# §15.6 Integral Representations

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

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

These representations are valid when $|\mathop{\mathrm{ph}\/}\nolimits\!\left(1-z\right)|<\pi$, except (15.6.6) which holds for $|\mathop{\mathrm{ph}\/}\nolimits\!\left(-z\right)|<\pi$. 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 $\mathop{\mathrm{ph}\/}\nolimits t=\pi$ and $\mathop{\mathrm{ph}\/}\nolimits\!\left(1+t\right)=0$.

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 $\mathop{\mathrm{ph}\/}\nolimits t=\pi$ and $\mathop{\mathrm{ph}\/}\nolimits\!\left(1-t\right)=0$.

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 $\mathop{\mathrm{ph}\/}\nolimits t$ and $\mathop{\mathrm{ph}\/}\nolimits\!\left(1-t\right)$ 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 $\mathop{\Gamma\/}\nolimits\!\left(a+t\right)$ and $\mathop{\Gamma\/}\nolimits\!\left(b+t\right)$ from those of $\mathop{\Gamma\/}\nolimits\!\left(-t\right)$, and $(-z)^{t}$ has its principal value.

In (15.6.7) the integration contour separates the poles of $\mathop{\Gamma\/}\nolimits\!\left(a+t\right)$ and $\mathop{\Gamma\/}\nolimits\!\left(b+t\right)$ from those of $\mathop{\Gamma\/}\nolimits\!\left(c-a-b-t\right)$ and $\mathop{\Gamma\/}\nolimits\!\left(-t\right)$, 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.