23, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 22
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 2 ) ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 23
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 2 ) ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 1 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 24
where δ 22 , δ 23 , and δ 24 are the same as given in Corollary 8.
Theorem 24.
0 1 0 1 ν 1 2 ( α 1 α 2 m + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 m 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 + m ) × F 1 2 α 1 , α 3 2 α 3 n | 1 ν 1 μ ν d μ d ν = Ψ 8
provided ( α 2 ) > 0 , ( α 1 α 2 m + 1 ) > 0 , ( 2 α 3 α 1 α 2 m 2 n + 1 ) > 0 , and Ψ 8 is the same as given in (25).
Corollary 24.
In Theorem 24, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = n = 1 , then we respectively obtain the following double integrals:
0 1 0 1 ν 1 2 ( α 1 α 2 + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 1 | 1 ν 1 μ ν d μ d ν = δ 22
0 1 0 1 ν 1 2 ( α 1 α 2 ) ( 1 μ ) 1 2 ( α 1 α 2 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 | 1 ν 1 μ ν d μ d ν = δ 23
0 1 0 1 ν 1 2 ( α 1 α 2 ) ( 1 μ ) 1 2 ( α 1 α 2 2 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 2 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 1 | 1 ν 1 μ ν d μ d ν = δ 24
where δ 22 , δ 23 , and δ 24 are the same as given in Corollary 8.
Remark 2.
It is interesting to mention here that in Theorem 9, 13, 17 or 21, in Theorem 10, 14, 18 or 22, in Theorem 11, 15, 19, or 23, and in Theorem 12, 16, 20, or 24, if we set m = n = 0 , then we respectively obtain the following interesting double integrals:
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 25
where δ 25 is the same as given in Corollary 9.
0 1 0 1 ν α 3 [ ( 1 μ ) ( 1 ν ) ] α 3 1 ( 1 μ ν ) 1 2 α 3 2 F 1 α 1 , α 2 α 1 + α 2 + 1 2 | 1 ν 1 μ ν d μ d ν = δ 25
where δ 25 is the same as given in Corollary 9.
0 1 0 1 ν α 2 ( 1 μ ) α 2 1 ( 1 ν ) 1 2 ( α 1 α 2 1 ) ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) 2 F 1 α 1 , α 3 2 α 3 | ν ( 1 μ ) 1 μ ν d μ d ν = δ 26
where δ 26 is the same as given in Corollary 10.
0 1 0 1 ν 1 2 ( α 1 α 2 + 1 ) ( 1 μ ) 1 2 ( α 1 α 2 1 ) ( 1 ν ) α 2 1 ( 1 μ ν ) 1 2 ( 1 α 1 α 2 ) × 2 F 1 α 1 , α 3 2 α 3 | 1 ν 1 μ ν d μ d ν = δ 26
where δ 26 is the same as given in Corollary 10.
The proof of Theorems 10 to 24 is similar to that of the Theorem 9 and, hence, it is omitted here.

4. Evaluation of Laplace-Type Integrals Involving Hypergeometric Function

The eight new classes of Laplace- type integrals involving a hypergeometric function to be established in this section are asserted in the following theorems.
Theorem 25.
For m , n N , ( s ) > 0 , and ( α 1 ) > 0 , the following result holds true:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + m + 1 ) , 2 α 3 + n | s t d t = Γ ( α 1 ) s α 1 W m , n ( α 1 , α 2 , α 3 ) .
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (5).
Proof. 
To derive the result given in the Theorem 25, we express F 2 2 as a series, change the order of integration and summation, and then evaluate the integral. We finally arrive at the right-hand side of (42). □
Corollary 25.
In Theorem 25, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 + 1 | s t d t = 2 4 α 3 2 s α 1 Γ ( α 3 + 1 2 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 + 1 α 1 α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 Γ α 1 + 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 + 1 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 | s t d t = 2 4 α 3 3 s α 1 Γ ( α 3 + 1 2 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 ( α 1 α 2 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ α 1 2 + 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + 1 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 + 1 | s t d t = 2 4 α 3 2 s α 1 Γ ( α 3 + 1 2 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 ( α 1 α 2 ) Γ ( α 2 ) Γ ( 1 + 2 α 3 α 1 ) Γ ( 1 + 2 α 3 α 2 ) × ( 2 α 3 α 1 + α 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ ( 1 + α 3 α 2 2 ) ( 2 α 3 + α 1 α 2 ) Γ α 1 2 Γ 1 + α 2 2 Γ 1 + α 3 α 1 2 Γ α 3 + 1 α 2 2
In exactly the same manner, the integrals given in the following Theorems 26, 27, 28, 29, 30, 31, and 32 can be evaluated; thus, they are given here without proof.
Theorem 26.
For m , n N , ( s ) > 0 , and ( α 3 ) > 0 , the following result holds true:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + m + 1 ) , 2 α 3 + n | s t d t = Γ ( α 3 ) s α 3 W m , n ( α 1 , α 2 , α 3 ) .
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (5).
Corollary 26.
In Theorem 26, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 + 1 | s t d t = 2 2 α 3 1 s α 3 Γ ( 2 α 3 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 + 1 α 1 α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 Γ α 1 + 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 + 1 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 | s t d t = 2 2 α 3 2 s α 3 Γ ( 2 α 3 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 + 1 | s t d t = 2 2 α 3 1 s α 3 Γ ( 2 α 3 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 1 + 2 α 3 α 1 ) Γ ( 1 + 2 α 3 α 2 ) × ( 2 α 3 α 1 + α 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ ( 1 + α 3 α 2 2 ) ( 2 α 3 + α 1 α 2 ) Γ α 1 2 Γ 1 + α 2 2 Γ 1 + α 3 α 1 2 Γ α 3 + 1 α 2 2
Theorem 27.
For m , n N , ( s ) > 0 , and ( α 1 ) > 0 the following result holds true:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + m + 1 ) , 2 α 3 n | s t d t = Γ ( α 1 ) s α 1 W m , n ( α 1 , α 2 , α 3 )
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (6).
Corollary 27.
In Theorem 27, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 1 | s t d t = 2 4 α 3 6 s α 1 Γ ( α 3 1 2 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 1 + α 1 + α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 | s t d t = 2 4 α 3 3 s α 1 Γ ( α 3 + 1 2 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 ( α 1 α 2 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ α 1 2 + 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + 1 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 1 | s t d t = 2 4 α 3 5 s α 1 Γ ( α 3 1 2 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 ( α 1 α 2 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2
Theorem 28.
For m , n N , ( s ) > 0 , and ( α 3 ) > 0 , the following result holds true:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + m + 1 ) , 2 α 3 n | s t d t = Γ ( α 3 ) s α 3 W m , n ( α 1 , α 2 , α 3 )
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (6).
Corollary 28.
In Theorem 28, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 1 | s t d t = 2 2 α 3 4 s α 3 Γ ( 2 α 3 1 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 1 + α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 | s t d t = 2 2 α 3 2 s α 3 Γ ( 2 α 3 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 2 ) , 2 α 3 1 | s t d t = 2 2 α 3 3 s α 3 Γ ( 2 α 3 1 ) Γ 1 + α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π ( α 1 α 2 ) Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2
Theorem 29.
For m , n N , ( s ) > 0 , and ( α 1 ) > 0 the following result holds true:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 m + 1 ) , 2 α 3 + n | s t d t = Γ ( α 1 ) s α 1 W m , n ( α 1 , α 2 , α 3 )
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (7).
Corollary 29.
In Theorem 29, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 + 1 | s t d t = 2 4 α 3 2 s α 1 Γ ( α 3 + 1 2 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 + 1 α 1 α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 Γ α 1 + 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 + 1 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 ) , 2 α 3 | s t d t = 2 4 α 3 4 s α 1 Γ ( α 3 + 1 2 ) Γ α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 ) , 2 α 3 + 1 | s t d t = 2 4 α 3 2 s α 1 Γ ( α 3 + 1 2 ) Γ ( α 1 + α 2 2 ) Γ α 3 + 1 α 1 + α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 + 1 α 2 2
Theorem 30.
For m , n N , ( s ) > 0 , and ( α 3 ) > 0 , the following result holds true:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 m + 1 ) , 2 α 3 + n | s t d t = Γ ( α 3 ) s α 3 W m , n ( α 1 , α 2 , α 3 )
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (7).
Corollary 30.
In Theorem 30, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 + 1 | s t d t = 2 2 α 3 1 s α 3 Γ ( 2 α 3 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 + 1 α 1 α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 + 1 Γ α 3 α 2 2 + 1 Γ α 1 + 1 2 Γ α 2 + 1 2 Γ α 3 α 1 2 + 1 2 Γ α 3 α 2 2 + 1 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 ) , 2 α 3 | s t d t = 2 2 α 3 3 s α 3 Γ ( 2 α 3 ) Γ α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 ) , 2 α 3 + 1 | s t d t = 2 2 α 3 1 s α 3 Γ ( 2 α 3 ) Γ ( α 1 + α 2 2 ) Γ 1 + α 3 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 + 1 ) Γ ( 2 α 3 α 2 + 1 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ 1 + α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ 1 + α 3 α 1 2 Γ α 3 + 1 α 2 2
Theorem 31.
For m , n N , ( s ) > 0 , and ( α 1 ) > 0 the following result holds true:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 m + 1 ) , 2 α 3 n | s t d t = Γ ( α 1 ) s α 1 W m , n ( α 1 , α 2 , α 3 )
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (8).
Corollary 31.
In Theorem 31, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 1 | s t d t = 2 4 α 3 6 s α 1 Γ ( α 3 1 2 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 1 + α 1 + α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 ) , 2 α 3 | s t d t = 2 4 α 3 4 s α 1 Γ ( α 3 + 1 2 ) Γ α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 ) , 2 α 3 1 | s t d t = 2 4 α 3 7 s α 1 Γ ( α 3 1 2 ) Γ α 1 + α 2 2 Γ α 3 1 α 1 + α 2 2 π 3 2 Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × ( 2 α 3 α 1 + α 2 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 ( 2 α 3 + α 1 α 2 2 ) Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2
Theorem 32.
For m , n N , ( s ) > 0 , and ( α 3 ) > 0 , the following result holds true:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 m + 1 ) , 2 α 3 n | s t d t = Γ ( α 3 ) s α 3 W m , n ( α 1 , α 2 , α 3 )
where W m , n ( α 1 , α 2 , α 3 ) is the same as given in (8).
Corollary 32.
In Theorem 32, if we take m = 0 , n = 1 ; m = 1 , n = 0 ; and m = 1 , n = 1 , we obtain the following integrals:
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 1 | s t d t = 2 2 α 3 4 s α 3 Γ ( 2 α 3 1 ) Γ ( 1 + α 1 + α 2 2 ) Γ α 3 1 + α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × Γ α 1 2 Γ α 2 2 Γ α 3 α 1 2 Γ α 3 α 2 2 + Γ 1 + α 1 2 Γ 1 + α 2 2 Γ α 3 1 + α 1 2 Γ α 3 1 + α 2 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 ) , 2 α 3 | s t d t = 2 2 α 3 3 s α 3 Γ ( 2 α 3 ) Γ α 1 + α 2 2 Γ α 3 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 ) Γ ( 2 α 3 α 2 ) × Γ 1 + α 1 2 Γ α 2 2 Γ α 3 + 1 α 1 2 Γ α 3 α 2 2 + Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 + 1 α 2 2
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 ) , 2 α 3 1 | s t d t = 2 2 α 3 5 s α 3 Γ ( 2 α 3 1 ) Γ α 1 + α 2 2 Γ α 3 1 α 1 + α 2 2 π Γ ( α 1 ) Γ ( α 2 ) Γ ( 2 α 3 α 1 1 ) Γ ( 2 α 3 α 2 1 ) × ( 2 α 3 α 1 + α 2 2 ) Γ 1 + α 1 2 Γ α 2 2 Γ α 3 1 + α 1 2 Γ α 3 α 2 2 + ( 2 α 3 + α 1 α 2 2 ) Γ α 1 2 Γ 1 + α 2 2 Γ α 3 α 1 2 Γ α 3 1 + α 2 2
Remark 3.
It is interesting to mention here that in Theorem 25, 27, 29, or 31 and in Theorem 26, 28, 30, or 32, if we set m = n = 0 , we respectively obtain the following interesting integrals:
0 e s t t α 1 1 F 2 2 α 2 , α 3 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 | s t d t = Γ ( α 1 ) s α 1 W 0 , 0 ( α 1 , α 2 , α 3 )
where W 0 , 0 ( α 1 , α 2 , α 3 ) is given in (3).
0 e s t t α 3 1 F 2 2 α 1 , α 2 1 2 ( α 1 + α 2 + 1 ) , 2 α 3 | s t d t = Γ ( α 3 ) s α 3 W 0 , 0 ( α 1 , α 2 , α 3 )
where W 0 , 0 ( α 1 , α 2 , α 3 ) is given in (3).

5. Conclusions

In this paper, we have evaluated a new class of Eulerian-type integrals (single and double) and Laplace -type integrals involving a hypergeometric function by employing four generalizations of the classical Watson summation theorem with two integer parameters discovered by Chu. Several interesting special cases have also been given. Hypergeometric integrals are widely recognized for their significant role in various fields, including statistics, mathematical physics, and quantum chemistry. In statistics, they provide the foundation for representing probability distributions such as the beta and gamma distributions, along with density and cumulative distribution functions. In the realm of physics, these integrals appear in the study of elastic plate vibrations, heat conduction within cylinders, viscous fluid motion, and certain electrical networks. They also contribute to the analysis of heat flow in solids within conducting media and the exploration of specific non-linear oscillation phenomena. Quantum chemistry employs hypergeometric integrals in molecular calculations, while communications engineering utilizes them in contexts like Gaussian noise analysis. Additional applications include their use in electrical impedance theory and combinatorial analysis. Since all the results obtained in this paper are given in terms of the gamma function, they may be potentially useful in applied mathematics, engineering mathematics, and mathematical physics.
We conclude the paper by remarking that, following the same lines, Eulerian-type integrals (single and double) and Laplace -type integrals by employing generalizations of Dixon’s and Whipple’s summation theorems are under investigation and we will form two subsequent papers in this direction.

Author Contributions

Writing—original draft, P.J., D.L. and A.K.R.; Writing— review & editing, D.L., A.K.R. and A.K. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by a grant from 2023 Research Fund of Andong National University.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare that they have no conflicts of competing interests.

Abbreviations

The following abbreviations are used in this manuscript:
MDPIMultidisciplinary Digital Publishing Institute
DOAJDirectory of open access journals
TLAThree letter acronym
LDLinear dichroism

References

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Jayarama, P.; Lim, D.; Rathie, A.K.; Kilicman, A. Evaluation of Certain Definite Integrals Involving Generalized Hypergeometric Functions. Axioms 2024, 13, 887. https://doi.org/10.3390/axioms13120887

AMA Style

Jayarama P, Lim D, Rathie AK, Kilicman A. Evaluation of Certain Definite Integrals Involving Generalized Hypergeometric Functions. Axioms. 2024; 13(12):887. https://doi.org/10.3390/axioms13120887

Chicago/Turabian Style

Jayarama, Prathima, Dongkyu Lim, Arjun K. Rathie, and Adem Kilicman. 2024. "Evaluation of Certain Definite Integrals Involving Generalized Hypergeometric Functions" Axioms 13, no. 12: 887. https://doi.org/10.3390/axioms13120887

APA Style

Jayarama, P., Lim, D., Rathie, A. K., & Kilicman, A. (2024). Evaluation of Certain Definite Integrals Involving Generalized Hypergeometric Functions. Axioms, 13(12), 887. https://doi.org/10.3390/axioms13120887

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