on some multiple hypergeometric functions of several ...

ON SOME MULTIPLE HYPERGEOMETRIC FUNCTIONS OF SEVERAL MATRIX ARGUMENTS Lalit Mohan Upadhyaya* & H.S.Dhami** Department of Mathematics, University of Kumaun, Almora Campus, Almora (Uttaranchal) India -263601 AMS Subject Classification no. 33C65,33D99,44 ABSTRACT In continuation of our previous studies [6], here we propose to define the Exton's

(k) (n) and the K , K and K quadruple hypergeometric functions, the Exton's (1) E D 3 6 11 (k ) ( n ) (k ) ( n ) functions, the generalized Horn's function and the Chandel's E H4 (1) C (n ) (n ) generalized Srivastava H B and H C functions of matrix arguments. INTRODUCTION The systematic treatment of hypergeometric functions of four variables has been given in the work of Exton [1], where different functions have been defined as generalization of certain classes of quadruple hypergeoemtric functions. We have defined matrix arguments for some of them including the generalized Srivastava functions H function

(n ) (n) and HC and the B

(k ) ( n ) H 4 . Possible limiting forms, transformation relations and cases of

reducibility have been dealt with. All the matrices appearing in this paper are (p × p) real symmetric positive definite matrices. The meanings of all other symbols used are the same as in the works of Mathai [2-4].

* Department of Mathematics, Municipal Post Graduate College, Mussoorie, Dehradun (Uttaranchal), India, 248179. ** To whom all the correspondence may be addressed.

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1. Preliminary Definitions and Results In this section we first quote some of the definitions and a result of Mathai [3] which will be required by us in proving our results in this paper. References to the other results of Mathai, which will be used by us, will be made at the appropriate places in the paper. DEFINITION 1.1 :- The 2 -function of matrix arguments

= (a; c, c' ;− X,− Y) 2 2

is defined as that function for which the matrix transform (M-transform) is the following : M( 2) = [ ∫X > 0 ∫Y > 0 X !1 − (p + 1)/ 2 Y ! 2 − (p + 1)/ 2 × (a ; c,c'; − X, − Y)dXdY] 2 + (c)+ (c') + (a − ! − ! )+ (! )+ (! ) p p p 1 2 p 1 p 2 = .......(1.1) + p (c − !1)+ p (c' − ! 2 ) + (a) p

for Re (a − !1 − ! 2 ,c − !1,c' − ! 2 ,!1,! 2 ) > (p − 1)/ 2. where Re(.) denotes the real part of (.). DEFINITION 1.2 :- The Lauricella function

(n) (n) FC = FC (a,b;c 1,....,c n; − X1,...., − X n ) of matrix arguments is defined as that function for which the M-transform is the following :

(n) ! − (p +1)/ 2 ! − (p +1)/ 2 × M(FC ) =[ ∫X > 0 ... ∫X > 0 X1 1 ... X n n 1 n (n) FC (a,b;c ,...,c n ; − X1,..., − X n )dX1...dX n ] 1 n {  + (c )+ (! )} p j p j + (a − ! −− ! )+ (b −! −−! ) p 1 n p 1 n j=1 =[ ] ...(1.2) n + (a)+ (b) p p {  + (c − ! )} p j j j=1 for Re(a − ! −− ! ,b −! −−! ,c −! ,! ) > (p −1)/ 2,j=1,...,n. 1 n 1 n j j j

3

(n )

DEFINITION 1.3 : The 2 -function of matrix arguments

(n) (n) 2 = 2 (a;c1,...,c n ; − X1,..., − X n ) is defined as a function for which the M-transform is the following :

! − (p +1)/2 ! − (p +1)/2 (n) M( 2 ) =[ ∫X >0 ... ∫X >0 X1 1 .... X n n × 1 n (n) 2 (a;c1,....,c n ; −X1,..., − X n )dX1...dX n ] n n {  + (c )} + (a − ! −−! ){  + (! )} p j p 1 n p j j=1 j=1 .......... (1.3) = n + p (a) {  + (c − ! )} p j j j=1 for Re (a −!1 −− ! n ,c j −! j ,! j ) > (p −1)/2, j=1,..., n. (n)

DEFINITION 1.4 : The Φ2 -function of matrix arguments

(n) (n) - 2 =- 2 (b1,...,b n ;c;−X1,...,−X n ) is defined as a function for which the M-transform is the following :

! − (p + 1)/2 ! − (p + 1)/2 (n) × ) = [ ∫X > 0 ... ∫X > 0 X 1 .... X n 2 1 n 1 n (n) - (b ,...., b ; c;− X ,..., − X )dX ...dX ] 2 1 n 1 n 1 n n + (c){  + (b − ! )+ (! )} p j p j j =1 p j = ....(1.4) n {  + (b )}+ (c − ! −  − ! ) p 1 n j =1 p j for Re (c − ! −  − ! , b − ! , ! ) > (p − 1)/2, j = 1,..., n. 1 n j j j M(-

4

THEOREM 1.1 :

+ p (c)

a − (p +1)/2 I F (a ;c;− X) =[ × ∫0 Y 11 + p (a) + p (c − a) I − Y c − a − (p +1)/2 e − tr(XY) dY]

.........( 1.5)

for Re(a, c − a) > (p −1)/2 and for 0< Y < I.

2. The Exton’s Quadruple Hypergeometric Functions DEFINITION 2.1 : The function K = K (a, a, a, a; b , b , b , b ; c , c , c , c ;− X,− Y,− Z,− T) of matrix

3

3

1 1 2

2 1 2 2 1

arguments is defined as that function for which the M-transform is the following:

M(K 3 ) = [ ∫X > 0 ∫Y > 0 ∫Z > 0 ∫T > 0 X !1 − (p + 1)/2 Y ! 2 − (p + 1)/2 × Z ! 3 − (p + 1)/2 T ! 4 − (p + 1)/2 × K 3 (a, a, a, a; b1, b1, b 2 , b 2 ; c1, c 2 , c 2 , c1;− X,− Y,− Z,− T)dXdYdZdT ] =[

+ p (a − !1 − ! 2 − ! 3 − ! 4 ) + p (b1 − !1 − ! 2 ) + p (b 2 − !3 − ! 4 ) + p (a) + p (c1)

+ p (b1)

+ p (b 2 )

×

+ p (c 2 )

+ (! )+ (! )+ (! )+ (! )] + p (c1 − !1 − ! 4 ) + p (c 2 − ! 2 − !3 ) p 1 p 2 p 3 p 4 for Re(a − !1 − ! 2 − !3 − ! 4 , b1 − !1 − ! 2 , b 2 − !3 − ! 4 , c1 − !1 − ! 4 , c 2 − ! 2 − ! 3 , !i ) > (p − 1)/2, i = 1,2,3,4.

...(2.1)

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DEFINITION 2.2 :

K 6 = K 6 (a , a , a , a; b, b, c1, c 2 ; e, d, d, d;− X,− Y,− Z,− T )

M(K 6 ) =[ ∫X > 0 ∫Y > 0 ∫Z > 0 ∫T > 0 X !1 − (p +1)/2 Y ! 2 − (p +1)/2 × Z ! 3 − (p +1)/2 T ! 4 − (p +1)/2 K 6 (a,a,a,a;b,b,c1,c 2 ;e,d,d,d; − X, −Y, − Z, −T)dXdYdZdT ] + p (a −!1 −! 2 − !3 −! 4 ) + p (b −!1 −! 2 ) + p (c1 −! 3 ) + p (c 2 −! 4 ) =[ × + p (a) + p (b) + p (c1 ) + p (c 2 ) + p (e) + p (d) + (! )+ (! )+ (! )+ (! )] ...(2.2) + p (e −!1 ) + p (d −! 2 − !3 −! 4 ) p 1 p 2 p 3 p 4 for Re (a −!1 −! 2 − !3 −! 4 ,b −!1 −! 2 ,c1 − !3 , c − ! ,e −! ,d − ! − ! − ! ,! ) > (p −1)/2, i =1,2,3,4. 2 4 1 2 3 4 i DEFINITION 2.3 :

K (a, a, a, a; b , b , b , b ; c, c, c, d;− X,− Y,− Z,− T) 11 = 11 1 2 3 4 M(K 11 ) =[ ∫X > 0 ∫Y > 0 ∫Z >0 ∫T >0 X !1 −(p +1)/2 Y ! 2 −(p +1)/2 × Z ! 3 − (p +1)/2 T ! 4 − (p +1)/2 K (a,a,a,a;b ,b ,b ,b ;c,c,c,d; 11 1 2 3 4 − X,− Y, − Z, −T)dXdYdZdT ] K

=[

+ p (a −!1 −! 2 − !3 −! 4 ) + p (b1 −!1) + p (b 2 −! 2 ) + p (b 3 − !3 )

+ p (a) + p (b1) + p (b 2 ) + p (b 4 − ! 4 ) + p (c) + p (d) × + p (b 4 ) + p (c − !1 − ! 2 − !3 ) + p (d − ! 4 )

+ p (!1 )+ p (! 2 )+ p (! 3 )+ p (! 4 )]

+ p (b3 )

.....(2.3)

for Re(a − ! − ! − ! − ! , b − ! , c − ! − ! − ! , 1 2 3 4 i i 1 2 3 d − ! , ! ) > (p − 1)/2, i = 1,2,3,4. 4 i

×

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THEOREM 2.1 :

K (a,a,a,a;b ,b ,b ,b ;c ,c ,c ,c ;− X,− Y, − Z,− T) 3 1 1 2 2 1 2 2 1 − tr(S1 + S2 ) b1 − (p +1)/2 1 S × =[ ∫S > 0 ∫S > 0 e 1 2 + p (b1)+ p (b 2 ) 1 1

1 1 1 2 2 2 S2 2 (a;c1,c 2 ;−S1 XS1 −S2 TS 2 2 , 1 1 1 1 2 2 2 .......(2. 4) − S1 YS1 −S 2 ZS2 2 )dS1dS2 ] for Re (b ,b ) > (p −1)/2. 1 2 b 2 − (p +1)/2

PROOF : We take the M-transform of the right side of eq.(2.4) with respect to the variables X,Y,Z,T and the parameters ! , ! , ! , ! respectively to get ,

1 2 3 4 (p 1)/2 ! − + Y ! 2 − (p +1)/2 × ∫X > 0 ∫Y > 0 ∫Z > 0 ∫T > 0 X 1 1 1 1 1 (p 1)/2 (p 1)/2 − + − + ! ! 2 2 2 Z 3 T 4 2 (a;c1,c 2 ;−S1 XS1 −S 2 TS 2 2 , 1

1 1 1 2 2 2 − S1 YS1 −S2 ZS2 2 )dXdYdZdT

.........( 2.5)

Applying the transformations,

1 2 X = S XS 2 ; 1 1 1 1 1 2 T1 = S 2 TS 2 2 1

1 2 Y = S YS 2 ; 1 1 1 1

1 2 Z = S ZS 2 ; 2 1 2 1

(p + 1)/2 (p + 1)/2 (p + 1)/2 then, dX = S dX; dY = S dY; dZ = S dZ; 1 1 1 1 1 2 (p + 1)/2 dT = S dT; and, X = S X ; Y = S Y ; Z = S Z ; 1 2 1 1 1 1 1 2 T1 = S 2 T which renders the expression (2.5) as below,

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!1 − (p +1)/2 X ∫X > 0 ∫Y > 0 ∫Z > 0 ∫T > 0 1 2 1 1 1 1 ! − (p +1)/2 ! − (p +1)/2 ! 4 − (p +1)/2 Y 2 Z 3 T × 1 1 1

S 1

− !1 − ! 2

S

− ! 3 −! 4

2 (a;c1,c 2 ; − X1 − T1,− Y1 − Z1)dX1dY1dZ1dT1

.......... ..(2.6)

Now making use of another set of transformations,

X =X , X =X +T ; 2 1 3 1 1

Y =Y , Y =Y + Z 2 1 3 1 1

so that, from eq.(6.2) page 91 of Mathai [2], we have,

dX dT = dX dX 1 1 2 3

and

dY dZ = dY dY 1 1 2 3

X = X , T = X −X ; Y = Y , Z = Y − Y 1 2 1 3 2 1 2 1 3 2 where, 0< X < X and 0< Y < Y , 2 3 2 3

and,

which render the expression (2.6) as below,

− !1 − ! 2

− !3 − ! 4

∫0 < X < X ∫X > 0 ∫0 < Y < Y ∫Y > 0 × 2 3 3 2 3 3 ! − (p + 1)/2 ! − (p + 1)/2 ! − (p + 1)/2 X 1 X −X 4 Y 2 × 2 3 2 2 ! − (p + 1)/2 Y −Y 3 ....(2.7) 2 (a ; c1, c 2 ;− X 3 ,− Y3 )dX 2dX 3dY2dY3 3 2 Integrating out the variables X and Y by using a type-1 Beta integral in the expression 2 2 S1

S2

(2.7) and then using the definition (1.1) in the resulting expression we obtain,

+ p (!1) − !3 − ! 4 + p (a − !1 − ! 2 − !3 − ! 4 ) S1 S2 × + p (a) + p (c1 − ! 4 − !1) + p (! 2 )+ p (!3 )+ p (! 4 )+ p (c1)+ p (c 2 ) .........(2.8) + p (c 2 − !3 − ! 2 ) − !1 − ! 2

Substituting this expression on the right side of eq.(2.4) and integrating out the variables S and S by using a Gamma integral in the resulting expression ,we finally have

1 2 M(K ) as given by eq.(2.1). Therefore, the theorem is established. 3

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THEOREM 2.2 :

K 6 (a,a,a,a;b,b,c1,c 2 ;e,d,d,d;− X, − Y,− Z,−T)

=[

1

∫S > 0 ∫S > 0 ∫S > 0 e 2 3 + p (b)+ p (c1 )+ p (c 2 ) 1

− tr(S1 + S 2 + S3 )

×

1 1 c − (p +1)/2 c1 − (p +1)/2 2 2 S1 S2 S3 2 (a;e,d;−S1 XS1 2 , 1 1 1 1 1 1 2 2 2 2 2 .........( 2.9) − S1 YS1 − S 2 ZS2 −S3 TS 3 2 )dS1dS 2dS3 ] b − (p +1)/2

for Re (b,c1,c 2 ) > (p −1)/2. PROOF : Taking M-transform of the right side of eq.(2.9) with respect to the variables X,Y,Z,T and the parameters ! , ! , ! , ! respectively we get,

1 2 3 4

! − (p +1)/2 Y ! 2 − (p +1)/2 × ∫X > 0 ∫Y > 0 ∫Z > 0 ∫T > 0 X 1 1 1 (p 1)/2 (p 1)/2 ! ! − + − + 2 Z 3 T 4 2 (a;e,d;−S1 XS1 2 , 1 1 1 1 1 1 2 2 2 2 2 .........(2.10) −S1 YS1 −S2 ZS2 −S3 TS3 2 )dXdYdZdT Applying the transformations,

1

1 1 1 1 1 1 1 2 2 2 2 2 2 2 X1 = S1 XS1 ; Y1 = S1 YS1 ; Z1 = S2 ZS2 ; T1 = S3 TS3 2 (p + 1)/2 (p + 1)/2 (p + 1)/2 whence, dX1 = S1 dX; dY1 = S1 dY; dZ1 = S2 dZ ; (p + 1)/2 dT1 = S3 dT ; and X1 = S1 X ; Y1 = S1 Y ; Z1 = S2 Z ; T1 = S3 T which render the expression (2.10) as below,

9

S1

− !1 − ! 2

S2

− !3

S3

− !4

×

! − (p + 1)/2 ! 2 − (p + 1)/2 Y1 ∫X > 0 ∫Y > 0 ∫Z > 0 ∫T > 0 X1 1 × 1 1 1 1 ! − (p + 1)/2 ! 4 − (p + 1)/2 Z1 3 T1 × 2 (a ; e, d ; − X1 ,− Y1 − Z1 − T1 )dX1dY1dZ1dT1

.......(2.11)

Now making use of another set of transformations,

Y2 = Y1, Y3 = Y1 + Z1, Y4 = Y1 + Z1 + T1

so that, we have, from eq. (6.7) page 95 of Mathai [2],

dY1dZ1dT1 = dY2dY3dY4 and Y1 = Y2 , Z1 = Y3 −Y2 , T = Y − Y , 1 4 3 where, 0< Y2 < Y3 < Y4 . These transformations on substitution in the expression (2.11) convert it into the following form,

S1

− !1 − ! 2

S2

− !3

S3

− !4

×

! − (p + 1)/2 ! − (p + 1)/2 Y2 2 ∫X > 0 ∫0 < Y < Y ∫0 < Y < Y ∫Y > 0 X1 1 × 1 2 3 3 4 4 ! − (p + 1)/2 ! − (p +1)/2 Y3 − Y2 3 Y4 − Y3 4 × 2 (a ; e, d ; − X1 ,−Y4 )dX1dY2dY3dY4

.......(2 .12)

Now, integrating out the variables Y and Y

2

3 in the expression (2.12), one- by- one and

in order, by using a type-1 Beta integral and then using definition (1.1) in the resulting expression, we have,

S1

− !1 − ! 2

S2

− !3

− ! 4 + p (! 4 )+ p (!3 )+ p (! 2 )+ p (e)+ p (d)+ p (!1) S3 × + p (a)+ p (e − !1)+ p (d − ! 4 − !3 − ! 2 )

+ p (a − !1 − ! 4 − !3 − ! 2 )

...............(2.13)

10

Substituting this expression on the right side of eq.(2.9) and then integrating out the variables S , S , S in the resulting expression by using a Gamma integral,we finally have

1 2

3

M(K 6 ) as given by eq.(2.2), which proves the theorem. THEOREM 2.3:

K11(a ,a ,a ,a ;b1,b 2 ,b3 ,b 4 ;c,c,c,d;− X,− Y,− Z,−T ) + p (c)+ p (d) =[ ∫∫∫∫ × + p(b1)+ p(b 2)+ p(b 3)+ p (b 4)+ p(d − b 4)+ p (c− b1 − b 2 − b3) U b1 −(p +1)/2 V b 2 −(p +1)/ 2 W b3 −(p +1)/ 2 S b 4 −(p +1)/2 × I − U − V − W c −b1 −b 2 −b3 −(p +1)/ 2 I −S d − b 4 −(p +1)/2 × 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 I+ U XU + V YV + W ZW +S TS 2 −a × dUdVdWdS] ..........(2.14) where U >0, V >0, W >0,0 0 ∫Z > 0 ∫T > 0 Z !3 − (p +1)/2 T ! 4 − (p +1)/2 × 1

1 1 1 1 1 1 1 2 2 2 2 2 2 2 I+ U XU + V YV + W ZW + S TS 2 −a dXdYdZdT ....( 2.15) Appealing to the transformations,

1

1 1 1 1 1 1 1 2 2 2 2 2 2 2 X1 = U XU ;Y1 = V YV ;Z1 = W ZW ;T1 =S TS 2 , sothat ,dX1 = U (p+1)/ 2 dX ;dY1 = V (p+1)/ 2 dY;dZ1 = W (p+1)/ 2 dZ; dT = S (p+1)/ 2 dT and , X = U X ; Y = V Y ; Z = W Z ; T = S T , 1 1 1 1 1 which convert the exp ression (2.15)asbelow ,

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!1 − (p +1)/2 X × ∫ ∫ ∫ 1 > 0 Y1 > 0 Z1 > 0 T1 >0 1 −a ! − (p +1)/2 !3 − (p +1)/2 ! 4 −(p +1)/2 Y1 2 Z1 T1 I+ X + Y + Z + T 1 1 1 1 × U −!1 V −! 2 W −!3 S −! 4 ∫X

dX1dY1dZ1dT1

........( 2.16)

This expression, on integrating out the variables X , Y , Z , T by using a type –2

1 1 1 1

Dirichlet integral, reduces to,

U −!1 V −! 2 W −!3 S −! 4 + p (!1) + p (! 2 ) + p (!3 ) + p (! 4 ) + p (a − !1 −! 2 − !3 −! 4 )

.......(2. 17)

+ p (a)

Substituting this expression on the right side of eq.(2.14) and then integrating out the variables U,V,W and S in the resulting expression by using a type –1 Dirichlet integral and a type –1 Beta integral respectively we finally have M(K ) as given by eq. (2.3),which

11

proves the theorem. A limiting case of eq.(2.14) has the following form :

− X − Y − Z −T lim K (a,a,a,a;b ,b ,b ,b ;c,c,c,d; , , , ) 11 1 2 3 4 a a a a a →∞

=[

+ p (c)+ p (d) + p(b1)+ p (b 2)+ p(b 3)+ p (b 4)+ p(d − b 4 )+ p (c − b1 − b 2 − b3)

∫∫∫∫

×

U b1 −(p +1)/2 V b 2 −(p +1)/2 W b3 −(p +1)/2 S b 4 −(p +1)/2 × I − U − V − W c − b1 − b 2 − b3 − (p +1)/ 2 I −S d − b 4 − (p +1)/ 2 × e − tr ( UX + VY + WZ +ST ) dUdVdWdS ] .........( 2.18) + p (c) =[ × F (b ;d;−T ) ∫∫∫ × + p (b1)+ p (b 2)+ p (b 3)+ p (c− b1 − b 2 − b 3) 1 1 4 U b1 − (p +1)/ 2 V b 2 − (p +1)/ 2 W b3 −( p +1) /2 × I − U − V − W c − b1 − b 2 − b3 −( p +1) /2 e −tr ( UX +VY+ WZ)dUdVdW ] ...( 2.19 ) which follows from eq.(2.18) by the use of theorem (1.1).

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THEOREM 2.4: A case of reducibility-

− X − X − X −T lim K11 (a,a,a,a;b1,b 2 ,b3 ,b 4 ;c,c,c,d; , , , ) a a a a a →∞ .......(2. 20) =1F1(b 4 ;d;−T) . 1F1(b1 + b 2 + b3 ;c;− X)

PROOF: This theorem follows by putting Z=Y=X in eq.(2.19) and transformations

applying the

W = U, W = U + V, W = U + V + W , then dUdVdW = dW dW dW ,from 1 2 3 1 2 3 eq.(6.7) page 95 of Mathai [2], and,

U = W1 ; V = W2 − W1 ; W = W3 − W2 where , 0 < W1 < W2 < W3 < I, which renders the right side of eq.(2.19) as ,

+ p (c)

× 1F1( b 4 ;d;−T) ∫∫∫ × (b ) (b ) (b ) (c b b b ) +p 1 + p 2 + p 3 +p − 1− 2 − 3 b − (p +1)/ 2 b −( p +1) /2 b − ( p +1) /2 W1 1 W2 − W1 2 W3 − W2 3 × c − b1 − b 2 − b 3 −( p +1)/ 2 −tr ( W3 X) e dW1dW2 dW3 ] ...( 2.21) I − W3 Now, integrating out W and W in the expression (2.21), one-by-one and in order, by 1 2 [

using a type-1 Beta integral and then applying theorem (1.1) to the resulting expression, the desired result is obtained. THEOREM 2.5 : A transformation theorem –

K11(a ,a ,a ,a ;b1,b 2 ,b3 ,b 4 ;c,c,c,d;−X,− Y,− Z,−T ) a = I+ X − K11[a ,a ,a ,a;c − b1 − b 2 − b3 ,b 2 ,b3 ,b 4 ;c,c,c,d; −1 −1 −1 −1 (I + X) 2 X (I+ X) 2 ,− (I+ X) 2 ( Y − X)( I+ X ) 2 , −1 −1 −1 −1 2 2 2 ( I X ) ( Z− X)( I+ X ) ,− ( I + X ) T( I + X ) 2 ] − + where Y − X > 0and Z− X > 0.

.......(2.22)

13

= I + Y −a K11[a,a,a,a;b1,c − b1 − b 2 − b 3,b3,b 4;c,c,c,d; −1 −1 −1 −1 2 2 2 (X − Y)(I + Y) ,(I+ Y) Y(I + Y) 2, −(I + Y) −1 −1 −1 −1 2 2 2 (Z − Y)(I + Y) , − (I + Y) T(I + Y) 2 ] −(I + Y) where Z − Y > 0 and X − Y >0.

.......(2.23)

= I + Z −a K11[a,a,a,a;b1,b 2 ,c − b1 − b 2 − b 3 ,b 4 ;c,c,c,d;

−1 −1 −1 −1 −( I + Z) 2 (X − Z)(I + Z) 2 ,−(I + Z) 2 ( Y − Z)(I + Z) 2 , −1 −1 −1 −1 ( I + Z) 2 Z(I + Z) 2 ,−(I + Z) 2 T( I + Z) 2 ] .......(2.24) where X − Z> 0 and Y − Z >0. = I + T −a K11[a,a,a,a;b1,b 2 ,b 3 ,d − b 4 ;c,c,c,d; −1 −1 −1 −1 - (I + T) 2 X(I + T) 2 ,−(I + T) 2 Y(I + T) 2 , −1 −1 −1 −1 −(I + T) 2 Z(I + T) 2 ,(I + T) 2 T(I + T) 2 ] .......(2.25) = I + X + T − a K11[a,a,a,a;c − b1 − b 2 − b 3 ,b 2 ,b 3 ,d − b 4 ;c,c,c,d; −1 −1 −1 −1 (I + X + T) 2 X(I + X + T) 2 ,−(I + X + T) 2 (Y − X)(I + X + T) 2 , −1 −1 −1 −1 2 2 2 2 ].....(2.26) (I X T) (Z X)(I X T) , (I X T) T(I X T) − + + − + + + + + + where Y − X > 0and Z − X > 0.

14

PROOF :

To prove this theorem we define the function K

11through an integral

representation,

K11 (a ,a ,a ,a;b1,b 2 ,b 3 ,b 4 ;c,c,c,d;−X,−Y,− Z,−T) + p (c)+ p (d) =[ ∫∫∫∫ × + p (b1)+ p (b 2)+ p (b 3)+ p (b 4 )+ p (d − b 4)+ p (c − b1 − b 2 − b 3) U b1 − (p +1)/ 2 V b 2 − (p +1)/ 2 W b3 −( p +1) /2 S b 4 −( p +1) /2 × I − U − V − W c − b1 − b 2 − b3 −( p +1) /2 I −S d − b 4 −( p +1) /2 × 1

1 1 1 1 1 1 1 2 2 2 2 2 2 2 I + X UX + Y VY + Z WZ + T ST 2 −a × dUdVdWdS] .............(2.27 ) where U > 0, V > 0, W > 0,0 ( p −1)/ 2, i =1,2,3,4 . To prove eq.(2.22), we apply the transformations

U = I − U − V − W, V = V, W = W, so that , dU dV dW = dUdVdW , 1 1 1 1 1 1 to eq.(2.27) and observing that,

1

1 1 1 1 1 2 2 2 2 2 I +X ( I − U1 − V1 − W1) X + Y V1Y + Z W1Z 2 1 1 1 1 −1 −1 2 2 2 2 2 X U1X ( I + X ) 2 + + T ST = I + X . I − (I + X ) −1

1 1 −1 2 2 2 (I + X ) ( Y − X ) V1 (Y − X ) ( I + X ) 2 + 1 1 −1 −1 2 2 2 (I + X ) (Z − X) W (Z − X) (I + X) 2 + 1 1 −1 1 −1 2 2 2 (I + X ) T ST (I + X ) 2 the desired result follows after a suitable reinterpretation of eq.(2.27). The results in eqs.(2.23) and (2.24) follow similarly. To prove the result of eq.(2.25) we observe that,

15

1

1 1 1 1 1 1 1 2 2 2 2 2 2 2 I + X UX + Y VY + Z WZ + T ST 2 = I + T . I + ( I + T) (I + T )

−1

−1

1 1 1 1 −1 −1 −1 2 X 2 UX 2 (I + T ) 2 + (I + T ) 2 Y 2 VY 2 (I + T ) 2 +

1 1 1 −1 −1 1 −1 2 Z 2 WZ 2 (I + T ) 2 − ( I + T ) 2 T 2 ( I − S)T 2 (I + T ) 2

Applying this result along with the transformation S = I − S, so that, dS = dS, to 1 1 eq.(2.27) and suitably interpreting the resulting expression in the light of eq.(2.27), the result of eq.(2.25) follows immediately. The result of eq.(2.26) is a combination of the results of eqs.(2.22) and (2.25). It is established by applying the transformations

U1 = I − U − V − W, V1 = V , W1 = W , S1 = I − S, so that, dUdVdW = dU dV dW and dS = dS , to eq.(2.27) and observing that 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 I + X ( I − U1 − V1 − W1) X + Y V1Y + Z W1Z + T (I − S1)T 2 1 1 −1 −1 2 2 2 X U1X ( I + X + T ) 2 + = I + X + T . I − (I + X + T ) 1 1 −1 −1 2 2 2 (I + X + T ) ( Y − X ) V ( Y − X ) (I + X + T ) 2 + 1 −1

1 1 −1 2 2 2 (I + X + T ) ( Z − X ) W ( Z − X ) (I + X + T ) 2 1 1 −1 1 −1 2 2 2 T S T ( I + X + T) 2 − (I + X + T ) 1 and then suitably interpreting the resulting expression as per eq.(2.27) gives the result of eq.(2.26). Two similar results of the type of eq.(2.26) in the variables Y and T and Z and T also exist. THEOREM 2.6: Another case of reducibility-

(i)

K11(a ,a ,a,a ;b1,b 2 ,b3 ,b 4 ;c,c,c,d;− X, −Y,− Y,−T ) = FG (a ,a ,a ,b 4 ,b1,b 2 + b3;d,c,c;− T,− X,− Y) ........(2.28)

(ii)

K11(a ,a,a ,a;b1,b 2 ,b3 ,b 4 ;c,c,c,d;−Y,− Y,− Y,−T ) ............(2.29) = F2 (a,b 4 ,b1 + b 2 + b3;d,c;−T ,− Y)

16

PROOF :

(i) In eq.(2.14) we put Z=Y and observe that

1

1 1 1 1 1 1 1 2 2 2 2 2 2 2 I + U XU + V YV + W ZW + S TS 2 1 1 1 1 1 1 = I + U 2 XU 2 + (V + W) 2 Y(V + W) 2 + S 2 TS 2 Now, applying the transformations

V1 = V, W1 = V + W, so that , dVdW = dV1dW1 and 0 < V1 < W1, the eq.(2.14) transforms into,

K11 (a,a,a,a;b1,b 2 ,b 3 ,b 4 ;c,c,c,d;−X,−Y,−Y,−T) + p (c)+ p (d) =[ ∫∫∫∫ × + p (b1)+ p (b 2)+ p (b 3)+ p (b 4 )+ p (d − b 4 )+ p (c− b1 − b 2 − b3) b −( p +1) /2 b −( p +1) /2 b −( p +1) /2 U b1 − (p +1)/ 2 V1 2 W1 − V1 3 S 4 × c − b1 − b 2 − b 3 −( p +1) /2 I − U − W1 I −S d − b 4 −( p +1)/2 × 1

1 1 1 1 1 2 2 2 I + U XU + W1 2 YW1 2 +S TS 2 −a dUdV1dW1dS] .......... (2.30) Integrating out V in the above equation by using a type-1 Beta integral and

1

comparing the resulting expression with eq.(1.9) of the authors’ paper [6], the required result follows. (ii) This result follows by putting X=Y in eq.(2.28) and using the theorem (1.2) of the authors’ paper [6] .

17

(k ) ( n )

(k ) ( n )

3. The Exton’s (1) E D Function and The Chandel’s (1) E C DEFINITION (3.1) :

Function.

(k ) ( n ) ( k ) ( n ) (1) E D = (1) E D (a ,b1,,b n ;c,c';− X1,,− X n ) ! −( p+1)/ 2 ! −(p+1) / 2 ( k ) (n ) M[ (1) E D ]=[ ∫X 0 ∫X 0 X1 1 Xn n ×  1> n> (k ) (n ) E (a ,b1,,b n ;c,c'; −X1,,−X n )dX1dX n ] (1) D Γp (a − !1 −−! n )Γp (b1 −!1 )Γp (b n − ρ n )Γp (c)Γp (c') =[ × Γp (a ) Γp ( b1)Γp ( b n )Γp (c − !1 −− ! k )Γp (c'−! k 1 −−! n ) + .......... .(3.1) Γp (!1)Γp (! n )] for Re (a − ! −−! ,b − ! ,c −! −− ! ,c'−! , ) 1 n i i 1 k k +1 −−! n !i > ( p −1)/ 2,i =1,,n. DEFINITION (3.2) :

(k ) ( n ) ( k ) ( n ) (1) E C = (1) E C (a ,a ',b;c1,,c n ;− X1,,− X n ) ! −(p+1)/ 2 ! −( p+1)/ 2 (k) (n) M[ (1) E C ]=[ ∫X 0 ∫X 0 X1 1 Xn n ×  1> n> (k ) (n ) (1) E C (a ,a ',b;c1,,c n ;− X1,,− X n )dX1dX n ] Γp (a − !1 −−! k )Γp (a '−! k+1 −−! n )Γp (b −!1 −−! n )Γp (c1)Γp (c n ) =[ × Γp (a )Γp (a ') Γp ( b) Γp (c1 − !1)Γp (c n − ! n ) .......... .(3.2) Γp (!1)Γp (! n )] for Re (a − ! −−! ,a '−! ,b ,c , ) 1 k k +1 − − ! n − !1 −−! n i − !i ! i > ( p −1)/ 2, i =1,,n.

18

THEOREM 3.1 :

( k) ( n ) (1) E D (a , b1,  , b n ; c, c ' ;− X1, ,− X n ) Γp (c)Γp (c' ) =[ ∫ (n) ∫ × Γp (b1)  Γp (b n )Γp (c − b1 −  − b k )Γp (c'− b k 1 −  − b n ) + b − (p +1) / 2 b − ( p +1) / 2 c - b −  − b − (p +1) / 2 1 k U1 1 I − U1 −  − U k  Un n × c'-b b (p 1) / 2 1 1 k +1 −  − n − + I − Uk 1 −  − Un I + U1 2 X1U1 2 +  + + 1 1 a U 2 X U 2 − dU  dU dU dU ] .......... (3.3) n n n 1 k k +1  n for U > 0, 0 < U +  + U < I, 0 < U U I and i 1 k k +1 +  + n < for Re ( b , c − b −  − b , c'− b b ) ( p 1) / 2, i = 1,  , n. i 1 k k +1 −  − n > − PROOF : Taking the M-transform of the right side of eq.(3.3) with respect to the variables X ,  , X and the parameters ! ,  , ! , we have,

1

n

1

n ! − (p +1) / 2 ! − (p +1) / 2 1 1 I + U1 2 X1U1 2 +  +  Xn n ∫X > 0  ∫X > 0 X1 1 1 n 1 1 −a U 2X U 2 dX  dX .........( 3.4) n n n 1 n

Making use of the transformations,

( p+1) / 2 1 1 Yi = U i 2 X i U i 2 , then , dYi = U i dX i , Yi = U i X i for i = 1,  , n ,in eq.(3.4) and integrating out the variables Y (i = 1,  , n ) , in the resulting expression by i using a type-2 Dirichlet integral, we obtain,

U 1

− !1

− ! n Γp (!1)  Γp (! n )Γp (a − !1 −  − ! n )  Un Γp (a )

.......( 3.5)

which, on substitution on the right side of eq.(3.3) and integrating out the variables U ,  , U and U ,  , U in the resulting expression by using a type-1 Dirichlet

1

k

k +1

n (k ) (n ) integral finally gives M[ E ] as given by eq.(3.1), thereby proving the theorem. (1) D

19

A limiting form of this function can be seen to have the following form :

lim .→∞

− X1 − Xn (k ) ( n ) ) , , E ( ., b ,  , b ; c, c' ; (1) D 1 n . .

Γp (c)Γp (c' ) =[ ∫ (n) ∫ × Γp (b1)  Γp (b n )Γp (c − b1 −  − b k )Γp (c'− b k 1 −  − b n ) + b − (p +1) / 2 b − ( p +1) / 2 c - b −  − b − (p +1) / 2 n 1 k U1 1 U I U U × − − −  n 1  k c'-b b (p 1) / 2 − tr ( U X +  + U X ) k +1 −  − n − + 1 1 k k × I−U U e −  − k +1 n − tr ( U k 1X k 1 +  + U n X n ) + + e dU  dU ] .......... (3.6) 1 n THEOREM 3.2 : A case of reducibility-

lim .→∞

−X −X −Y −Y (k ) (n )     −  E , ( n k ) , ] [ , b , , b ; c , c ' ; , ( k ) , . (1) D 1 n . . . .

= 1F1 ( b1 +  + b k ; c;− X ).1 F1 (b k +1 +  + b n ; c' ;− Y ) ......( 3.7) in PROOF : To prove this theorem, we put X == X = X and X 1 k k +1 == X n = Y eq.(3.6) and apply the following sets of transformations on its right side,

V = U , V = U + U ,  , V = U +  + U , and W = U , 1 1 2 1 2 1 1 k k k +1 +U =U +  + Un W =U , , W 2 k +1 k+ 2 n −k k+1

so that from eq.(6.7) page 95 of Mathai [2],

dU1dU k = dV1dVk ,dU k +1dU n = dW1dWn −k and , U1 = V1 , U 2 = V2 − V1 , , U = V − V =W ,U = W − W ,, U = W −W ;U k k k −1 k +1 1 k +2 2 1 n n −k n−k −1 where 0 < V1 < V2 0 e ( a ) Γp 1 1 1 1 (n − k ) ( b k 1,  , b n ; c ' ;− U 2 X k 1U 2 ,  , − U 2 X k U 2 )Φ2 + + 1 1 ........( 3.9) − U 2 X n U 2 )dU ] for Re (a ) > ( p − 1) / 2. PROOF : Taking the M-transform of the right side of eq.(3.9) with respect to the variables X ,  , X and the parameters ! ,  , ! respectively, we get,

1

n

1

n

21

!1−(p+1) / 2 !k −( p+1) / 2 !k 1−(p+1) / 2 + X X X   × ∫X1>0 ∫X n >0 1 k k+1 1 1 1 1 !n −( p+1) / 2 ( k ) 2 2 2  Xn Φ (b ,  , b ; c; − U X U ,  ,− U X U 2 ) × 2 1 k 1 k 1 1 1 1 ( n −k ) 2 2 2 (b ,  , b n ; c' ;− U X U ,  ,− U X n U 2 )dX  dX n .....( 3.10) Φ 2 1 k+1 k +1 1

1

Applying the transformations, Y = U 2 X U 2 , so that , i i

dYi = U ( p +1) / 2 dX i and Yi = U X i for i = 1,  , n in the expression (3.10) and

using the definition(1.4) in the resulting expression gives,

U − !1 −  − ! n

Γp (b1 − !1)  Γp (b n − ! n )Γp (c)Γp (c ' )Γp (!1)  Γp (! n ) ....(3.11) Γp ( b1)  Γp ( b n )Γp (c − !1 −  − ! k )Γp (c '− ! k 1 −  − ! n ) +

Using this expression on the right side of eq.(3.9) and integrating out the variable U in the resulting expression by using a Gamma integral we are led to M[

( k ) (n ) as given by (1) E D ]

eq.(3.1), which concludes the proof.

THEOREM 3.4 :

(k) (n) E (a, a' , b; c ,  , c ;− X ,  ,− X ) (1) C 1 n 1 n 1 =[ ∫U> 0 ∫V> 0 e− tr(U + V) U a − (p +1)/2 V a'− (p +1)/2 × + p (a) + p (a' ) 1 1 1 1 1 1 (n) 2 2 2 2 2 2 (b; c1,  , c n ;− U X1U ,  ,− U X k U ,− V X k 1V 2 ,  , + 1 1 2 .......(3. 12) − V X n V 2 )dUdV] for Re (a, a' ) > (p − 1)/2.

22

(n )

PROOF : Taking the M-transform of the 2 -function on the right side of eq.(3.12) with respect to the variables X ,  , X and the parameters ! ,  , ! , we have,

1 n 1 n !1−(p +1)/2 ! k −(p+1)/2 ! k 1−(p+1)/2 + X X X ×   ∫X >0 ∫X >0 1 k k 1 + 1 n 1 1 1 1 ! −(p+1)/2 (n) 2, 2 2 2 (b; c , , c ; U X U , , U X U X n n  −  − 2 1 n 1 k 1 1 1 1 2 2 2 − V X k 1V ,,− V X n V 2 )dX1dX k dX k 1dX n .....(3.13 ) + + 1

1

1

1

Making use of the transformations Y = U 2 X U 2 and Y = V 2 X V 2 , so that, i i j j

dYi = U ( p+1) / 2 dX i , dY j = V (p +1) / 2 dX j , and Yi = U X i , Y j = V X j for i = 1,  , k; j = k + 1,  , n in the expression (3.13) and using the definition (1.3) in the resulting expression, we are led to Γp ( b − !1 −  − ! n )Γp (c1 )  Γp (c n ) ! ! ! !   − − − − − − U 1 k V k +1 n × Γp (b)Γp (c1 − !1 )  Γp (c n − ! n )

....( 3.14 ) Γp (!1)  Γp (! n ) Using this expression on the right side of eq.(3.12) and then integrating out the variables U and V in the resulting expression by using a Gamma integral, the outcome is ( k ) (n ) M[ (1) E C ] as given by eq.(3.2), thus finishing the proof.

23

4. The Generalized Horn’s Function

(k ) ( n ) H4

DEFINITION 4.1 :

(k ) ( n ) ( k ) (n ) , ,b ;c , ,c ; X , , X ) H = H (a ,b k +1  n 1  n − 1  − n 4 4 M[

!1−(p+1)/ 2 ! n −( p+1)/ 2 ( k ) (n ) H ]=[ ∫ X X ×   4 X1>0 ∫X n >0 1 n (k) (n) H 4 (a ,b k 1,,b n ;c1,,c n ; − X1,,− X n )dX1dX n ] +

Γp (a − 2!1 −− 2! k −! k+1 −− ! n )Γp (b k +1 − ! k+1)Γp ( b n −! n ) =[ × Γp (a )Γp ( b k 1)Γp ( b n )Γp (c1 − !1)Γp (c n −! n ) + ......( 4.1) Γp (c1 )Γp (c n )Γp (!1)Γp (! n )] for Re(a − 2! −− 2! −! ,b , ,b , 1 k k +1 −−! n k +1 − ! k +1  n − ! n c − ! ,! ) > ( p −1/ 2, i =1,,n. i i i

24

THEOREM 4.1 :

(k ) ( n ) ,  , b n ; c ,  , c n ;− X ,  ,− X n ) H (a , b 4 k +1 1 1 )  Γp (c n ) Γp (c k +1 =[ ∫ (n − k)  )  Γp (b n )Γp (c )  Γp (c n − b n ) Γp (b −b k+1 k +1 k+1 b −(p+1) / 2 c −b −( p+1) / 2 bn −( p+1) / 2 k +1 k +1 k +1 U U I U − ×  ∫ k +1 n k+1 1 1 1 1 −a cn −bn −( p+1) / 2 2 2 2 I+U X U +  + U n X n U n2  I − Un k+1 k +1 k +1 (k) F [(a + 1) / 2 , ( 2a + 1) / 4; c ,  , c ; C 1 k 1 1 1 1 U 2 +  + U n 2 X n U n 2 ) −1 × − 4 (I + U 2 X k+1 k +1 k+1 1 1 1 1 2 2 2 X (I + U X U +  + U n X n U n 2 ) −1,  , 1 k +1 k+1 k +1 1 1 1 1 1 1 1 − 2 2 2 2 2 X U X U 2 ++ − 4 (I + U +  + U n X n U n ) X (I + U k+1 k +1 k+1 k k +1 k+1 k +1 1 1 2 U n X n U n 2 ) −1 ] dU ........( 4.2)  dU n ] k +1

for p = 2, for 0 < U < I , j = k + 1 ,  , n and j for Re ( b ,  , bn , c ,  , c n − b n ) > ( p − 1) / 2. −b k +1 k +1 k+1

PROOF : We take the M-transform of the right side of eq.(4.2) with respect to the variables X ,  , X and the parameters ! ,  , ! to obtain,

1

n

1

! −(p+1)/ 2 X k ∫X >0 ∫X >0 X1 1 1 n X

n

! n −( p+1)/ 2

n ! k −( p+1)/ 2

−( p+1)/ 2 ! X k 1 k+1 × +

1

1 1 1 2 2 2 I+ U X U U X U 2 −a × k+1 k +1 k +1 ++ n n n Continued to the next page…………..

25

(k ) F [( a +1) / 2 ,( 2a +1) / 4; c ,, c ; C 1 k 1 1 1 1 2 2 2 2 ) −1 × U X U 4 ( I U X U − + k +1 k +1 k +1 + + n n n 1

1 1 1 2 2 2 X (I + U X U U X U 2 ) −1 ,, + +  1 k +1 k +1 k +1 n n n 1 1 1 1 1 2 − 4( I + U k 1X k +1U k 21 + + U n 2 X n U n 2 ) − X k × + + 1 1 1 1 2 2 2 ( I + U k 1X k 1U k 1 + + U n X n U n 2 ) −1 ]dX 1dX k dX k 1dX n + + + +

....( 4 .3)

Making use of the transformations,

1 1 1 2 2 2 U X U 2 )−1 X × X U Y = 4( I + U i i k +1 k +1 k +1 +  + n n n 1 1 1 1 2 X U 2 +  + U 2 X U 2 ) −1 ; i = 1 ,  , k (I + U n n n k +1 k +1 k +1 1 1 2 Y = U X U 2 ; j = k + 1,  , n j j j j so that , Y = 4( I + Y Y )−1 X (I + Y Y ) −1   + + + + i k +1 n i k +1 n − ( p+1) dYi = 4p (p +1) / 2 I + Yk 1 +  + Yn dX i ; i = 1,  , k + (p +1) / 2 dX ; j = k + 1,  , n dY = U j j j 1

and Y = 4p I + Y Y i k +1 +  + n

−2

X ; i = 1,  , k i

Y = U X ; j = k + 1,  , n j j j in the expression (4.3) and then using the definition (1.2) in the resulting expression and integrating out the variables Y ,  , Y by using a type-2 Dirichlet integral,we have,

k +1

n

26

4

− p(!1+  + ! k )

Γp (c1 )  Γp (c k ) −! k 1 −! n + U  Un × k +1 Γp (c1 − !1)  Γp (c k − ! k )

Γp [( a + 1) / 2 − !1 −  − ! k ] Γp [( 2a + 1) / 4 − !1 −  − ! k ] Γp (!1)  Γp (! n ) × Γp [( a + 1) / 2] Γp [( 2a + 1) / 4] Γp (a − 2!1 −  − 2! k − ! k 1 −  − ! n ) + .......( 4.4) Γp (a − 2!1 −  − 2! k ) Using this expression on the right side of eq.(4.2) and integrating out the variables U ,  , U in the resulting expression by using a type-1 Beta integral and observing

k +1

n

that,

4 =

− p(!1 +  + ! k ) Γp 1 Γp (a )

[(a + 1) / 2 − ! −  − ! ] Γp [( 2a + 1) / 4 − !1 −  − ! k ] 1 k Γp [( a + 1) / 2] Γp [( 2a + 1) / 4]Γp (a − 2!1 −  − 2! k )

for p = 2

.......( 4.5)

from eq.(6.13) page 84 of Mathai [3], finally we have M[

(k ) ( n ) H 4 ] as given by eq.(4.1),

thereby completing the proof. This result is different from the corresponding result in the scalar case.

(n)

(n )

5. The Generalized Srivastava H B and H C DEFINITION 5.1 :

Functions

(n ) (n ) = H (. ,  , . ;  ,  ,  ;− X ,  ,− X ) B B 1 n 1 n 1 n ! − ( p+1) / 2 ! − ( p+1) / 2 (n ) × X n M[H ] = [ ∫X > 0  ∫X > 0 X 1 B 1 n 1 n (n ) H (. ,  , . ;  ,  ,  n ;− X1,  ,− X n )dX1  dX n ] B 1 n 1 H

Continued in the next page………..

27

Γp (.1 − !1 − ! n )Γp (. 2 − !1 − ! 2 )  Γp (. n − ! n −1 − ! n )Γp ( 1)  Γp ( n ) × =[ Γp (.1 )Γp (. 2 )  Γp (. n )Γp (1 − !1)  Γp (  n − ! n ) .......( 5.1) Γp (!1)  Γp (! n )] for Re (. − ! − ! , . − ! − ! ,  , . − ! , , ) ( p 1) / 2, 1 1 n 2 1 2 n n −1 − ! n i − !i !i > − i = 1,  , n DEFINITION 5.2 :

(n ) (n ) H C = H C (.1,  , . n ; ;− X1,  ,− X n ) M[ H

! − ( p+1) / 2 ! − ( p+1) / 2 (n ) ] = [ ∫X 0  ∫X 0 X 1 X n  × C 1 n 1> n>

(n ) ( , , ; ; X , , X )dX  dX ] C .1  . n  − 1  − n 1 n Γp (. − ! − ! n )Γp (. − ! − ! )  Γp (. n − ! − ! n )Γp ( ) 1 1 2 1 2 n −1 =[ × Γp (. )Γp (. )  Γp (. n )Γp (  - ! −  − ! n ) 1 2 1 .......( 5.2) Γp (! )  Γp (! n )] 1 for Re (. − ! − ! n , . − ! − ! ,  , . n − ! − !n ,  − ! −  − !n , ! ) 1 1 2 1 2 n−1 1 i > ( p − 1) / 2, i = 1,  , n. H

THEOREM 5.1 :

(n ) H B (.1,  , . n ; 1,  , n ;− X1,  ,− X n ) =[

1

 ∫T > 0 e ∫ Γp (.1 )Γp (. 2 )  Γp (. n ) T1 > 0 n

− tr (T1 +  + Tn ) 1

. − ( p+1) / 2 T1 1 ×

1 1 1 2 2 2T 2 ) × T2 F ( ; ; T T X T  Tn −  0 1 1 2 1 11 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 0 F1( ;  2 ;−T3 T2 X 2T2 T3 )0 F1( ; n ;− T1 Tn X n Tn T1 ) × dT  dT ] .......... .(5.3) 1 n for Re (.i ) > (p − 1) / 2 , i = 1,  , n. . 2 − ( p+1) / 2

. n − ( p+1) / 2

28

PROOF : Taking the M-transform of the right side of eq.(5.3) with respect to the variables X ,  , X and the parameters ! ,  , ! we obtain,

1

n

1

n ! 2 − (p +1)/2

! − (p +1)/2 ! − (p+1)/2 ∫X > 0  ∫X > 0 X 1 X X n ×  1 2 n 1 n 1 1 1 1 1 1 1 1 2 2 2 2 2 2 F ( ; ; T T X T T ) F ( ;  ;− T T X T 2 T 2 ) × 2 2 3 2 3 2 11 2 0 1 0 1 1 − 2 1 1 1 1 1 2 2 ......(5.4 ) 0 F1( ;  n ;− T1 Tn X n Tn 2 T1 2 )dX1  dX n ] Making use of the transformations

1

1 1 1 1 1 1 1 2 2 2 2 2 2 Y1 = T2 T1 X1T1 T2 , Y2 = T3 T2 X 2T2 2 T3 2 , 1 1 1 1 (p +1) / 2 ( p+1) / 2 2 2 Y = T T X T 2 T 2 ; so that , dY = T T dX , n 1 n n n 1 1 2 1 1 ( p +1) / 2 (p +1) / 2 (p +1) / 2 ( p+1) / 2 dY = T T dX ,  , dY = T T dX 2 3 2 2 n 1 n n and , Y1 = T2 T1 X1 , Y2 = T3 T2 X 2 ,  , Yn = T1 Tn X n in the expression (5.4), and using the definition of M-transform of a F -function from 0 1 eq.(2.3.5) page 38 of Mathai [3] , we have,

T 1 

− !1 − ! n

T 2

− !1 − ! 2

− ! n −1 − ! n + p (1 )+ p (!1) + p ( 2 )+ p (! 2 )  Tn × + p (1 − !1 ) + p (2 − ! 2 )

+ p ( n )+ p (! n ) + p ( n − ! n )

.........( 5.5)

which, on substitution on the right side of eq.(5.3) and integrating out the variables

( n) T1,  , Tn in the resulting expression by using a Gamma integral leads to M[ H B ] , as given by eq.(5.1) , thus concluding the proof.

29

THEOREM 5.2 :

(n ) ( , , ; ; X , , X ) C .1  . n  − 1  − n −tr (T1++Tn ) .1−(p+1) / 2 1 e T1 =[ ×  ∫ ∫T >0 Γp (.1)Γp (. n ) T1>0 n

H

1 1 1 1 1 1 1 2 2 2 2 2 2 F ( ; ; T T X T T −T T X T 2 T 2  Tn 0 1 − 2 1 11 2 3 2 2 2 3 1 1 1 1 2 2 −− T1 Tn X n Tn 2 T1 2 )dT1dTn ] .......( 5.6) for Re(. ) > ( p −1)/ 2, i =1,,n. i . n −(p+1)/ 2

1

PROOF: Taking the M- transform of the right side of eq.(5.6) with respect to the variables X ,  , X and the parameters ! ,  , ! ,we have,

1

n

1 n !1−(p+1)/ 2 ! n −( p+1)/ 2 X X   × ∫X >0 ∫X >0 1 n 1 n 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 0 F1( ;;− T2 T1 X1T1 T2 − T3 T2 X 2T2 T3 − 1

1 1 1 2 − T1 Tn 2 X n Tn 2 T1 2 )dX1dX n

.......... .(5.7)

Making use of the transformations,

1

1 1 1 1 1 1 1 2 2 2 2 2 2 Y = T T X T T , Y = T T X T 2T 2 , , 1 2 1 11 2 2 3 2 2 2 3 1 1 1 1 2 Y = T T 2X T 2T 2 , n n 1 n 1 n ( p+1) / 2 ( p+1) / 2 (p +1) / 2 ( p+1) / 2 so that , dY1 = T2 T1 dX1; dY2 = T3 T2 dX 2 ; (p +1) / 2 (p +1) / 2 T dX ;  ; dYn = T1 n n and , Y = T T X ; Y = T T X ;  ; Y = T T X 1 1 2 1 2 3 2 2 n 1 n n in the expression (5.7) and then using the M-transform of a F function by applying 0 1 theorem 3.3 page 55 of Mathai [3] , we obtain,

30

T 1

− !1 − ! n

T 2

− !1 − ! 2

− ! n − 1 − ! n Γp (  Γp (!1 )  Γp (! n ) ....( 5.8) T n Γp (  − !1 −  − ! n )

Using this expression on the right of eq.(5.6) and integrating out the variables T , ……,

1 (n ) T in the resulting expression by using a Gamma integral, finally yields M[H ] , as C n

given by eq.(5.2), whence the theorem is proved. References 1. Exton H.(1976).Multiple Hypergeometric Functions and Applications. Ellis Horwood Limited, Publishers, Chichester. 2. Mathai A.M. (1992). Jacobians of Matrix Transformations I. Centre for Mathematical Sciences,Trivandrum, India. 3. Mathai A.M. (1993). Hypergeometric Functions of Several Matrix Arguments. Centre for Mathematical Sciences, Trivandrum,India. 4. Mathai A.M., Pederzoli G. (1996). Some Transformations for Functions of Matrix Arguments. Indian Journal of Pure and Applied Mathematics; 27(3), pp. 277-284. 5. Srivastava H.M. and Karlsson P.W.(1985).Multiple Gaussian Hypergeometric Series. Ellis Horwood Limited, Publishers, Chichester. 6. Upadhyaya Lalit Mohan and Dhami H.S.(Nov.2001). Matrix Generalizations of Multiple Hypergeometric Functions. # 1818 IMA Preprints Series, University of Minnesota, Minneapolis, U.S.A.