# §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 $\frac{1}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}(% 1-t)^{c-b-1}}{(1-zt)^{a}}\mathrm{d}t,$ $\Re c>\Re b>0$.
 15.6.2 $\frac{\Gamma\left(1+b-c\right)}{2\pi\mathrm{i}\Gamma\left(b\right)}\int_{0}^{(% 1+)}\frac{t^{b-1}(t-1)^{c-b-1}}{(1-zt)^{a}}\mathrm{d}t,$ $c-b\neq 1,2,3,\dots$, $\Re b>0$.
 15.6.3 $e^{-b\pi\mathrm{i}}\frac{\Gamma\left(1-b\right)}{2\pi\mathrm{i}\Gamma\left(c-b% \right)}\int_{\infty}^{(0+)}\frac{t^{b-1}(t+1)^{a-c}}{(t-zt+1)^{a}}\mathrm{d}t,$ $b\neq 1,2,3,\dots$, $\Re(c-b)>0$.
 15.6.4 $e^{-b\pi\mathrm{i}}\frac{\Gamma\left(1-b\right)}{2\pi\mathrm{i}\Gamma\left(c-b% \right)}\int_{1}^{(0+)}\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^{a}}\mathrm{d}t,$ $b\neq 1,2,3,\dots$, $\Re(c-b)>0$.
 15.6.5 $e^{-c\pi\mathrm{i}}\Gamma\left(1-b\right)\Gamma\left(1+b-c\right)\*\frac{1}{4% \pi^{2}}\int_{A}^{(0+,1+,0-,1-)}\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^{a}}\mathrm% {d}t,$ $b,c-b\neq 1,2,3,\dots$.
 15.6.6 $\frac{1}{2\pi\mathrm{i}\Gamma\left(a\right)\Gamma\left(b\right)}\int_{-\mathrm% {i}\infty}^{\mathrm{i}\infty}\frac{\Gamma\left(a+t\right)\Gamma\left(b+t\right% )\Gamma\left(-t\right)}{\Gamma\left(c+t\right)}(-z)^{t}\mathrm{d}t,$ $a,b\neq 0,-1,-2,\dots$.
 15.6.7 $\frac{1}{2\pi\mathrm{i}\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c-a% \right)\Gamma\left(c-b\right)}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty}% \Gamma\left(a+t\right)\Gamma\left(b+t\right)\Gamma\left(c-a-b-t\right)\Gamma% \left(-t\right)(1-z)^{t}\mathrm{d}t,$ $a,b,c-a,c-b\neq 0,-1,-2,\dots$.
 15.6.8 $\frac{1}{\Gamma\left(c-d\right)}\int_{0}^{1}\mathbf{F}\left(a,b;d;zt\right)t^{% d-1}(1-t)^{c-d-1}\mathrm{d}t,$ $\Re c>\Re d>0$.
 15.6.9 $\int_{0}^{1}\frac{t^{d-1}(1-t)^{c-d-1}}{(1-zt)^{a+b-\lambda}}\mathbf{F}\left({% \lambda-a,\lambda-b\atop d};zt\right)\mathbf{F}\left({a+b-\lambda,\lambda-d% \atop c-d};\frac{(1-t)z}{1-zt}\right)\mathrm{d}t,$ $\lambda\in\mathbb{C}$, $\Re c>\Re d>0$. ⓘ Symbols: $\mathbb{C}$: complex plane, $\mathrm{d}\NVar{x}$: differential of $x$, $\in$: element of, $\int$: integral, $\Re$: real part, $\mathbf{F}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $\mathbf{F}\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: Olver’s hypergeometric function, $z$: complex variable, $a$: real or complex parameter, $b$: real or complex parameter and $c$: real or complex parameter Notes: With $\lambda=a+b$, this equation specializes to (15.6.8). Referenced by: (15.6.8), §15.6, §15.6, 3rd item Permalink: http://dlmf.nist.gov/15.6.E9 Encodings: TeX, pMML, png Addition (effective with 1.0.18): The constraint $\lambda\in\mathbb{C}$ was added. See also: Annotations for §15.6 and Ch.15

These representations are valid when $|\operatorname{ph}\left(1-z\right)|<\pi$, except (15.6.6) which holds for $|\operatorname{ph}\left(-z\right)|<\pi$. 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.