Help with the integral $\int_{0}^{\infty}\frac{x^{s}}{\Gamma(s)}s^{z-1}ds$ Consider the integral : 
$$\int_{0}^{\infty}\frac{x^{s}}{\Gamma(s)}s^{z-1}ds$$
Where $x\in \mathbb{R}^{+}$  , $z\in \mathbb{C}$
Using the Hankel contour representation of the reciprocal gamma function, we have:
$$\int_{0}^{\infty}\frac{x^{s}}{\Gamma(s)}s^{z-1}ds=\frac{i}{2\pi}\oint \int_{0}^{\infty} \left(-\frac{x}{t}\right)^{s}s^{z-1}e^{-t}dsdt$$
$$=\Gamma(z)\frac{i}{2\pi}\oint \left(-\log\left(-\frac{x}{t}\right)\right)^{-z}e^{-t}dt$$
But i have no idea on how to do this integral. any help is highly appreciated
 A: It looks like Ramanujan's master theorem (see [1] pg. 298-300). Instead of Ramanujan's formula we use Hardy's theorem which is also in [1] pg. 299-300. 
THEOREM 1. (Ramanujan-Hardy)
Let $s=\sigma+it$, $\sigma,t$ both real. Let $H(\delta)=\{s:\sigma\geq-\delta\}$, $0<\delta<1$. If $\psi(s)$ is analytic on $H(\delta)$ and exist constants $C,P,A$, with $A<\pi$ such that
$$
|\psi(s)|\leq Ce^{P\sigma+A|t|}\textrm{, }\forall s\in H(\delta), 
$$
for $x>0$ and $0<c<\delta$, we define
$$
\Psi(x)=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\pi}{\sin(\pi s)}\psi(-s)x^{-s}ds.
$$ 
If $0<x<e^{-P}$, then
$$
\Psi(x)=\sum^{\infty}_{k=0}\psi(k)(-x)^k.
$$
For $0<\sigma<\delta$, we have
$$
\int^{\infty}_{0}\Psi(x)x^{s-1}dx=\frac{\pi}{\sin(\pi s)}\psi(-s).
$$
Here (in your case) if we set 
$$
\psi(x)=\frac{\phi(x)}{\Gamma(x+1)},
$$ 
and
$$
\Psi(x)=\frac{a^x}{\Gamma(x)}=\sum^{\infty}_{k=0}\psi(k)(-x)^k=\sum^{\infty}_{k=0}\frac{\phi(k)}{k!}(-x)^k,
$$
then
$$
\int^{\infty}_{0}\frac{a^x}{\Gamma(x)}x^{s-1}dx=\frac{\pi}{\sin(\pi s)}\psi(-s)=\frac{\pi}{\sin(\pi s)}\frac{\phi(-s)}{\Gamma(1-s)}
$$
Using now the formula
$$
\frac{\pi}{\sin(\pi s)\Gamma(1-s)}=\Gamma(s),
$$
we arrive to
$$
I(s):=\int^{\infty}_{0}\frac{a^x}{\Gamma(x)}x^{s-1}dx=\Gamma(s)\phi(-s).
$$
Hence if $(Mf)(s)$ is the Mellin transform of $f$, then
$$
\left(Mf\right)(s)=\int^{\infty}_{0}f(x)x^{s-1}dx
$$
and
$$
\left(M\Psi\right)(s)=I(s)=\int^{\infty}_{0}\frac{a^x}{\Gamma(x)}x^{s-1}dx
=\Gamma(s)\phi(-s).
$$
But from 2 we have the following
THEOREM 2. (for the conditions see 2)
$$
\int^{\infty}_{-\infty}\left(M\Psi\right)(\sigma+it)f(t)dt=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}f(i(\sigma+k)).
$$
Hence in our case with $\Psi(t)=a^t/\Gamma(t)$ and $f(t)=e^{-itx}(M\Psi)(\sigma-it)$, we have 
$$
\int^{\infty}_{-\infty}(M\Psi)(\sigma+it)f(t)dt=
$$
$$
=\int^{\infty}_{-\infty}(M\Psi)(\sigma+it)e^{-itx}(M\Psi)(\sigma-it)dt=
$$
$$
=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}f(i(\sigma+k))=
$$
$$
=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}e^{-ii(\sigma+k)x}\Gamma(\sigma-ii(\sigma+k))\phi(-\sigma+ii(\sigma+k))=
$$
$$
=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}e^{(\sigma+k)x}\Gamma(2\sigma+k)\phi(-2\sigma-k).
$$
Hence with $\sigma=1/2$:
$$
\int^{\infty}_{-\infty}\left|(M\Psi)\left(\frac{1}{2}+it\right)\right|^2e^{-itx}dt=2\pi e^{x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}\Gamma(k+1)\phi(-k-1)e^{kx}.
$$
Hence we can write
$$
\int^{\infty}_{-\infty}\left|(M\Psi)\left(\frac{1}{2}+it\right)\right|^2e^{itx}dt=2\pi e^{-x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}\left(\int^{\infty}_{0}\frac{a^t}{\Gamma(t)}t^{k}dt\right)e^{-kx}\textrm{, }x>0
$$
Which is a sampling formula, recovering the absolute value of the Mellin transform of $\Psi(x)$, from the values of $(M\Psi)(x)$ at $x=k+1$, where $k$ belongs to the non negative integers $k=0,1,2,\ldots$. As someone can see the result can be generalized very easily for $\Psi(x)$ analyitic around $0$ and entire in $\textbf{C}$ such that $\int^{\infty}_{0}|\Psi(t)t^{k}|dt<\infty$.
NOTES.
1) The general formula that rises is
$$
\int^{\infty}_{-\infty}\left|(M\Psi)\left(\frac{1}{2}+it\right)\right|^2e^{itx}dt=2\pi e^{-x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}(M\Psi)(k+1)e^{-kx}\textrm{, }x>0
$$
As an example of evaluation take $\Psi(x)=e^{-x}$. Then
$$
\int^{\infty}_{0}e^{-t}t^{k}dt=k!.
$$
Hence we get the next integral
$$
\int^{\infty}_{-\infty}\left|\Gamma\left(\frac{1}{2}+it\right)\right|^2e^{itx}dt=2\pi\frac{e^{x/2}}{e^x+1}\textrm{, }x>0.
$$
2)
Also from the inverse Fourier theorem with $\Psi(x)=\frac{a^x}{\Gamma(x)}$, then
$$
\left|(M\Psi)\left(\frac{1}{2}+iw\right)\right|^2=\left|\int^{\infty}_{0}\frac{a^t}{\Gamma(t)}t^{-1/2+iw}dt\right|^2=
$$
$$
=\int^{\infty}_{-\infty}e^{-x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}\left(\int^{\infty}_{0}\frac{a^t}{\Gamma(t)}t^{k}dt\right)e^{-kx}e^{-i x w}dx
$$
[1]: Bruce. C. Berndt. "Ramanujan's Notebooks Part 1". Springer-Verlang. New York, Berlin, Heidelberg, Tokyo. 1985.
2: N.D. Bagis."Numerical Evaluations of Functions Series and Integral Transforms with New Sampling Methods". Thesis. Aristotele University of Thessaloniki, Greece (2007), (in Greek from Researchgate here)
