H-Function with Complex Parameters: Evaluation

Transcription

1 H-Function with Complex Parameters: Evaluation F. Al-Musallam and Vu Kim Tuan Dept. of Math. & Comp. Sci., Kuwait University P.O.Box 5969 Safat, 3060 Kuwait s: Abstract Sufficient conditions for computation of the H-function with complex parameters by means of residues are derived. Examples and applications to integral transforms are given. Introduction Let P and Q be positive integers; {α j : j =,,..., P }, {β j : j =,,..., Q}, {p j : j =,,..., P }, and {q j : j =,,..., Q} be sets of complex numbers, where α j s and β j s being nonzero. Let c be a real number such that c Re p j α j, for all j with Reα j = 0. Let Λ c denote a contour from c i to c + i and not passing through any pole of the function P ks = Γp j + α j s Q Γq j + β j s, and such that. If Reα j > 0, then all the points are to the left of Λ c. s = m α j p j α j, m = 0,,,..., 3. If Reα j < 0, then all the points 3 are to the right of Λ c. The work of this author was partially supported by Kuwait University Research Administration under project SM 67.

2 3. If Reα j = 0, and Re p j α j < c, then all the points 3 are to the left of Λ c. 4. If Reα j = 0, and Re p j α j > c, then all the points 3 are to the right of Λ c. The authors in [] introduced the H-function with complex parameters via the Mellin- Barnes integral [] as follows, H P Q z, c p, α P q, β Q = P Γp j + α j s πi Q Λ c Γq j + β j s z s ds. 4 Here, and elsewhere in this paper, p, α P = p, α, p, α,..., p P, α P, and similarly for q, β Q. When α P and β Q are real vectors, the H-function with complex parameters reduces to the Fox H-function [5], and when α P and β Q are integer vectors, it reduces to the Meijer G-function []. The G- and H-functions play an important role in statistics and physical sciences [7, 8, 9, 0]. Let N = M = P Reα j Q Reβ j, 5 P [Imα j ln α j + ArgsignReα j α j Reα j ] Q [Imβ j ln β j + ArgsignReβ j β j Reβ j ], c = P P Q + Rep j Q P Req j + c[ Reα j Q Reβ j ]. In [], the authors derived necessary and sufficient conditions, under which the integral 4 defining the H-function converges absolutely. Theorem Corollary, [] The integral 4 converges absolutely if, and only if, P Imα j Q Imβ j = 0, and M π N, and either i M argz < π N, if M < π N, or ii argz = M ± π N, and c <. These conditions, naturally, reduce to those of the classical case where the α j s and β j s are real. Also in [] some well-known integrals that are not special cases of the classical Fox H-function were expressed as H-function with complex parameters. The present paper is a continuation of the work []. In this paper we determine sufficient conditions that enable one to compute the H-function with complex parameters 4 as a sum of residues at the right or the left poles of ksz s. We also compute the H-function 4 in particular cases and give an application to integral transforms.

3 Evaluation by Means of Residues We say that fz and gz are asymtotically equivalent, as z, within the sector α argz β, and we write this fact as if there is a function rz such that fz gz, as z, α argz β, fz = gz[ + rz], and rz converges uniformly to zero as z, within the sector α argz β. On the other hand we write fz gz, as z, within the sector α argz β, if there are positive constants W and W such that W gz fz W gz, α argz β. In what follows we shall use Argw to denote the principal argument of w, i.e, π < Argw π, and adapt the convention that Arg0 = 0. Grouping and renaming the parameters, the integrand ks can be written as ks = A Γa j + γ j s B Γb j δ j s C Γc j + iζ j s D Γd j iλ j s E Γe j + µ j s F Γf j σ j s G Γg j + iυ j s H Γh j iω j s, 6 where ζ j, λ j, υ j, ω j are positive numbers, and γ j, δ j, µ j, σ j are complex numbers with positive real parts. Clearly, P = A + B + C + D, Q = E + F + G + H, {α,..., α P } = {γ,..., γ A, δ,..., δ B, iζ,..., iζ C, iλ,..., iλ D }, {β,..., β P } = {µ,..., µ E, σ,..., σ F, iυ,..., iυ G, iω,..., iω H }, {p, p,..., p P } = {a,..., a A, b,..., b B, c,..., c C, d,..., d D }, {q, q,..., q Q } = {e,..., e E, f,..., f F, g,..., g G, h,..., h H }. According to [], Lemma, the asymptotic expansion of ks, defined by 6, as 3

4 s, Res 0, is F ks K sk s s 0 Υs sin πf j σ j s B sin πb j δ j s P Q exp ln s Res Reα j Reβ j P exp ln s Ims Imα j P exp ResArgs Imα j C π exp RessignIms ζ j exp ResL A L B L E + L F P exp ImsArgs Reα j Q Imβ j Q Imβ j D λ j G υ j + H ω j Q Reβ j exp ImsM, 7 where K s, K s, Υ, L A, L B, L E, and L F are defined as [ ] [ A K s = π A+B E F γ j +Rea ] j B δ j +Reb j [ ] [ E µ j +Ree ] 8 j F σ j +Ref j A B exp Ima j [Argγ j + Args] + Imb j [Argδ j + Args] exp E exp Ime j [Argµ j + Args] K s = π C+D G H [ C ζ +Rec j j [ G υ +Reg j j π C signims Args Imc j + F Imf j [Argσ j + Args], ] [ D λ +Red j j ] [ H ω +Reh j j D Imd j ] ] 9 G Img j H Imh j, Υs = +signims G sin πg j + iυ j s C sin πc j + iζ j s + signims H sin πh j iω j s D sin πd j iλ j s, 0 4

5 L A = A [Imγ jargγ j Reγ j ln γ j ], Put L B = B [Imδ jargδ j Reδ j ln δ j ], L E = E [Imµ jargµ j Reµ j ln µ j ], L F = F [Imσ jargσ j Reσ j ln σ j ]. L = P [Imα j Argα j Reα j ln α j ] Q [Imβ j Argβ j Reβ j ln β j ], and T = P Reα j Q Reβ j. 3 Let Ω c denote a contour from c i to c + i and not passing through any pole of the function ks, defined by. Denote by P O, P O r, and P O l the set of poles of ksz s, the set of poles of ksz s that are to the right of Ω c, and the set of poles of ksz s that are to the left of Ω c, respectively. Theorem Let Ω c ksz s ds converge absolutely, i.e, the conditions of Theorem are satisfied. a If T 0, then ksz s ds = πi Ω c s j P O r Resksz s, s j ; and in case T = 0, z must also satisfy z > exp L + π Reα j <0 Imα j. b If T 0, then Ω c ksz s ds = πi s j P O l Resksz s, s j ; and in case T = 0, z must also satisfy z < exp L π Reα j >0 Imα j. ProofProof. Using the change of variable s s c, one immediately sees that it is sufficient to prove part a of the theorem for the contour Ω 0 which is the horizontal 5

6 shifting of the contour Ω c by c units, and the function ks, as defined in, is replaced by k c s, P k c s = Γp j + α j c + α j s Q Γq j + β j c + β j s. 4 Observe that the asymptotic expansion of k c s is given by 7 where 0 is replaced by c and a j, b j, c j, d j, e j, f j, g j, and h j are chosen so that {p + α c, p + α c,..., p P + α P c} = {a,..., a A, b,..., b B, c,..., c C, d,..., d D }, {q + β c, q + β c,..., q P + β P c} = {e,..., e E, f,..., f F, g,..., g G, h,..., h H }. Furthermore, since the path of integration coincides with the imaginary axis at infinity and does not pass through any singular point of the integrand, it follows that Imc j, and Imd j are different from zero for all j. The poles of k c s are laying on finitely many half-lines, and equidistant on each of these half-lines. Let l, l,..., l J be those half-lines. Note that J P. Let z i, and y i be the initial point and the distance between consecutive poles on l i, respectively, i =,,..., J. For each positive integer r, the annulus r s r, contains finitely many poles of k c s that are on l i, say M i r, i =,,..., J. If Z i r and Z i r are the points of intersections of l i with the circles s = r and s = r +, respectively, then, Now for some r and r, and since M i r y i Z i r Z i r. Z i r = z i + r e iθ i, Z i r = z i + r e iθ i, r z i = Z i r zi r, and r Z i r + zi = r + + z i, it follows that, Z i r Z i r = r r r + + z i r z i = + z i. Therefore, J M i r i= J i= + Z i y r Zr i J + i J i= y i + z i = v, which means that the total number of poles in the annulus r s r is bounded by the constant v which is independent of r. Let s, s,..., s Jr be the poles of k c s within or on the annulus r s r. Then J r v. Assume further that r = s 0 s s... s Jr s Jr+ = r, and s i0 {s 0, s,..., s Jr, s Jr+} such that s i0 + s i0 = max 0 i J r { s i+ s i }. Clearly, s i0 + s i0 v +. 6

7 Let R r = s i s i0. If s is such that s = R r then Let w s R r w s i 0 + s i0 I Cr = k c sz s ds. πi C r, for all w P O. v + The integral is taken in a clockwise direction around the contour C r, consisting of the contours Cr and Cr, where Cr is a large circular arc, of center at the origin and radius R r, laying to the right of Ω 0 and originating from and ending on Ω 0 ; and Cr = {s Ω 0 : s R r }. It is clear from the definirion of R r that C r does not pass through any pole of k c s. Now we can split I Cr up into the sum of two integrals I Cr = I C r + I C r. The integral I Cr is equal to the negative of the sum of all the residues of k c sz s at its poles within the contour C r due to the negative orientation of the contour C r, that will cover P O r as r, and since πi Ω 0 k c sz s ds = lim r I C r, part a of Theorem would be proved if we show that lim r I C r = 0. Since IC r max skc sz s, for large r, and dp O, Cr s C r + v, it follows that lim I C r r = 0, if we can show that lim sk c sz s = 0, for Res 0, s ds, P O. Here da, B denotes the distance between the two sets A and B. +v Observe that F sin πf j + σ j s F B sin πb j + δ j s eπ Imσ js F B B = exp π[ Imσ j s Imδ j s ] = ϕs, eπ Imδ js whenever ds, P O, and +v Υs τs =. π exp Res + signims[ G υ j C ζ j] + signims[ H ω j D λ j]. Since F Imσ j s = F Imσ j Res + Reσ j Ims Ims F F Reσ j + Res Imσ j, 7

8 and B Imδ j s = B Imδ j Res + Reδ j Ims Ims B B Reδ j Res Imδ j, it follows that F Imσ js B Imδ js Res[ F Imσ j + B Imδ j ] + Ims [ F Reσ j B Reδ j]. Now, = C π signims[ ζ j +π[ F Imσ j + D λ j G υ j + H ω j ] L A L B L E + L F B Imδ j ] + π G + signims[ υ j + π H signims[ ω j + [ [ L A L B π L E L F + π D λ j ] B Imδ j + π C [ ζ j + F Imσ j + π G [ υ j + ] D λ j ] ] H ω j ], C ζ j ] and since, Arg α = Argα πsignimα, and if α is real αsignα = α we see that L B π B Imδ j = B [Imδ j argδ j Reδ j ln δ j ] π B Imδ j = B [Imδ j [arg δ j πsignim δ j ] Reδ j ln δ j ] π B Imδ j = B [Im δ j arg δ j + Reδ j ln δ j ] π B Imδ j. Hence L A L B π B Imδ j + π [ C ζ j+ D λ j] = A [Imγ j argγ j Reγ j ln γ j ] + B [Im δ j arg δ j Re δ j ln δ j ]+ π [ C ζ j+ D λ j] π B Imδ j = P [Imα j argα j Reα j ln α j ] π B Imδ j. Also, L E L F +π F Imσ j + π[ G υ j+ H ω j] = Q [Imβ j argβ j Reβ j ln β j ]. Our previous analysis shows that π Υs exp RessignIms[ C ζ j D λ j G υ j + H ω j] exp ResL A L B L E + L F exp Res L π B Imδ j Q F sin πf j σ j s Q B sin πb j δ j s F + π Ims 8 Reσ j B Reδ j,

9 where L is defined as in. Since, z s = z Res e argzims = exp Res ln z + argzims. it follows that, kc sz s K sk s s c exp π Ims [ exp F Reσ j Res[T ln s ln z L + π exp Ims[argz T Args M]. B Reδ j ] B Imδ j ] Observe that K sk s is bounded for Res 0, so there is a K > 0 such that Also, K sk s K, for all Res 0. T = P Q P Q Reα j Reβ j = Reα j Reβ j +[ F Reσ j B F Reδ j ] = N + [ Reσ j B Reδ j ], and B Imδ j = Reα j <0 Imα j. Let Y s = Args π T, Θs = π N signimsargz M. 5 Since Res 0 and T 0, then Y s 0. Because the conditions of Theorem are satisfied, Θs 0, and therefore, kc sz s K s c Ims [Y s+θs] e exp Res[T ln s ln z L + π Imα j ], Reα j <0 provided that Res 0, and ds, P O. +v Let R = {s : ds, P O, π Args π}, R +v 4 = {s : ds, P O Args < π }, and observe that 4 Ims = s sinargs s sin π 4 = s, for all s in R, +v, while Res = s cosargs s, for all s in R. 9

10 If T < 0, then lim T ln s L + π Reα s j <0 Imα j ln z =, and hence there exists a constant U > 0 such that exp Res[T ln s L + π Imα j ln z ] U for all Res 0. Reα j <0 Similarly, since Y s + Θs 0, we have exp Ims [Y s + Θs]. Also Y s = Args π T π 4 T > 0, for all s R. Thus kc sz s U K s c e π 4 s T, for all s in R, and s large. and kc sz s K s c exp s [ T ln s + ln z + L π Reα j <0 Imα j ], for all s in R. Since lim s c+ exp s s [ T ln s + ln z + L π Reα j <0 Imα j ] = 0, and lim s s c+ e π 4 s T = 0, it follows that lim s sk c sz s = 0. If T = 0, then Y s = 0. Thus, if Θs = 0, then necessarily argz M = π N, and hence, by the convergence conditions, c <. In this case kc sz s K s c exp Res[ ln z L + π Imα j ]. Reα j <0 Since s c+ 0 as s and exp Res[ ln z L + π Reα j <0 Imα j ] is bounded for z exp L + π Reα j <0 Imα j, we see that lim sk c sz s = 0. s If Θs > 0 for all s, then Θs min{ π N ± argz M} = δ > 0. Thus, e Ims Θs is bounded for all s, and moreover, Since exp e Ims Θs e δ s, for all s in R. Res[ ln z L + π Reα j <0 Imα j ] is bounded for z exp L + π Reα j <0 Imα j, it follows that kc sz s U s c e δ s, for all s in R, 0

11 for some constant U > 0. On the other hand, since ln z + L π Reα j <0 Imα j 0, we have exp Res[ ln z L + π Reα j <0 Imα j ] exp s [ln z + L π Reα j <0 Imα j ], for all s in R. Since e Ims Θs for all s, we see that there is a constant U 3 > 0 such that kc sz s U3 s c exp s [ln z + L π Imα j ], for all s in R. Reα j <0 The previous analysis shows that lim sk c sz s = 0, for Res 0, ds, P O s. +v This completes the proof of part a of the theorem. Now we proceed to prove part b of the theorem. Let α j = α j, j =,,..., P ; β j = β j, j =,,..., Q, and define, P k s = Γp j + α js Q Γq j + β j s. Observe that k s = k s, and under the change of variables s s, s s ksz s ds = k s ds = k s ds. Ω c Ω c z Ω c z We can now apply part a to k s s z and the contour Ω c which is the reflection of the contour Ω c with respect to the origin. For this purpose, note that P [ M = Imα j ln α j + ArgsignReα j α jreα j ] L = Q [ Imβ j ln β j + ArgsignReβj β j Reβ j ] = M, P [ Imα j Argα j Reα jln ] α j = + Q [ Imβ j Argβ j Reβ jln ] β j P [Imα j Argα j πsignimα j Reα j ln α j ] Q [Imβ j Argβ j πsignimβ j Reβ j ln β j ] [ P = L + π Imα j ] Q Imβ j = L,

12 since, by the conditions of Theorem, P Imα j Q Imβ j = 0. Moreover, T = P Q Reα j Reβ j = T, and hence, T 0 if, and only if, T 0. Also M arg z = M argz, for any z. It is clear that s P O l if, and only if, s is a pole of k s s z that is to the right of Ω c. Furthermore, if s 0 is a pole of k s s z of order m, then there is a ρ > 0 such that k s s = fs + m z where a m 0 and f is analytic at s 0. Hence and therefore Thus, Ω c ksz s ds = a j s s 0 j, for s s 0 < ρ, a j ksz s = f s + m j s + s 0, for s + s 0 < ρ, j Resksz s, s 0 = Resk s = πi Ω c k s s ds = πi z s j P O l Resksz s, s j. s, s 0. z s j P O l Resk s In case T = T = 0, then z must also satisfy z > exp L + π Reα j <0 Imα j = exp L + π or, equivalently, z < exp L π This completes the proof of Theorem. Reα j >0 Reα j >0 Imα j. s, s j z Imα j,

13 3 Some special cases of the H-function In general, if w and α 0 are complex numbers, then Res s= m+w α Γw + αs = m, m =,,... αm! Example If Rep 0, Rep + < Req, and Imp > 0, then 0, if x H x, 0 p, i = q, i ix ip x i q p, if 0 < x <. Since N = 0 = M and 0 = Rep Req <, the H-function exists for x > 0. Since T = 0 = L, the function can be computed by the residues of the right poles for x, and by those of the left poles for 0 < x. The poles are at s = im + ip, m = 0,,,.... Since Imp > 0, all the poles are on the left, and hence H x, 0 p, i = 0, if x, q, i and for 0 < x, H x, 0 p, i q, i [ = Ress=im+p Γp + is ] Γq p m x im ip m = i m!γq p m x im ip = ix ip x i m Γq p mm! = ix ip + p q m x i m = ix ip x i q p. Γq p m! Example If p and q are complex numbers such that Rep, Req 0, and Rep+ Req < 0, then x, 0 p, i, q, i 0, if ImpImq < 0 = isgnimpγp + q, if ImpImq > 0 x iq + x i p+q for all x > 0. It is easy to check that M = N = 0, and 0 = + Rep + Req <. Hence, x, 0 p, i, q, i = i Γp + isγq isx s ds, πi i 3

14 exists for all x > 0. Since T = 0, and L = π, the function can be computed by the right poles of ys = Γp + isγq isx s for x e π and by its left poles for 0 < x e π. The poles of Γp + is are at s = im + ip, while those of Γq is are at s = im iq, and Res s=im+p Γp + is = i m m! i The case Imp > 0, Imq < 0., Res s= im+q Γq is = i m. m! In this case all the poles of ys are left poles and hence x, 0 p, i, q, i = 0, for x e π, and for 0 < x e π, x, 0 p, i, q, i = [ Ress=im+p Γp + is ] Γp + q + mx im ip + [ Ress= im+q Γq is ] Γp + q + mx im+iq = iγp + q m p + q m! m x im ip + iγp + q m p + q m! m x im+iq = iγp + qx ip + x i p q + iγp + qx iq + x i p q = 0. ii The case Imp < 0, Imq > 0. In this case all the poles are right poles and hence, x, 0 p, i, q, i = 0, for 0 < x e π, and for x e π, the exact computation as above shows that, x, 0 p, i, q, i = 0, for x e π. iii The case Imp < 0, Imq < 0. In this case, the poles at s = im + ip, are right poles, and the poles at s = im iq, are left poles. Thus, for x e π, x, 0 p, i, q, i = [ Ress=im+p Γp + is ] Γp + q + mx im ip = iγp + q m p + q m! m x im ip = iγp + qx ip + x i p q = iγp+q, x iq +x i p+q and for 0 < x e π, x, 0 p, i, q, i = [ Ress= im+q Γq is ] Γp + q + mx im+iq = iγp + q m p + q m! m x im+iq = iγp + qx iq + x i p q. Therefore, x, 0 p, i, q, i = iγp + qx iq + x i p q, for all x > 0. 4

15 iv The case Imp > 0, Imq > 0. In this case, the poles at s = im + ip are left poles, and the poles at s = im iq are right poles. Thus, for 0 < x e π, x, 0 p, i, q, i = [ Ress=im+p Γp + is ] Γp + q + mx im ip = iγp + q m p + q m! m x im ip = iγp + qx ip + x i p q = iγp + qx iq + x i p q, and for x e π, x, 0 p, i, q, i = [ Ress= im+q Γq is ] Γp + q + mx im+iq = iγp + q m p + q m! m x im+iq = iγp + qx iq + x i p q. Therefore, x, 0 p, i, q, i = iγp + qx iq + x i p q, for all x > 0. Remark The special case x =, of the previous example, is evaluated in [6], formula 6.4. Example 3 If a is a complex number, then H z, c 0, = ia, i, ia, i π eπa eπz e πa e πz, c > 0, for argz < π. It is easy to check that N = and M = 0 and the conditions for existence are satisfied for argz < π. Since T = > 0, we can compute the function using the left poles at s = m, m = 0,,,.... Thus, H z, c 0, = ia, i, ia, i c+i Γs πi c i Γia isγ ia+is z s ds = m m! Γia+imΓ ia im zm = m sin πia+m π z m m! = m sinh πa+m π z m = eaπ e π z m m! π e aπ e π z m m! π m! = π eπa eπz e πa e πz. Example 4 If c > 0 and a is a complex number, Rea > c, then H 3 0 z, c 0,, ia, i, ia, i = π z m m! sinh πa + m, argz < π. Now, N =, and M = 0, and the conditions of convergence are satisfied for argz < π. Since T = > 0, we can compute the value of the function using the residues at the left poles of the function Γia isγ ia + isγsz s. Because Rea > c > 0, the only left poles we have are the poles s = m of Γs. Hence, 5

16 H 3 0 z, c 0,, ia, i, ia, i = = m = π πi m! Γia + imγ ia imz m = z m m! sinh πa+m. c+i c i Γia isγ ia + isγsz s ds m m! π sin πia+m zm Example 5 Let a, a,..., a P + ; b, b,..., b P be complex numbers such that P + Rea j < P Reb j. Then P + P +F P a P + ; b P ; P z = isignimz Γ a j H P + z i, c P Γb j 6 0, i a j, i P +, b j, i P z =, where c > 0, if Imz 0 and c < 0 if Imz < 0, and P F Q is the generalized hypergeometric function [, 4, ]., Consider the function x, c 0, i a j, i P +, b j, i P H P + = c+i πi c i Γis P + Γ a j + is P Γb j is x s ds. It is easy to see that N = 0, M = 0, and P + c = P Re a j P P + Reb j = + Rea j P Reb j <, and the conditions of Theorem are satisfied. Since L = 0 and T = 0 we can compute the value of the function using the residues at the poles of Γis P + Γ a j + is P Γb j is x s, which are at s = im, m = 0,,,.... If c > 0, then the poles are left poles. Hence, computing with left poles, for 0 < x, HP + x, c 0, i = i m x a j, i P +, b j, i Q im Q m! P + P Γ a P j m Γb j+m = i Q mp P + a j Q m x m! P + Q im Q Γ a P j Γb P j b j m i = Q Q P + Γ a P j Γb j P + F P a P + ; b P ; P x i. Let z = x i, then z =, x = z i, and Imz 0, and we obtain 6. If c < 0, then the poles are right poles. Hence, computing with right poles, for x >, H P + x, c 0, i a j, i P +, b j, i P = Let z = x i, then z =, x = z i, and Imz < 0, and we obtain 6. i P + Γ a j P Γb j P + F P a P + ; b P ; P x 6

17 Example 6 Let a and b be complex numbers such that Rea + < Reb, and Ima >, then H x, a, i b, i i Γb a x ia x i b a, if 0 < x = 0, if x It is easy to see that N = 0 = M, / = Rea Reb <, and the conditions of Theorem are fulfilled. Moreover, T = 0, L = 0, and the poles are at s = ia + im, m = 0,,,.... Since Ima >, there are no right poles, and therefore H x, a, i = 0, for x. b, i. For 0 < x, H x, a, i b, i = i m x ia m!γb a m x ia im = i Γb a i = Γb a x ia x i b a. + a b m x im m! 4 H-function as a Fourier kernel Let the conditions of Theorem be fulfilled, and moreover, Assume further, that M = N = 0, / =. 7 Re p j α j >, if Reα j > 0, Re p j α j <, if Reα j < 0, Re p j α j, if Reα j = 0. Conditions 8 guarantee that the vertical line i, + i can be taken as the contour of integration in the definition of HQ P z, p, α P. q, β Q 8 Theorem 3 The H-integral transform gx = x d dx HQ P 0 xy, p, α P q, β Q fydy 9 is a bounded integral transform from L R + onto itself, and the inverse transform has the form fx = x d H Q+ dx P + xy, q + β, β Q,,, 0, gydy. 0 p + α, α P, 3,,, 0 7

18 ProofProof. Under the assumptions 7 ks s, and therefore, ks L i, + i. Hence, HP Q x, p, α P exists for x > 0 q, β Q and belongs to L R +, since it is the inverse Mellin transform of ks [3]. Consequently, the integral 9 converges absolutely if f L R +. By the Mellin convolution theorem [3] we obtain d gx = x πi dx +i ksf sx s ds, i where f is the Mellin transform of f. Due to the assumption on ks on i, + i and the fact that f s L i, + i, it follows Hence, ss ksf s L i, + i. gx = πi +i i ss + ksf sx s ds, where the integral is understood in the sense of L i, + i. Therefore, g s = ss + ksf s, or Consequently, f s = = s sk s g s s s Q Γq j + β j β j s P Γp j + α j α j s g s. fx = πi = x πi dx +i Γ sγs i Γ3 sγ + s d = x d dx +i 0 Q Γq j + β j β j s P Γp j + α j α j s ss + g sx s ds Q Γq j + β j β j s P Γp j + α j α j s g sx s ds Γ sγs i Γ3 sγ + s H Q+ P + xy, q + β, β Q,,, 0, p + α, α P, 3,,, gydy. 8

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