Gamma Identities from Inverse Transform I have been working on a transform (the notes are a little rough and the hosting website has some fraction formatting issues recently) but the idea is there:
Basically, the transform $\mathcal{I}_x[f(x)](s)$, will extract the coefficients of the scaled inverse function, $f^{-1}(x)/x$, in a manner analogous to the Mellin transform.
I have found a pattern that seems to generate Gamma function identities that Mathematica cannot simplify or verify. This was done by acting on simple polynomial expressions for which I previously found that the principal inverse (via series reversion) was a generalised hypergeometric function which led to the following results:
\begin{align}
\mathcal{I}[x+x^2](s) = \frac{\Gamma(1-\frac{2s}{1})\Gamma(\frac{s}{1}) }{\Gamma(2-s)}\\
\mathcal{I}[x+x^3](s) = \frac{\Gamma(1-\frac{3s}{2})\Gamma(\frac{s}{2})}{2 \Gamma(2-s)} \\
\mathcal{I}[x+x^4](s) = \frac{8 \cdot 3^{s-\frac{5}{2}}\pi \Gamma(-4s/3)\Gamma(s/3)}{\Gamma(2/3-s/3)\Gamma(4/3-s/3)\Gamma(-s/3)} \stackrel{?}{=}\frac{\Gamma(1-\frac{4s}{3})\Gamma(\frac{s}{3})}{3 \Gamma(2-s)}
\end{align}
the last one appears to be true, but Mathematica won't simplfy it.
In short the generalised conjecture is that the Mellin transform of this generalised hypergeometric function has a very simple result
$$
\mathcal{M}_x\left[\;_{(m-1)}F_{(m-2)}\left(\left\{\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}{m}\right\};\left\{\frac{2}{m-1},\cdots,\frac{m-2}{m-1},\frac{m}{m-1}\right\};-\frac{m^mx^{m-1}}{(m-1)^{m-1}}\right)\right](s) = \frac{\Gamma\left(1-\frac{m s}{m-1}\right)\Gamma\left(\frac{s}{m-1}\right)}{(m-1)\Gamma(2-s)} = \mathcal{I}_x[x+x^m](s)
$$
which only makes sense for integer $m$. Any way to prove this general statement? How do we simplify the identity that Mathematica couldn't?
 A: For the Gamma function identity, we use the multiplication formula for the Gamma function, choosing $n=3, z=\frac{2}{3}-\frac{s}{3}$:
\begin{align}
\Gamma\left(nz\right)&=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right)\\
\Gamma\left(2-s\right)&=\frac{1}{2\pi}3^{3/2-s}\Gamma\left( \frac{2}{3}-\frac{s}{3} \right)\Gamma\left(1-\frac{s}{3} \right)\Gamma\left( \frac{4}{3}-\frac{s}{3} \right)
\end{align}
Now, with the functional relation $ \Gamma\left( 1+u \right)=u\Gamma(-u) $ applied with $u=-s/3$ and $u=-4s/3$
\begin{equation}
\frac{\Gamma(1-\frac{4s}{3})\Gamma(\frac{s}{3})}{3 \Gamma(2-s)}=\frac{8\pi\Gamma(-\frac{4s}{3})\Gamma(\frac{s}{3})}{3^{5/2-s}\Gamma\left( \frac{2}{3}-\frac{s}{3} \right)\Gamma\left(-\frac{s}{3} \right)\Gamma\left( \frac{4}{3}-\frac{s}{3} \right)}
\end{equation}
as expected.
For the Mellin transform question, we evaluate the inverse transform of the function
\begin{equation}
f(s)=\frac{\Gamma(1-\frac{ms}{m-1})\Gamma(\frac{s}{m-1})}{ \Gamma(2-s)}
\end{equation}
assuming that the domain of analicity is $0<\Re (s) <\frac{m-1}{m}$.  It can be noticed that the poles in the left complex half-plane are $s_p=-p(m-1)$, with $p=0,1,2\ldots$ Their residues are
\begin{equation}
r_p=\frac{(m-1)\Gamma(mp+1)}{\Gamma\left(2+p(m-1) \right)}\frac{(-1)^p}{p!}
\end{equation}
In the following we use the multiplication formula above with $z=1/n$,
\begin{equation}
\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)=(2\pi)^{(n-1)/2}n^{-1/2}
\end{equation}
to write
\begin{align}
\Gamma\left(np\right)&=(2\pi)^{(1-n)/2}n^{np-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(p+\frac{k}{n}\right)\\
&=(2\pi)^{(1-n)/2}n^{np-(1/2)}\Gamma(p)\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)\left( \frac{k}{n} \right)_p\\
&=n^{np-1}\Gamma(p)\prod_{k=1}^{n-1}\left( \frac{k}{n} \right)_p
\end{align}
Using the functional relation this residue can be converted into a ratio of Gamma functions:
\begin{align*}
r_p&=\frac{mp(m-1)}{p(m-1)\left( 1+p(m-1) \right)}\frac{\Gamma(mp)}{\Gamma\left(p(m-1) \right)}\frac{(-1)^p}{p!}\\
&=\frac{m}{\left( m-1 \right)\left(p+\frac{1}{m-1} \right)}
\frac{m^{mp-1}\Gamma(p)\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{(m-1)^{(m-1)p-1}\Gamma(p)\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\frac{(-1)^p}{p!}\\
&=\frac{\Gamma\left(p+\frac{1}{m-1} \right)}{\Gamma\left(p+1+\frac{1}{m-1} \right)}
\frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}\\
&=\frac{\left(\frac{1}{m-1} \right)_p\Gamma\left( \frac{1}{m-1} \right)}{\left(1+\frac{1}{m-1} \right)_p\Gamma\left(1+ \frac{1}{m-1} \right)}
\frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}\\
&=\left( m-1 \right)\frac{\left(\frac{1}{m-1} \right)_p}{\left(1+\frac{1}{m-1} \right)_p}
\frac{\prod_{k=1}^{m-1}\left( \frac{k}{m} \right)_p}{\prod_{k=1}^{m-2}\left( \frac{k}{m-1} \right)_p}
\left(- \frac{m^m}{(m-1)^{m-1}} \right)^{p}
 \end{align*}
The inverse transform can thus be expressed as a hypergeometric series:
\begin{align*}
\mathcal{M}_x^{-1}\left[ \frac{1}{m-1}f(s)\right]&=\frac{1}{m-1}\sum_{p=0}^\infty r_px^{p(m-1)}\\
&=\;_{(m-1)}F_{(m-2)}\left(\left\{\frac{1}{m},\frac{2}{m},\cdots,\frac{m-1}{m}\right\};
\left\{\frac{2}{m-1},\cdots,\frac{m-2}{m-1},\frac{m}{m-1}\right\};-\frac{m^mx^{m-1}}{(m-1)^{m-1}}\right)\\
\end{align*}
which is the proposed identity.
