Inverse Laplace transform of the Riemann zeta function I wonder whether the inverse Laplace transform of the Riemann zeta function be obtained on the domain $s \in \mathbb{R}_{>1}$. Are any results on this known?
In order to solve the problem, I tried obtaining the inverse Laplace transform of the individual terms comprising the series expression $$\zeta(s) = \sum_{n=1}^{\infty} n^{-s} ,$$
but it seems $\mathcal{L}^{-1}_{x}[n^{-x}](t) $ does not exist. However, I can imagine other expressions for the Riemann zeta function could be used to obtain such a transform - including one of its integral representations.
The underlying motivation for this question is that I'd like to obtain exact evaluations of sums involving the Riemann zeta function by means of (inverse) Laplace transforms, as laid out in, for example, these papers.
 A: $\mathcal{L}$ means the bilateral Laplace transform.
For each $\sigma\ne 1$, $t\mapsto \zeta(\sigma+it)$ is a tempered distribution, so its inverse Fourier transform makes sense in the sense of distributions.

*

*For $\sigma > 1$, $g_\sigma(u)=\sum_{n\ge 1} \delta(u-\log n) n^{-\sigma}$ (series of shifted/weighted Dirac delta) is a tempered distribution and its Fourier transform is clearly $\zeta(\sigma+it)$ so $$\mathcal{L}^{-1}[\zeta(s)]_{\Re(s)=\sigma} = \sum_{n\ge 1} \delta(u-\log n) $$


*For $\sigma \in (0,1)$, $f_\sigma(u)=-e^{u-\sigma u}+\sum_{n\ge 1} \delta(u-\log n) n^{-\sigma}$ is a tempered distribution and its Fourier transform is $\zeta(\sigma+it)$ so
$$\mathcal{L}^{-1}[\zeta(s)]_{\Re(s)=\sigma} =-e^u+ \sum_{n\ge 1} \delta(u-\log n) $$


*$f_\sigma(u)$ stays a tempered distribution for $\sigma\le 0$, this is the tricky part as $\lim_{A\to \infty} \int_{-A}^A f_\sigma(u) e^{-itu}du$ doesn't converge anymore. My approach is to say that (limit in the sense of distributions) $$f_\sigma(u)=\lim_{r\to 0}\frac1{2\pi}\int_{-\infty}^\infty \zeta(\sigma+it) e^{-r t^2} e^{itu} dt$$ Using the Cauchy integral theorem and the polynomial growth of $\zeta(s)$ on vertical strips, it is
$$=\lim_{r\to 0}\frac1{2\pi}\int_{-\infty}^\infty \zeta(\sigma+i(t-i(1/2-\sigma)) e^{-r (t-i(1/2-\sigma))^2} e^{i(t-i(1/2-\sigma))u} dt
= e^{(1/2-\sigma)u} f_{1/2}(u)$$
So that
$$\mathcal{L}^{-1}[\zeta(s)]_{\Re(s)=\sigma} =
\mathcal{L}^{-1}[\zeta(s)]_{\Re(s)=1/2}=
-e^u+ \sum_{n\ge 1} \delta(u-\log n) $$
A: Note that
$$F(s)=s \int\limits_1^\infty f(x)\, x^{-s-1}\,dx=s \int\limits_0^\infty f(e^x)\, e^{-x s}\,dx=s\, \mathcal{L}_x\left[f\left(e^x\right)\right](s)\tag{1}$$
so
$$\mathcal{L}_s^{-1}\left[\frac{F(s)}{s}\right](x)=f(e^x)\tag{2}$$
and for
$$f(x)=\lfloor x\rfloor=\sum\limits_{n=1}^x 1\tag{3}$$
which can also be evaluated as
$$f(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \theta(x-n)\right)\tag{4}$$
assuming the Heaviside step function $\theta(x)$ is defined as
$$\theta(x)=\left\{\begin{array}{cc}
 0 & x<0 \\
 1 & x\geq 0 \\
\end{array}\right.\tag{5}$$
then
$$F(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{1}{n^s}\right)=\zeta(s),\quad\Re(s)>1\tag{6}$$
and
$$\mathcal{L}_s^{-1}\left[\frac{F(s)}{s}\right](x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\theta(e^x-n)\right)=\sum\limits_{n=1}^{e^x} 1=\lfloor e^x\rfloor\,.\tag{7}$$

This answer illustrates the relationship between the Laplace transform and the Riemann zeta function $\zeta(s)$. If the motivation for this question is evaluation of $\zeta(s)$ at integer values of $s$, then for negative integer values $s=-n$ one has
$$\zeta(-n)=\frac{(-1)^n\, B_{n+1}}{n+1}\tag{8}$$
and for positive even integer values $s=2 n$ one has
$$\zeta (2 n)=\frac{(-1)^{n+1} (2 \pi)^{2 n}\, B_{2 n}}{2 (2 n)!}\tag{9}$$
where $B_n$ is the Bernoulli_number. The Wikipedia article Particular values of the Riemann zeta function provides additional information.
