Can integration contour of an inverse Mellin transform be deformed at will in fundamental strip?

Let $m(x)$ be inverse Mellin transform of $M(s)$:

$m(x)=\frac{1}{2\pi i} \int\limits_{c-i\infty}^{c+i\infty}x^{-s}M(s)ds$

Mellin transform $M(s)$ is analytic on fundamental strip $a<\Re(s)<b$ and all the usual conditions of Mellin transformations apply and are satisfied on the fundamental strip. Hence, $c$ is any real number from the fundamental strip: $a<c<b$ and the integration contour $C$ is the straight line $\Re(s)=c$.

$\bf{Question:}$ Can the integration contour $C$ be deformed continuously without self-intersecting at will on the fundamental strip to another curve $\gamma$, or under what conditions, so that the result $\int_\gamma x^{-s}M(s)ds$ remain bounded for all $x$, not necessarily equal to the original function $m(x)$?

Thanks.

• The related question is can: the fourier integral contour $\mathbb{R}$ be deformed without affectiong the result... – anonymous Jun 10 '18 at 13:42