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

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.

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.

• First of all thank you for the answer and within it the wanted proof. Now I can be sure, that the formula is right but that was not what I desired. I have to reformulate my original question to clarify it : I would be really interested in a derivation of the formula, so that I can see how to get from the Mellin Integral of the Gaussian Hypergeometric Function to the terms of the Beta Function. – mrtaurho Jul 3 '18 at 16:37
• Generally speaking, the Mellin transform of the Meijer G-function is a rational combination of linear gamma functions, and ${_2F_1}$ is a special case of the G-function. You can write ${_2F_1}$ as the integral of $t^{\alpha-1} (1-t)^{\gamma-\alpha-1} (1+x t)^{-\beta}$ (assuming that the integral converges) and change the order of integration, then integrating wrt $x$ will again give an elementary function of $t$, but it'll require some calculations. – Maxim Jul 3 '18 at 18:14
• I am not that familiar with the Meijer G-function. Could you maybe add this calculation you were speaking of? It would be really interesting for me to see how to do this. – mrtaurho Jul 4 '18 at 15:32
• With the change of variables $x=(1/\xi-1)/t$, $${_2F_1}(\alpha,\beta;\gamma;-x) = \frac {\Gamma(\gamma)} {\Gamma(\alpha) \Gamma(\gamma-\alpha)} \int_0^1 t^{\alpha-1} (1-t)^{\gamma-\alpha-1} (1+xt)^{-\beta} dt, \\ \int_0^\infty x^{s-1} (1+xt)^{-\beta} dx = t^{-s} \int_0^1 \xi^{-s-1+\beta} (1-\xi)^{s-1} d\xi = t^{-s} B(s, \beta-s),\\ \int_0^1 t^{\alpha-s-1} (1-t)^{\gamma-\alpha-1} dt = B(\gamma-\alpha, \alpha-s).$$ I don't think this way is better; we're still using an integral representation of ${_2F_1}$, but one that's not directly related to the Mellin transform. – Maxim Jul 5 '18 at 13:26
• I think Temme's Special Functions: An Introduction to the Classical Functions of Mathematical Physics is quite good. – Maxim Jul 7 '18 at 15:54

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}

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