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15 Hypergeometric FunctionProperties

§15.6 Integral Representations

The function \mathbf{F}\left(a,b;c;z\right) (not F\left(a,b;c;z\right)) has the following integral representations:

15.6.1 \mathbf{F}\left(a,b;c;z\right)=\frac{1}{\Gamma\left(b\right)\Gamma\left(c-b%
\right)}\int_{0}^{1}\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^{a}}\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; \Re c>\Re b>0.
15.6.2 \mathbf{F}\left(a,b;c;z\right)=\frac{\Gamma\left(1+b-c\right)}{2\pi\mathrm{i}%
\Gamma\left(b\right)}\int_{0}^{(1+)}\frac{t^{b-1}(t-1)^{c-b-1}}{(1-zt)^{a}}\,%
\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; c-b\neq 1,2,3,\dots, \Re b>0.
15.6.3 \mathbf{F}\left(a,b;c;z\right)={\mathrm{e}}^{-b\pi\mathrm{i}}\frac{\Gamma\left%
(1-b\right)}{2\pi\mathrm{i}\Gamma\left(c-b\right)}\int_{\infty}^{(0+)}\frac{t^%
{b-1}(t+1)^{a-c}}{(t-zt+1)^{a}}\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; b\neq 1,2,3,\dots, \Re\left(c-b\right)>0.
15.6.4 \mathbf{F}\left(a,b;c;z\right)={\mathrm{e}}^{-b\pi\mathrm{i}}\frac{\Gamma\left%
(1-b\right)}{2\pi\mathrm{i}\Gamma\left(c-b\right)}\int_{1}^{(0+)}\frac{t^{b-1}%
(1-t)^{c-b-1}}{(1-zt)^{a}}\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; b\neq 1,2,3,\dots, \Re\left(c-b\right)>0.
15.6.5 \mathbf{F}\left(a,b;c;z\right)={\mathrm{e}}^{-c\pi\mathrm{i}}\Gamma\left(1-b%
\right)\Gamma\left(1+b-c\right)\*\frac{1}{4\pi^{2}}\int_{A}^{(0+,1+,0-,1-)}%
\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^{a}}\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; b,c-b\neq 1,2,3,\dots.
15.6.6 \mathbf{F}\left(a,b;c;z\right)=\frac{1}{2\pi\mathrm{i}\Gamma\left(a\right)%
\Gamma\left(b\right)}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty}\frac{\Gamma%
\left(a+t\right)\Gamma\left(b+t\right)\Gamma\left(-t\right)}{\Gamma\left(c+t%
\right)}(-z)^{t}\,\mathrm{d}t,|\operatorname{ph}\left(-z\right)|<\pi; a,b\neq 0,-1,-2,\dots.
15.6.7 \mathbf{F}\left(a,b;c;z\right)=\frac{1}{2\pi\mathrm{i}\Gamma\left(a\right)%
\Gamma\left(b\right)\Gamma\left(c-a\right)\Gamma\left(c-b\right)}\int_{-%
\mathrm{i}\infty}^{\mathrm{i}\infty}\Gamma\left(a+t\right)\Gamma\left(b+t%
\right)\Gamma\left(c-a-b-t\right)\Gamma\left(-t\right)(1-z)^{t}\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; a,b,c-a,c-b\neq 0,-1,-2,\dots.
15.6.8 \mathbf{F}\left(a,b;c;z\right)=\frac{1}{\Gamma\left(c-d\right)}\int_{0}^{1}%
\mathbf{F}\left(a,b;d;zt\right)t^{d-1}(1-t)^{c-d-1}\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; \Re c>\Re d>0.
15.6.9 \mathbf{F}\left(a,b;c;z\right)=\int_{0}^{1}\frac{t^{d-1}(1-t)^{c-d-1}}{(1-zt)^%
{a+b-\lambda}}\mathbf{F}\left({\lambda-a,\lambda-b\atop d};zt\right)\mathbf{F}%
\left({a+b-\lambda,\lambda-d\atop c-d};\frac{(1-t)z}{1-zt}\right)\,\mathrm{d}t,|\operatorname{ph}\left(1-z\right)|<\pi; \lambda\in\mathbb{C}, \Re c>\Re d>0.

In all cases the integrands are continuous functions of t on the integration paths, except possibly at the endpoints. Note that (15.6.8) can be rewritten as a fractional integral. In addition:

In (15.6.1) all functions in the integrand assume their principal values.

In (15.6.2) the point \ifrac{1}{z} lies outside the integration contour, t^{b-1} and (t-1)^{c-b-1} assume their principal values where the contour cuts the interval (1,\infty), and (1-zt)^{a}=1 at t=0.

In (15.6.3) the point \ifrac{1}{(z-1)} lies outside the integration contour, the contour cuts the real axis between t=-1 and 0, at which point \operatorname{ph}t=\pi and \operatorname{ph}\left(1+t\right)=0.

In (15.6.4) the point \ifrac{1}{z} lies outside the integration contour, and at the point where the contour cuts the negative real axis \operatorname{ph}t=\pi and \operatorname{ph}\left(1-t\right)=0.

In (15.6.5) the integration contour starts and terminates at a point A on the real axis between 0 and 1. It encircles t=0 and t=1 once in the positive direction, and then once in the negative direction. See Figure 15.6.1. At the starting point \operatorname{ph}t and \operatorname{ph}\left(1-t\right) are zero. If desired, and as in Figure 5.12.3, the upper integration limit in (15.6.5) can be replaced by (1+,0+,1-,0-). However, this reverses the direction of the integration contour, and in consequence (15.6.5) would need to be multiplied by −1.

In (15.6.6) the integration contour separates the poles of \Gamma\left(a+t\right) and \Gamma\left(b+t\right) from those of \Gamma\left(-t\right), and (-z)^{t} has its principal value.

In (15.6.7) the integration contour separates the poles of \Gamma\left(a+t\right) and \Gamma\left(b+t\right) from those of \Gamma\left(c-a-b-t\right) and \Gamma\left(-t\right), and (1-z)^{t} has its principal value.

In each of (15.6.8) and (15.6.9) all functions in the integrand assume their principal values.

See accompanying text
Figure 15.6.1: t-plane. Contour of integration in (15.6.5). Magnify