Notes on Whittaker & Watson, Chapter XIV, part 2

Whittaker & Watson, A Course of Modern Analysis. Chapter XIV, The Hypergeometric Function.

14.5 Barnes' contour integral for the hypergeometric function.

Consider the integral $$ \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds $$ with $\arg(-z)<\pi$, on the path from ${-i\infty}$ to ${i\infty}$ such that the poles of $\Gamma(a+s)$ and $\Gamma(b+s)$ at $s=-a-n,-b-n, n=1,2,\dots$ are to the left of the path, and the poles of $\Gamma(-s)$ at $s=1,2,\dots$ are to the right of the path. This path exists when $a,b$ are not negative integers.

The integrand is analytic in $z$ for $|\arg z|\le\pi-\delta$, and by the asymtotic expansion of $\log\Gamma(s+a)$ (§13.6) it is $O\left(|s|^{a+b-c-1}\exp(-\arg(-z)\Im s-\pi|\Im s|)\right)$ in the limit $s\to\infty$.

To close the contour, we integrate on the path of the semicircle of radius $N+\frac12$ on the right side of the complex plane $\Re s\ge0$. With the use of the reflection formula of the Gamma function, the integral becomes $$ \frac{1}{2\pi i}\int_{\substack{|s|=N+\frac12\\\Re s\ge0}}\frac{\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)\Gamma(1+s)}\frac{(-z)^s}{\sin s\pi}ds $$ in which the integrand is $O(N^{a+b-c-1})\cdot\frac{(-z)^s}{\sin s\pi}$ on the semicircle. Let $\theta\equiv\arg s$ and $|z|<1$, and use $|\sin z|\ge|\sinh \Im z|\ge \exp(-|\Im z|))$, we thus have $$ \begin{align*}\left|\frac{(-z)^s}{\sin s\pi}\right| & \le\exp\left(\left(N+\frac12\right)\cos\theta\log|z|-\left(N+\frac12\right)\sin\theta\arg(-z)-\left(N+\frac12\right)\pi|\sin\theta|\right) \\ & \le\exp\left(\left(N+\frac12\right)\cos\theta\log|z| - \left(N+\frac12\right)\delta|\sin\theta|\right)\to0. \qquad N\to\infty,|z|<1 \end{align*} $$ Therefore the integral on the path from ${-i\infty}$ to ${i\infty}$ can be calculated by summing the residues at $s=1,2,\dots$ $$ \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds=\sum_{n=0}^{\infty}\frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)n!}z^n $$ The LHS defines an analytic function for $\arg(z)<\pi$ and $|z|<1$, and the hypergeometric function can be then defined by $$ \frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}F(a,b;c;z)=\frac{1}{2\pi i}\int_{-\infty i}^{\infty i}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds. $$ 14.51 The continuation of the hypergeometric function.

For $|z|>1$, the integral on the large circle tends to zero on the left side of the complex plane ($\cos\theta<0,\log|z|>0$). Thus the interal is the sum of the residues at $s=-a-n,-b-n$ for $ n=1,2,\dots$ $$ \begin{align*} & {} \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds\\ =& {} \sum_{n=1}^{\infty}\frac{\Gamma(a+n)\Gamma(1-c+a+n)}{\Gamma(1+n)\Gamma(1-b+a+n)}\frac{ \sin(c-a-n)\pi}{\cos n\pi\sin(b-a-n)\pi}(-z)^{-a-n}\\ +& {}\sum_{n=1}^{\infty}\frac{\Gamma(b+n)\Gamma(1-c+b+n)}{\Gamma(1+n)\Gamma(1-a+b+n)}\frac{ \sin(c-b-n)\pi}{\cos n\pi\sin(a-b-n)\pi}(-z)^{-b-n}, \end{align*} $$ which gives the continuation $$ \begin{align*}\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}F(a,b;c;z) & =\frac{\Gamma(a)\Gamma(a-b)}{\Gamma(a-c)}(-z)^{-a}F(a,1-c+a;1-b+a;z^{-1}) \\ & + \frac{\Gamma(b)\Gamma(b-a)}{\Gamma(b-c)}(-z)^{-b}F(b,1-c+b;1-a+b;z^{-1}) \end{align*} $$ for $|z|>1$ and $\arg(-z)<\pi$.

14.52 Barnes' lemma that the integral $$ \frac{1}{2\pi i}\int_{-\infty i}^{\infty i}\Gamma(\alpha+s)\Gamma(\beta+s)\Gamma(\gamma-s)\Gamma(\delta-s)ds = \frac{\Gamma(\alpha+\gamma)\Gamma(\alpha+\delta)\Gamma(\beta+\gamma)\Gamma(\beta+\delta)}{\Gamma(\alpha+\beta+\gamma+\delta)}. $$ on the path to which the poles of $\Gamma(\alpha+s)\Gamma(\beta+s)$ are located on the left and the poles of $\Gamma(\gamma-s)\Gamma(\delta-s)$ are on the right. The integral can be calculated by considering the contour consisting of the path from ${-i\infty}$ to ${i\infty}$ and the large right semicircle with centre at the origin. The integral on the semicircle tends to 0 as the radius tends to $\infty$. Thus the integral on the path from ${-i\infty}$ to ${i\infty}$ can be calculated by summing of residues at the poles of $ \Gamma(\gamma-s)\Gamma(\delta-s)$, which proves the result for $\Re(\alpha+\beta+\gamma+\delta-1)<0$, and extends to the rest of the domain by analytic continuation.

14.53 The connexion between hypergeometric functions of $z$ and $1-z$.

By Barnes' lemma, the integral $$ \begin{align*} & \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds \\ = & {} \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{1}{2\pi i}\frac{\Gamma(-s)(-z)^sds}{\Gamma(c-a)\Gamma(c-b)}\int_{-k-i\infty}^{-k+i\infty}\Gamma(a+t)\Gamma(b+t)\Gamma(s-t)\Gamma(c-a-b-t)dt \\ = & {} \frac{1}{2\pi i}\int_{-k-i\infty}^{-k+i\infty}\frac{1}{2\pi i}\frac{\Gamma(a+t)\Gamma(b+t)\Gamma(c-a-b-t)dt}{\Gamma(c-a)\Gamma(c-b)}\int_{-i\infty}^{i\infty}\Gamma(s-t)\Gamma(-s)(-z)^sds\\ = & {} \frac{1}{2\pi i}\int_{-k-i\infty}^{-k+i\infty}\frac{1}{2\pi i}\frac{\Gamma(a+t)\Gamma(b+t)\Gamma(c-a-b-t)}{\Gamma(c-a)\Gamma(c-b)}\Gamma(-t)(1-z)^tdt. \quad(1-z)^t=F(-t,\beta,\beta,z) \end{align*} $$ In the similar manner that Barnes' lemma is proved, it can be shown that $$ \begin{align*} & \Gamma(c-a)\Gamma(c-b)\Gamma(a)\Gamma(b)F(a,b;c;z) \\ & =\Gamma(c)\Gamma(a)\Gamma(b)\Gamma(c-a-b)F(a,b;a+b-c+1;1-z) & & \\ & + \Gamma(c)\Gamma(c-a)\Gamma(c-b)\Gamma(a+b-c)(1-z)^{c-a-b}F(c-a,c-b;c-a-b+1;1-z) \end{align*} $$ for $\arg(-z)<\pi$ and $\arg(1-z)<\pi$.