On Solution Of Some Double Integrals Involving H-Function

On Solution Of Some Double Integrals Involving H-Function

By

S. N. Singh* and Raj Mehta**

1. Abstract

   In this paper, we shall establish some double integrals involving H-Function.

   Keyword Argument, contour. Introduction:

   Integrals are useful in the study of certain boundary value problems. It is also helpful in obtaining the expansion formulae. Integrals are also used in the study of statistical distribution, probability and integral equation. Here, we have developed and solved integrals involving hyper-geometric functions.

   Gauss hyper geometric function 2F1 [a, b; c; z] has been generalized by the introduction of p parameters of nature of a, b and q parameters of the nature of c.

   This ensuring series

   a1, , ap; (a1)n (ap)n zn pFq z =

   b1, , bq; n = 0 (b1)n (bq)n n!

   p

   (ai)n zn

   =

   q

   n = 0

   (bj)n n!

   is known as the generalized hypergeometric series and the function pFq is called generalized hypergeometric function of variable z. pFq is not defined if any denominator parameter bq is a negative integer or zero. If any numerator parameter ap is zero or a negative integer, the series terminates. If pFq does not terminate, it converges

   1. for all finite z if p q;

   2. for |z| < 1 if p = q + 1;

      j=1

      j=1

   3. for |z| = 1 if p = q + 1 and R q

j=1

j=1

bj p

aj > 0

and diverges for all z 0 if p > q + 1.

The class of the hypergeometric series and functions considered above are of single variable. The great success of the theory of hypergeometric series in one variable has stimulated the development of corresponding theory in two and more than two variables.

In this section, we will establish the following double integrals:

/2 /2 2 + +1 ei x (cosx)+ ei + y (siny)1(cosy)1 ×

0 0

p,q

p,q

× Hm,n [z 2ei x+y cosx siny 2ei yx cosx cosy ]dxdy

i

= e 2

i

Hm,n+3 [ze 2

| ,+ , 1, , 1, ,(aj ,j )1,p ], (1)

2 + +1

p+3,q+3 2 +

bj ,j 1,q , 1,+ ,(,),(,),

Provided that Re ( + ) > -1, Re() > 0, Re() > 0, 0 and 0, |arg z| <

½ A , where A is given in equation

n p m q

i i + i i A > 0,

j=1

j=n+1

j=1

j=m+1

p q

i i < 0

j=1 j=1

t /2 2 + +1 x1(t x)1ei y (cosy)+ ×

0 0

p,q

p,q

× Hm,n [z 2xeiy cosy 2(t x)eiy cosy ]dxdy

= t +1 Hm,n+3 [z(t/2)+ | ,+ , 1, , 1, ,(aj,j )1,p ] ,

j j

j j

2 + +1

p+3,q+3

b ,

1,q

, 1,+ ,(,),(,),

(2)

Where Re ( + ) > – 1, Re() > 0, Re() > 0, 0 and 0, |arg z| < ½ A

, where A is given in equation

n p m q

i i + i i A > 0,

j=1

j=n+1

j=1

j=m+1

p q

i i < 0

j=1 j=1

/2 2 + +1 x1exei y (cosy)+ ×

0 0

p,q

p,q

× Hm,n [z 2xeiy cosy 2eiy cosy ]dxdy

= Hm,n+2 [ z

| ,+ , 1, ,(aj,j )1,p ], (3)

2 + +1 p+2,q+2 2 +

bj ,j 1,q ,(,),(,),

Provided that Re ( + ) > -1, Re() > 0, 0 and 0, |arg z| < ½ A, where A is given in equation

n p m q

i i + i i A > 0,

j=1

j=n+1

j=1

j=m+1

p q

i i < 0

j=1 j=1

t /2 x1 (t x)1ei + y (siny)1(cosy)1 ×

0 0

p,q

p,q

× Hm,n [z 2xeiy siny (t x)eiy cosy ]dxdy

= e t+1Hm,n+4 [zt+ e | ]

= e t+1Hm,n+4 [zt+ e | ]

i i 1, , 1, , 1, , 1, ,(aj ,j )1,p

2 p+4,q+3

2

b ,

, 1,+ ,(1,+)

, (4)

j j 1,q

where Re () > 0, Re () > 0, Re() > 0, Re() > 0, 0 and 0, |arg z| < ½

A , where A is given in equation

n p m q

i i + i i A > 0,

j=1

j=n+1

j=1

j=m+1

p q

i i < 0

j=1 j=1

/2 x1 exei + y (siny)1(cosy)1 ×

0 0

p,q

p,q

× Hm,n [z xeiy siny eiy cosy ]dxdy

i

m,n+3

i

1, , 1, , 1, (aj,j )1,p

= e 2 Hp+3,q+1[ze 2 | b ,

,(,+)

], (5)

j j 1,q

provided that Re () > 0, Re () > 0, Re() > 0, 0 and 0, |arg z| < ½

A, where A is given in equation

n p m q

i i + i i A > 0,

j=1

j=n+1

j=1

j=m+1

p q

i i < 0

j=1 j=1

Proof: To prove (1), we express the H-function of one variable on the left hand side as contour integral. Then interchanging the order of integration (which is justifiable due to given condition), we get

I = 1

2i

L (s) zs ×

2

2

× [ 2 + s + + s +1 eix +s +s cosx +s + +s dx]

0

0

0

× [ 2 eiy +s + +s siny +s 1 cosy +s 1dy]ds

Now using the results (5), (6) and interpreting it with the help of

, 1,

, 1,

H, | = 1 5 i = -1,

,

, 1,

2

1 +

=

1 +

we get right hand side of (1).

Proceeding as above we can prove (2) to (5) with the help of A, B, C,

and D

/2 2 + +1 ei() (cos)+ d = (++1) ,

0

2 + +1(+1)(+1)

(A)

Re( + ) > 1.

/2 0

ei(+) (sin)1(cos)1d = e i /2 ()(),

(B)

(B)

(+)

Re() > 0, () > 0.

0

0

x1exdx = ,

(C)

t x1(t x)1dx = t+1 () ,

(D)

0 (+)

Re() > 0, > 0.

PARTICULAR CASE:

Collecting real and imaginary part of (4), we get

t /2 x1 (t x)1 cos + y (siny)1(cosy)1 ×

0 0

p,q

p,q

× Hm,n [z 2xeiy siny (t x)eiy cosy ]dxdy

= cos t+1Hm,n+4 [zt+ ei | 1, , 1, , 1, , 1, ,(aj ,j )1,p ]

2 p+4,q+3

2

b ,

, 1,+ ,(1,+)

(6)

j j 1,q

and

t /2 x1 (t x)1 sin + y (siny)1(cosy)1 ×

0 0

p,q

p,q

× Hm,n [z 2xeiy siny (t x)eiy cosy ]dxdy

= sin t+1Hm,n+4 [zt+ ei | 1, , 1, , 1, , 1, ,(aj ,j )1,p ]

2 p+4,q+3

2

b ,

, 1,+ ,(1,+)

(7)

j j 1,q

INTEGRAL REPRESENTATION:

x1 ai ;A 1 ,,A r

: 11,1 , c 1 ,C 1

: 1r ,1 , c r ,C r

H0,n:m 1 ,n1+1: mr ,nr +1 | i

i 1 n

i i 1 .01

i i 1 .0r

p,q:p1+1,q1 :pr +1,qr

x bi ;B 1 ,,B r :, d 1 ,D 1

: d r ,CD r

r i i 1 n i

i 1 .01

i i 1 .0r

1 1

1

1 1 1

=1

0

0 =1

x1t1

ai ;A 1 ,,A r

: 11 ,1 , c 1 ,C 1

: 1r ,1 , c r ,C r

H0,n:m 1 ,n1+1: mr ,nr +1 | i

i 1 n

i i 1 .01

i i 1 .0r

p,q:p1+1,q1 :pr +1,qr

x t bi ;B 1 ,,B r

:, d 1 ,D 1

: d r ,CD r

r r i

i 1 n i

i 1 .01

p>i i 1 .0r

dt1 . dtr (8)

Proof:

We consider

1 1 + +

1

1

0,:

,

+1:: , +1

1 1

,: 1

1

+1,

1 1

1

1 +1,

x1t1

ai ;A 1 ,,A r

: 11 ,1 , c 1 ,C 1

: 1r ,1 , c r ,C r

| i

i 1 n

i i 1 .01

i i 1 .0r

x t bi ;B 1 ,,B r

:, d 1 ,D 1

: d r ,D r

r r i

i 1 n i

i 1 .01

i i 1 .0r

= 1 , ,

+

2

1

=1

11 + +

+ 1

+ 1

1

1

1

1

1 1 1

1

1

r

xi 1 H0,n:m 1 ,n1+1: mr ,nr +1

i

i=1

p,q:p1+1,q1 :pr +1,qr

x1 ai ;A 1 ,,A r

| i i 1 n

: 11 ,1 , c 1 ,C 1

i i

i i

1 .01

: 1r ,1 , c r ,C r

i i 1 .0r

x bi ;B 1 ,,B r :, d 1 ,D 1

: d r ,CD r

r i i 1 n i

i 1 .01

i i 1 .0r

which on using

v U

xa v

= 1

v

x x y v1U y dy

(E)

a

a

gives

x1 ai ;A 1 ,,A r

: 11,1 , c 1 ,C 1

: 1r ,1 , c r ,C r

H0,n:m 1 ,n1+1: mr ,nr +1 | i

i 1,p

i i 1 .p 1

i i 1 .pr

p,q:p1+1,q1 :pr +1,qr

x bi ;B 1 ,,B r

:, d 1 ,D 1

: d r ,CD r

r i i

1 ,q i i 1 .q 1

i i 1 .qr

r

xi 1 H0,n:m 1 ,n1+1: mr ,nr +1

i

i=1

p,q:p1+1,q1:pr +1,qr

r 11+ +

xi 1 11 10,:1,1+1::,+1

1

1

i

i i

i i

i=1

11

1

,:1+1, +1,

x1t1

ai ;A 1 ,,A r

: 11,1 , c 1 ,C 1

: 1r ,1 , c r ,C r

| i

i 1 p

i i 1 .p 1

1 .pr

x t bi ;B 1 ,,B r

:, d 1 ,D 1

: d r ,D r

r r i

i 1 q i

i 1 .q 1

i i 1 .qr

1 1

1 1 1 11111

=1

0 0 1

11

y1 ai ;A 1 ,,A r

H0,n:m 1,n1 +1: mr ,nr +1 | i i 1 p

: 11 ,1 , c 1 ,C 1

i i

i i

1 .p 1

: 1r ,1 , c r ,C r

i i

i i

1 .pr

p,q:p1 +1,q1:pr +1,qr

y bi ;B 1 ,,B r :, d 1 ,D 1

: d r ,CD r

r i i 1 q i

i 1 .q 1

i i 1 .qr

1

Now putting yi = xiti, where i = 1, 2, , r, we establish the following particular case.

PARTICULAR CASES:

1. If we take r = 2 in (8) it gives

   x1 ai ;A 1 ,,A 2

   H0,n:m 1 ,n1+1: m 2,nr +1 | i i 1 p

   : 12,1 , c 1 ,C 1

   i i

   i i

   1 .p 1

   : 12,1 , c 2 ,C 2

   i i

   i i

   1 .p 2

   p,q:p1+1,q1 :p2+1,q2

   x bi ;B 1 ,,B 2 :, d 1 ,D 1

   : d 2 ,CD 2

   2 i i 1 q i

   i 1 .q 1

   i i 1 .q 2

   2 1 1 1 2

   =

   1 1 1

   =1

   0

   0 =1

   x1t1

   ai ;A 1 ,,A 2

   : 11 ,1 , c 1 ,C 1

   : 12 ,1 , c 2 ,C 2

   i i

   i i

   H0,n:m 1 ,n1+1: m 2,n2 +1 | i

   i 1 p

   i i 1 .p 1

   1 .p 2

   p,q:p1+1,q1 :p2+1,q2

   x t bi ;B 1 ,,B 2

   :, d 1 ,D 1

   : d 2 ,CD 2

   2 2 i

   i 1 q i

   i 1 .q 1

   i i 1 .q 2

   dt1 . dt2 (9)

   where | arg x1 | < ½ M, | arg x2 | < ½ N

   p

   M = Ai 1

   q

   m 1 q1

   n1 p1

   Bi 1 di 1 Di 1 + ci 1 Ci 1 > 0,

   p

   N = Ai 2

   q

   m 2 q2

   n2 p2

   Bi 2 di 2 Di 2 + ci 2 Ci 2 > 0,

2. Again if we take n = p = q = 0, r = 1 in (8), we easily arrive at the following integral representation of the H-function:

1 ,1 +1

1|

11 ,1 , 1 , 1

1, 1

1 +1,1

1 , 1

1, 1

= 1 1 1 1 1 1 1 1 1 1 ×

1 ,1 +1

1|

11 ,1 , 1 , 1

1, 1 1 10)

1 1 0

where

1

1

1

1 +1,1

1

1 , 1

1, 1

1 1 + 1 1 > 0,

=1

=1 +1

=1

= 2 +1

|arg x1| < ½ A

REFERENCE

1. Watanabe, y.: Notes on the generalized derivative of Riemann Liuville and its applications to Leibnitzs formula I and II, Tohoku Math. J., 34 (1931), 8-27 and 28-41.

2. Erdelyi, A: Table of Integral Transform, Vol. II, McGraw-Hill, New York, 1954.

3. MacRobert, T. M. : Beta function formulae and integrals involving E-furnctio, Match., Ann., 142 (1961). p. 450-452.

4. Samko, S. C., Kilbas, A. A. and Maricev, O. I.: Integrals and derivatives of fractional order and some of their applications, Nauka tekhnika, Minsk, (1987) (in Russian).

5. MacRobert, T. M.: Beta function formulae and integrals involving E-function, Math., Ann., 142 (1961), p. 450-452.

*Head, Department of Mathematics Jamtara College, Jamtara (Jharkhand).

**Deptt . of Mathematics, Guru Ramdas Khalsa Institute of Science & Technology, Barela. Distt. Jabalpur (M.P.) 483001, INDIA .

***