After I did calculations to get the Mellin transform $$\mathcal{M} \left\{ e^{-\lfloor x\rfloor} \right\}(s), $$ where with $\lfloor x\rfloor$ we are denoting the floor function I asked me about if my result was right, and where converges the function.
After that I've combined the definition of Mellin transform and floor function, my result was $$\sum_{k=0}^\infty\frac{k^{s-1}}{e^k},$$ and Wolfram Alpha said to me that it is $\Phi(\frac{1}{e},1-s,0)$, that is a particular value of a Lerch Transcendent.
Question. Can you verify if my result was right? If the result is right only is required a yes here. What is the domain of convergence for my example? How compute the domain of convergence? Many thanks.
I tried also repetead the exercise with the fractional part function $\operatorname{frac}(x)=\{x\}$ instead of the floor function, but it is more difficult. Do you know if it was in the literature?
If $\text{Re}(s) > 0$ (so the integral converges at $0$) $$\eqalign{{\mathcal M}\left\{e^{-\lfloor x \rfloor}\right\}(s) &= \int_0^\infty e^{-\lfloor x \rfloor} x^{s-1}\; dx \cr &= \sum_{k=0}^\infty \int_k^{k+1} e^{-k} x^{s-1}\; dx \cr &= \sum_{k=0}^\infty e^{-k} \frac{(k+1)^{s}-k^s}{s}\cr}$$ The series converges absolutely, so we can rearrange: $$ \sum_{k=1}^\infty (e^{1-k}-e^{-k}) \frac{k^s}{s} = \frac{e-1}{s} \sum_{k=1}^\infty e^{-k} k^s$$
If you replace $\lfloor x \rfloor$ by $\left\{ x\right\}$, it never converges: $e^{-1} \le e^{-\left\{ x\right\}} \le 1$, and you need $\text{Re}(s) > 0$ for convergence at $0$ but $\text{Re}(s) < 0$ for convergence at $\infty$.