Laplace transformation for floor(exp(x))

Question

I have seen a question where I was asked to calculate the Laplace transformation for floor(exp(x)). Now, I have tried dividing floor(exp(x)) into Heaviside functions in order to calculate the sum of integrals, however, I don't know what each interval is equal to. I would be grateful to be given any further hints to be able to solve it.

Answer 1

The answer is $$\frac{\zeta(s)}s$$

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Derivation: $$\begin{align} F(s):=\int^\infty_0\operatorname{floor}(e^x)e^{-sx}dx &=\int^\infty_1\operatorname{floor}(t)t^{-s-1}dt\qquad t=e^x \\ &=\sum^\infty_{n=1}n\int^{n+1}_n t^{-s-1}dt \\ &=\frac1s\sum^\infty_{n=1}n\bigg(n^{-s}-(n+1)^{-s}\bigg) \\ &=\frac1s\sum^\infty_{n=1}\bigg(n^{-s+1}-(n+1-1)(n+1)^{-s}\bigg) \\ &=\frac1s\sum^\infty_{n=1}\bigg(\underbrace{n^{-s+1}-(n+1)^{-s+1}}_{\text{telescopic sum}}+(n+1)^{-s}\bigg) \\ &=\frac1s\left(1+\sum^\infty_{n=1}(n+1)^{-s} \right)\\ &=\frac1s\bigg(1+\zeta(s)-1\bigg)\\ &=\color{red}{\frac{\zeta(s)}{s}} \end{align} $$ for $\text{Re }s>1$.

This abscissa is expected, as the integrand behaves like $e^{-(s-1)x}$ as $x\to\infty$. Noteworthily, $F(s)$ admits a meromorphic continuation to $\mathbb C$, with simple poles at $0$ and $1$ and residues $-\frac12$ and $1$ respectively.