Mellin transform of a Gaussian Hypergeometric Function with negative $x$-argument

Question

I am quite fascinated by the formula for the Mellin transform of the Gaussian Hypergeometric Function, which is given by:

$$\mathcal M [_2F_1(\alpha,\beta;\gamma;-x)] = \frac {B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)}$$

Source : Table of Integral Transforms page $336$, $6.9 (3)$

I have found this within a table of integral transforms of various functions and I would be really interested in a proof for this formula.

Answer 1

The inverse Mellin transform is given by $$\mathcal M^{-1}[F] = \frac 1 {2 \pi i} \int_{\sigma -i \infty}^{\sigma + i \infty} F(s) x^{-s} ds.$$ For $F(s) = \Gamma(s) \Gamma(\alpha - s) \Gamma(\beta - s) / \Gamma(\gamma - s)$, the line $\operatorname{Re} s = \sigma$ should separate the poles of $\Gamma(s)$ from the poles of $\Gamma(\alpha - s) \Gamma(\beta - s)$. For $0 < x < 1$, the sequence of integrals over left semicircles centered at $\sigma$ with radii $\sigma + k + 1/2$ tends to zero and the inverse transform can be calculated as the sum of the residues at $s = -k$: $$\mathcal M^{-1}[F] = \sum_{k=0}^\infty \operatorname{Res}_{s = -k} \frac {\Gamma(s) \Gamma(\alpha - s) \Gamma(\beta - s)} {\Gamma(\gamma - s)} x^{-s} = \\ \sum_{k=0}^\infty \frac {\Gamma(\alpha + k) \Gamma(\beta + k)} {\Gamma(\gamma + k)} \frac {(-x)^k} {k!} = \\ \frac {\Gamma(\alpha) \Gamma(\beta)} {\Gamma(\gamma)} {_2F_1}(\alpha, \beta; \gamma; -x).$$

Since both the integral and ${_2F_1}$ are analytic functions of $x$ when $0 < \operatorname{Re} x$, we conclude that the identity holds for $0 < x$, giving your formula.

Answer 2

Additum

Recently I have come across Ramanujan's Master Theorem. This Theorem provides an elegant way to show the given relation. Therefore lets write the Gaussian Hypergeometric Function as infinite power series

$$_2F_1(\alpha,\beta;\gamma;-x)=\sum_{k=0}^{\infty}\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}\frac{\Gamma(\beta+k)}{\Gamma(\beta)}\frac{\Gamma(\gamma)}{\Gamma(\gamma+k)}\frac{(-x)^k}{k!}=\sum_{k=0}^{\infty}\phi(k)\frac{(-x)^k}{k!}$$

For an analytic function $f(x)$ which is in the form of the last sum - especially with some $\phi(k)$ and a negative $x$ argument - the Mellin Transform of this function is given by

$$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\phi(-s)$$

From hereon by plugging in $_2F_1(\alpha,\beta;\gamma;-x)$ as $f(x)$ we get

$$\begin{align} \int_0^{\infty}x^{s-1}~_2F_1(\alpha,\beta;\gamma;-x)dx~&=~\Gamma(s)\phi(-s)\\ &=~\Gamma(s)\frac{\Gamma(\alpha-s)}{\Gamma(\alpha)}\frac{\Gamma(\beta-s)}{\Gamma(\beta)}\frac{\Gamma(\gamma)}{\Gamma(\gamma-s)}\\ &=~\frac{\Gamma(s)\Gamma(\alpha-s)}{\Gamma(\alpha)}\frac{\Gamma(s)\Gamma(\beta-s)}{\Gamma(\beta)}\frac{\Gamma(\gamma)}{\Gamma(s)\Gamma(\gamma-s)}\\ &=~\frac{B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)} \end{align}$$

$$\therefore~\mathcal M [_2F_1(\alpha,\beta;\gamma;-x)] = \frac {B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)}$$