Inverse of Mellin transform with lower bound at $1$ 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?
g(s) in the first formula and g(s) in second one are not the same.
For instance if f(x)=Exp(-x) the first gives g(s)=Gamma(s) - Euler Gamma and the second gives Incomplete Gamma g(s)=Gamma(s,1). Inversion formula is identical for both, because they are both Mellin transforms. In the first case it gives Exp(-x) from zero to infinity as a result and in the second case we obtain 0 from zero to one and Exp(-x) from one to infinity. For x = 1 we obtain Exp(-1)/2.
g(s) in the first formula and g(s) in second one are not the same.
For instance if f(x)=Exp(-x) the first gives g(s)=Gamma(s) - Euler Gamma and the second gives Incomplete Gamma g(s)=Gamma(s,1). Inversion formula is identical for both, because they are both Mellin transforms. In the first case it gives Exp(-x) from zero to infinity as a result and in the second case we obtain 0 from zero to one and Exp(-x) from one to infinity. For x = 1 we obtain Exp(-1)/2.