Mellin transform of powers of gamma function

Question

If $a>0$, the Cahen-Mellin integral gives $$\DeclareMathOperator{\Res}{\operatorname{Res}} \frac{1}{2\pi i}\int\limits_{a-i\infty}^{a+i \infty}\varGamma(z) u^{-z}dz=e^{-u}=\sum\limits_{m=0}^{\infty}\Res_{z=-m}\Gamma(z)u^{-z} $$ My question is the following: if $k$ $\in$ $\mathbb{N}$, $$ \frac{1}{2\pi i}\int\limits_{a-i\infty}^{a+i \infty}\varGamma^k(z) u^{-z}dz=\sum\limits_{m=0}^{\infty}\Res_{z=-m}\Gamma^k(z)u^{-z}=f(u) $$ Does the series converge for all $u$? What kind of function $f(u)$ (increasing, decreasing) do I have?

Answer 1

For $k\in\mathbb{N}$ one has the inverse Mellin transform $$\frac{1}{2\pi i}\int\limits_{a-i\infty}^{a+i \infty}\varGamma^k(z) u^{-z}dz=G_{0,k}^{k,0}\left(u\left| \begin{array}{c} 0^{\otimes k} \\ \end{array} \right.\right),$$ with $G$ the Meijer-G function and $0^{\otimes k}$ is the string $0,0,0\ldots 0$ of length $k$. For $k=1$ this is the exponential $e^{-u}$ and for $k=2$ this is a Bessel function (modified Bessel function of the second kind), $$\frac{1}{2\pi i}\int\limits_{a-i\infty}^{a+i \infty}\varGamma^2(z) u^{-z}dz=2 K_0\left(2 \sqrt{u}\right).$$ I don't know of a more explicit representation of the Meijer-G function for $k\geq 3$. Some plots suggest it is a decaying function of $u>0$ for all $k$.