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. 
 A: 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 $x$.
A: 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.
