© Hindawi Publishing Corp. H-FUNCTION WITH COMPLEX PARAMETERS II: EVALUATION

Sufficient conditions for computation of the H-functions with complex param- eters by means of residues are derived and some examples are given.


Introduction. Let
(1.1) Here P and Q are nonnegative integers, p j (1 ≤ j ≤ P ), q j (1 ≤ j ≤ Q) are complex numbers, and α j (1 ≤ j ≤ P ), β j (1 ≤ j ≤ Q) are nonzero complex numbers. Throughout the present paper an empty product is interpreted as unity and an empty sum as zero.
Let c be a real number such that c ≠ Re(−p j /α j ), for all j with Re α j = 0, and Λ c denote a contour from c −i∞ to c +i∞ not passing through any pole of k(s) and such that (1) if Re α j > 0, then all the points s = − m α j − p j α j , m = 0, 1, 2,..., (1.2) are to the left of Λ c .
(2) If Re α j < 0, then all the points (1.2) are to the right of Λ c .
The following theorem gives the conditions, derived in [1], under which the integral (1.3) defining the H-function converges absolutely.
These conditions, naturally, reduce to those of the classical case where the α j 's and β j 's are real. Also in [1] 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 [1]. In this paper, we determine sufficient conditions that enable one to compute the H-function with complex parameters (1.3) as a sum of residues at the right or the left poles of k(s)z −s . It should be mentioned that Braaksma [3] was the first to apply the residue theorem to evaluate the integral (1.3) in the special case when α j = ±1 and β j = ±1. See also [17,18], where the residue theorem was used to compute the integral (1.3), with real α j 's and β j 's, for two other contours. It is striking that under some conditions, the H-function with complex parameters can be computed, in certain annulus of the complex plane, by the residues at the right poles as well as the residue at the left poles. This phenomenon has no parallel when the vectors (α j ) P and (β j ) Q are real vectors. We apply our result to compute the H-function (1.3) in some particular cases.

Evaluation by means of residues.
We need the following remark.
Remark 2.1. If w is a complex number, that is, not a zero of sin z, and z 0 is the zero of sin z closest to w, then To see this, observe that | sin x| ≥ (2/π )|x|, for all −π/2 ≤ x ≤ π/2. Since z 0 is the zero of sin z closest to w, then necessarily, | Re(w − z 0 )| ≤ π/2, and moreover, If | Im w| > ln 2, Let Ω c denote a contour from c − i∞ to c + i∞ not passing through any pole of the function k(s), defined by (1.1). Denote by PO, (PO) r , and (PO) l the set of poles of k(s), the set of poles of k(s) that are to the right of Ω c , and the set of poles of k(s) that are to the left of Ω c , respectively. and in case T = 0, z must also satisfy |z| > exp(−L + 2π Re α j <0 | Im α j |).
and in case T = 0, z must also satisfy |z| < exp(−L + π( where the contour of integration Ω 0 is the horizontal shifting of the contour Ω c by c units. The poles of k c (s) are precisely the horizontal shifting of the poles of k(s) by c units, and have the same position with respect to Ω 0 as the corresponding poles of k(s) with respect to Ω c ; and moreover, if s 0 is a pole of k(s), then Res s=s 0 k(s) = Res s=s 0 −c k c (s). Thus, without a loss of generality we may assume that c = 0. We construct a family of circles with centers at the origin and whose radii R r approaching infinity. Moreover, we insure that the family stays a positive distance away from the set PO. The poles of k(s) are lying on finitely many half-lines, and equidistant on each of these half-lines. Let l 1 ,l 2 ,...,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 = 1, 2,...,J. For each positive integer r , the annulus r − 1 ≤ |s| ≤ r , contains finitely many poles of k(s) that are on l i , say M i (r ), i = 1, 2,...,J. If Z i 1 (r ) and Z i 2 (r ) are the points of intersections of l i with the circles |s| = r − 1 and |s| = r , respectively, then for some r 1 and r 2 , and since it follows that which means that the total number of poles in the annulus r − 1 ≤ |s| ≤ r is bounded by the constant v which is independent of r . Let s 1 ,s 2 ,...,s Jr be the poles of k(s) on or within the annulus r − 1 ≤ |s| ≤ r . Then Clearly, and hence, the promised family of circles may be taken as all circles centered at the origin and whose radii are R r , r ≥ 1. Let The integral is taken in a clockwise direction around the contour C r , consisting of the contours C 1 r and C 2 r , where C 1 r is a large circular arc, with center at the origin and radius R r , lying to the right of Ω 0 and originating from and ending on Ω 0 ; and C 2 r = {s ∈ Ω 0 : |s| ≤ R r }. It is clear from the definition of R r that C r does not pass through any pole of k(s). Now we split I Cr up into a sum of two integrals The integral I Cr is equal to the negative of the sum of all the residues of k(s)z −s at its poles within the contour C r (due to the negative orientation of the contour C r ), that will cover (PO) r as r → ∞, and since Theorem 2.2(a) would be proved if we show that lim r →∞ I C 1 r = 0. Since , (2.19) it follows that lim r →∞ I C 1 r = 0, if we can show that lim |s|→∞ |sk(s)z −s | = 0, for Re(s) ≥ 0, d(s, PO) ≥ 1/(2(1+v)). Here d(A, B) denotes the distance between the two sets A and B. For this purpose we use the asymptotic expansion of |k(s)| given in Lemma 1.1.
We deduce that Re α j =0, Im α j >0 2 sin π p j + α j s Since Im β j s − Im α j s ≤ (Re s) Im β j +| Im s| Re β j − | Im s| Re α j − (Re s) Im α j = Im β j + Im α j Re s + Re β j − Re α j | Im s|, we see that, through the use of (2.21) and the fact that | sin z| ≤ e | Im z| , for all complex z, Re β j <0 2 sin π q j + β j s Re α j <0 2 sin π p j + α j s Re α j . (2.23) Here and throughout the proof, C denotes a universal positive constant, that may be distinct in different instances. Since Im(q j +β j s) = Im q j +(Im β j ) Re s, if Reβ j = 0, we see that Re β j =0, Im β j >0 2 sin π q j + β j s Re α j =0, Im α j >0 2 sin π p j + α j s ≤ C exp π(Re s)

Remark 2.3.
If µ = 0, N > 0, and T ≠ 0, then the H-function with complex parameters is analytic in the sector | arg z − M| < (π/2)N. This is the case since the series of the residues at the left poles, if T > 0, and the series of the residues at the right poles, if T < 0, represent analytic functions.
(2.53) Thus, if T = 0, then the H-function with complex parameters can be computed for any z in the annulus r 1 < |z| < r 2 by the residues at the right poles as well as the residues at the left poles. This situation has no analogy in the real case, that is, when (α) P and (β) Q are real vectors.
In fact for the function it is easy to see that N = 0, M = 0, and Re b j < −1, (3.4) so the conditions of Theorem 1.2 are satisfied.
Since L = 0 and T = 0, we can compute the value of the function using the residues at the poles s = im, m = 0, 1, 2,....
Remark 3.2. It seems to the authors that (3.2) provides the first integral representation for the generalized hypergeometric geometric function P +1 F P (z) on the unit circle.
Example 3.3. If Re p ≠ 0, 1 + Re p < Re q, and Im p > 0, then Since N = 0 = M and ∆ 0 = Re p − Re q < −1, the H-function exists for x > 0. Since T = 0 = L, the function can be computed by the residues at the right poles for x > 1, and by the residues at the left poles for 0 < x < 1. The poles are at s = im + ip, m = 0, 1, 2,.... Since Im p > 0, all the poles are on the left, and hence and for 0 < x < 1, (3.9) Example 3.4. If p and q are complex numbers such that Re p, Req ≠ 0, and Re p+ Re q < 0, then (3.10) for all x > 0.
It is easy to check that M = N = 0, and ∆ 0 = −1 + Re p + Re q < −1. Hence, exists for all x > 0. Since T = 0 and L = π , the function can be computed using the residues at the right poles of Γ (p + is)Γ (q − is)x −s for x > e −π and the residues at the left poles for 0 < x < e π . Thus in the interval (e −π ,e π ), the function can be computed using the residues at the right as well as at the left poles.
The poles of Γ (p +is) are at s = im+ip, while those of Γ (q −is) are at s = −im−iq, and (3.12) (i) The case Im p > 0, Im q < 0. In this case all the poles are left poles and hence and for 0 < x < e π , (3.14) (ii) The case Im p < 0, Im q > 0. In this case all the poles are right poles and hence, H 2 0 x, 0 (p, i), (q, −i) = 0, for 0 < x < e π , (3.15) and for x > e −π , the exact computation as above shows that, x −iq 1 + x i p+q , (3.17) and for 0 < x < e π , (3.18) Therefore, (iv) The case Im p > 0, Im q > 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 , and for x > e −π , (3.21) Therefore, For 0 < x < 1, (3.29)