Mellin-Barnes transform of $\frac{1}{\Gamma}$ Let $\varphi \in L_1(c+i\mathbb R)$, where $c > 0$. Then we can define the Mellin-Barnes transform (or the inverse Mellin transform) of the function $\varphi$ by the formula
$$
    \mathcal M_c^{-1} \varphi(x) = \frac{1}{2\pi i} \int\limits_{c-i\infty}^{c+i\infty} x^{-z} \varphi(z) \, dz, \quad x \geq 0.
$$
As in the case of the Fourier transform the Mellin-Barnes transform can be continued to the space of linear continuous functionals on smooth rapidly decaying functions on $c+i\mathbb R$. My question is how to find the  Mellin-Barnes transform of $\varphi(z) = \frac{1}{\Gamma (z)}$.
I tried to use the fact that the inverse Mellin transform of product of two functions is the convolution of their inverse Mellin transforms. The Mellin-Barnes transform of $\Gamma(z)$ is $e^{-t}$, the Mellin-Barnes transform of $1$ is $\delta(t-1)$. Hence the Mellin-Barnes transform of $\frac{1}{\Gamma(z)}$ convoluted with $e^{-t}$ must give $\delta(t-1)$ so that the desired function is the convolutional inverse of $e^{-t}$ where convolution is taken in the sense:
$$
   f*g(s) = \int\limits_0^{+\infty} f(t)g(st^{-1})t^{-1}dt. 
$$
But I don't know if this can help.
 A: Please double check:
Gradshteyn/Ryzhik  6.4.23 3 with $\alpha = 0$;
$$\int_{0}^{\infty}{\frac{x^{1-s}}{\Gamma(x+1)}dx}=\mu(1,1-s)\Gamma(-s),$$
where $\text{Re}(s)\leq 2$.
Following Gradshteyn/Ryzhik back we get to https://authors.library.caltech.edu/43491/10/Volume%203.pdf
section 18.3 (page 222), Eq 17 with $s=0,\beta=1-s$ which disagrees with G&R (: Possibly a notational error?)
I think I will drop back and just develop DLMF http://dlmf.nist.gov/5.7.E1 which, in its own way, is also gruesome.
$\frac{1}{\Gamma\left(z\right)}=\sum_{k=1}^{\infty}c_{k}z^{k}$
$(k-1)c_{k}=\gamma c_{k-1}-\zeta\left(2\right)c_{k-2}+\zeta\left(3\right)c_{k-3}$ $-\dots+(-1)^{k}\zeta\left(k-1\right)c_{1},k\geqq3$
For the Mellin transform we must establish the “critical strip”
$\alpha\leqq Re\,s\,\leqq\beta$
Where ${\displaystyle \lim_{z\rightarrow\infty}}\left(\left|\frac{1}{\Gamma\left(z\right)}\right|\right)\leqq\mathcal{O}\left(z^{-\beta}\right),{\displaystyle \lim_{z\rightarrow0}}\left(\left|\frac{1}{\Gamma\left(z\right)}\right|\right)\leqq\mathcal{O}\left(z^{-\alpha}\right)$
Using the asymptotic from DLMF comments http://www.math.sfu.ca/~cbm/aands/
$\Gamma\left(z\right)\sim\sqrt{2\pi}e^{-z+z\cdot ln\left(z\right)-\left(\frac{1}{2}\right)ln\left(z\right)}=\sqrt{2\pi}e^{z\cdot\left(ln\left(z\right)-1\right)-\left(\frac{1}{2}\right)ln\left(z\right)}$
$\frac{1}{\Gamma\left(z\right)}\sim\frac{1}{\sqrt{2\pi}}e^{z-z\cdot ln\left(z\right)+\left(\frac{1}{2}\right)ln\left(z\right)}$
We get $\beta=+\infty$ and from the formula we get $\alpha=0$
Thus, the Mellin xform exists and we can use the correspondance principle
${\displaystyle \mathbb{\mathcal{M}}_{z\rightarrow s}\left(\frac{1}{\Gamma\left(z\right)}\right)=\sum_{k=1}^{\infty}c_{k}\frac{1}{s+k}}$
