I've seen two definitions of the Mellin transform: more commonly, $g$ is the Mellin transform of $f$ if $$g(s)=\int_0^\infty x^{s-1}f(x)\; dx,$$ or secondly, and more rarely, defined by the same integral with the lower bound at $1$ in place of $0$. The second type seems to be easier to work with in some cases because it avoids a possible singularity at $0$. Now, the inverse Mellin transform is well defined for the first definition. However, for the second, what is the formula for the inverse?
If $$g(s)=\int_1^\infty x^{s-1} f(x)\; dx=\int_0^\infty e^{st}f(e^t)\;dt,$$ then applying the Laplace inversion formula we have $$f(s)=\frac 1{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}s^{-x}g(x)\; dx$$ (for an appropriate constant $\gamma$) which is exactly the same as for the first definition of the Mellin transform. Clearly this is wrong; the two transforms are not identical. What is wrong here and what is the actual formula?