Assume the following definitions:
$\zeta(s)$ - Riemann zeta function
$\delta(x)$ - Dirac delta function
My question is:
What is the theoretical convergence of the following formula for $\zeta(s)$? Does this formula only converge for $Re(s)>1$?
$\zeta(s)=\int_{1-\epsilon}^{\infty}x^{-s}\sum_n \delta(x-n)\ dx$
This is nothing more than the definition of the Dirac delta distribution $\delta$ : $$\int_{-\infty}^\infty \delta(x-a)f(x)dx = \lim_{\epsilon \to 0} \int_{-\infty}^\infty \frac{1_{|x-a| < \epsilon}}{2\epsilon}f(x)dx = f(a)$$ whenever $f$ is continuous at $x=a$.
From this you can say that $$\int_{b}^c \delta(x-a)f(x)dx = \int_{-\infty}^\infty \delta(x-a)f(x)1_{|x-a| < \epsilon}dx=\begin{cases}f(a) \text{ if } a \in (b,c)\\ 0 \text{ if } a \not\in [b,c] \end{cases}$$ (since $f(x)1_{|x-a| < \epsilon}$ is continuous at $x=a$)
Finally by definition of the improper integral $$\int_{1-\epsilon}^\infty \sum_{n=-\infty}^\infty \delta(x-n) x^{-s}dx = \lim_{N \to \infty}\int_{1-\epsilon}^{N+\epsilon} \sum_{n=-\infty}^\infty \delta(x-n) x^{-s}dx$$ $$ = \lim_{N \to \infty}\sum_{n=-\infty}^\infty\int_{1-\epsilon}^{N+\epsilon} \delta(x-n) x^{-s}dx=\lim_{N \to \infty} \sum_{n=1}^N n^{-s}$$ Converging of course to $\zeta(s)$ only when $Re(s) > 1$.
Now you are supposed to ask : so if this is "by definition" and "just a notation" then what is it useful for ?
Well it is interesting because you can generalize this to many other Dirichlet series, you can integrate by parts and get the Abel summation formula, you can use the theory of Fourier series for writing that formally $$\sum_{n=-\infty}^\infty \delta(x-n) = \sum_{n=-\infty}^\infty e^{2i \pi n x}$$ which leads to the functional equation $\zeta(s) = 2^s \pi^{s-1} \sin(\pi s/2) \Gamma(1-s) \zeta(1-s)$, and do many other things.
In other words : "$\zeta(s)$ is the Mellin transform of the Dirac comb $\sum_{n=-\infty}^\infty \delta(x-n)$" is important from the theoretical perspective.