# Inverse Mellin transform of $f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}$

Given the function $$f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)},$$

can we find and inverse Mellin transform for $f(s)$? That is, $$\frac{1}{2 \pi i}\int_{- i\infty}^{ i\infty} 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)} x^{-s-1 } ds$$ for $x>0$.

I was wondering if the integral can be expressed in terms of hypergeometric functions? For example, this is very similar to the Mellin–Barnes integral \begin{align} {}_2F_1(a,b;c;z) =\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\,ds \end{align}

However, I am not sure how to connect it to my problem!

Thanks.

• this is related to Mellin-Barnes integrals – tired Jan 10 '17 at 17:26
• @tire I was thinking about it, but we have a factor of $2^{ \frac{it+1}{6}}$ in front. How to deal with that? – Boby Jan 10 '17 at 17:28
• can't this just be absorbed into the exponential from the ILT? – tired Jan 10 '17 at 17:42
• @tired am not very sure what ILT is? – Boby Jan 10 '17 at 18:56
• @tired when you have time can you put more details on your thoughts about Mellin-Barnes integral? – Boby Jan 11 '17 at 14:12

Using residues, this isn't very hard. Essentially

$$g(x) = \frac{1}{2\pi i}\int x^{-s} f(s)\,ds = \sum_{k} Res_{s=s_k} x^{-s}f(s)$$

where the $s_k$ are the poles of $\Gamma(2s/3)$ (which occur when $s = -3k/2$ for $k \ge 0$) which aren't cancelled by the zeroes of $\Gamma(s/2)$ (which occur when $s = -2k$). Since the poles are simple, the residues aren't too hard to find.

The poles have principal part

$$\frac{3}{2}\frac{(-1)^n}{n!(s+3n/2)}$$

Therefore

$$g(x) = \sum_{n=0}^\infty \frac{3}{2}\frac{(-1)^n}{n!\Gamma(-3n/4)}2^{(1-3n/2)/6}x^{3n/2}$$

That should do it. Some of these terms disappear because $\frac{1}{\Gamma(-3n/4)}$ sometimes vanishes, I'm too lazy to siphon out the terms that do or don't appear :). Note this is an analytic continuation, and the functions $g(\sqrt[3]{x^2})$ and $g(x^2)$ are entire in $x$.

• Thanks. I was thinking that there is an expression in terms of hypergeometric functions? – Boby Jan 10 '17 at 20:59
• That's a good question. I despise hypergeometric functions and how clunky and ugly they are, so I got nothing to add there, lol. – user335907 Jan 10 '17 at 21:44
• I looked around and seems that your approach might be the best. Can you explain to me a few things? 1) Can you tell me about polls of $\Gamma()$ why are they occurring at $s=-3k/2$ ? – Boby Jan 13 '17 at 15:08
• $\Gamma(s)$ has poles at the non positive integers $s=-n$ for $n \ge 0$, Their principal part is $\frac{(-1)^n}{n!(n+s)}$. Therefore when you take $\Gamma(2s/3)$ its principal parts are $\frac{(-1)^n}{n!(n+2s/3)}$, which when written in a more manageable form equals $\frac{3}{2}\frac{(-1)^n}{n!(s+3n/2)}$. I can't really say it simpler, all you need to do is to read a little bit up on $\Gamma$ function. – user335907 Jan 13 '17 at 17:12
• Thanks. Now I understand this part. Can you give more details on how you compute the Residue? – Boby Jan 13 '17 at 18:29

Let me offer a partial solution, which I might finish later. Repeatedly using the Gauss multiplication formula yields the following identity:

$$2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}=\frac1{\sqrt{\pi } \sqrt[6]{2}}\frac{\Gamma \left(\frac{s}{6}+\frac{11}{12}\right) \Gamma \left(\frac{s}{6}+\frac{2}{3}\right) \Gamma \left(\frac{s}{6}+\frac{5}{12}\right)}{\Gamma\left(\frac{s}{6}+\frac{1}{2}\right) \Gamma \left(\frac{s}{6}+\frac{5}{6}\right)}\left(\sqrt{\frac38}\right)^{-s}$$

and now the inverse Mellin transform can be directly converted into a Meijer $G$-function, yielding

$$\frac{3\sqrt[6]{32}}{\sqrt{\pi}}G_{2,3}^{3,0}\left(\frac{27z^6}{512}\middle| {{\frac12,\frac56}\atop{\frac{5}{12},\frac23,\frac{11}{12}}}\right)$$

This can then be further expanded into a sum of $3$ ${}_2 F_2$ hypergeometric functions, using this formula.