I would like to learn to calculate inverse Mellin tranforms. My example (!) is for the following entire function $$\varphi(s)=\frac{1-e^{2s}}{e^{-s}-1},$$ (see if you want the Wikipedia's entry for the Faulhaber's formula) where $s=\sigma+it$ is our complex varible.
Wikipedia's article for Mellin inversion theorem is saying to me that is required some conditions with the purpose to apply the Method for my analytic function (if it is possible for the example (!) that I am exploring: that is if our function satisfies such conditions in this inversion theorem).
Question. It's possible to apply the Method explained to the function $\varphi(s)$ in a strip? Many thanks.
I believe that $\varphi(s)=\frac{1-e^{2s}}{e^{-s}-1}$ doesn't tend to zero uniformly as $\Im s\to\pm\infty$, but currently I don't know how do a mathematical reasoning (I believe that the supremum of the complex modulus is bounded in each strip independently of the imaginary part), thus I believe that it is not possible apply the Method. What are the conditions that fail?
$1-e^{2s} = (1-e^s)(1+e^{s})$ thus $$\varphi(s) = e^s \frac{1-e^{2s}}{1-e^s} = e^s(1+e^s) = e^s + e^{2s}$$ and it is the Mellin transform of the (tempered) distribution $$T(x) = e\, \delta(x-e)+e^2\,\delta(x-e^2)$$ i.e. $\varphi(s) = \int_0^\infty x^{s-1}T(x)dx$ And the procedure for computing the inverse Mellin transform of the Mellin transform of a (tempered) distribution is more complicated than a contour integral. You should look instead at the Fourier transform of Schwartz functions and tempered distributions.