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.