I have some very serious question (which truly isn't a joke) about convergence of definite integral in the form of $\int_{x}^{\infty}f (t)dt$, and its analytic continuation. To start, let's say that $f (t)$ is analytic.
Question: isn't it a sufficient condition to make that mystic analytic continuation (in most, if not all of cases), just by assuming that integral is convergent at infinity (even if that's not true)?
What I mean is, instead of $\int_{x}^{\infty}f (t)dt $ just write $-F (x)$
And thats realy works. Just try it for example on Abel-Plana formula. In case of Abel-Plana formula, if we assume that series of some function is convergent, its integral have to be convergent too. Just watch. $$ \begin{split} \sum_{k=x}^{\infty}f (k) &= \lim\limits_{y \rightarrow \infty} \Bigg [\int_{x}^{y}f (t)dt+\frac {1}{2}f (x)+\frac {1}{2}f (y)\\ &\qquad\qquad+\int_{0}^{\infty} \frac{f (x+it)-f (x-it)-f (y+it)+f (y-it)}{e^{2\pi t}-1} dt \Bigg]. \end{split} $$ Now I assume that for $\lim\limits_{y \rightarrow \infty} $ integral $\int_{x}^{\infty}f (t)dt$ is convergent (so $f (\infty) $ too), independetly about fact if that's true. $$ \sum_{k=x}^{\infty}f (k)=-F(x)+\frac {1}{2}f (x)+\int_{0}^{\infty} \frac{f (x+it)-f (x-it)}{e^{2\pi t}-1} dt. $$ I can't say if it always works, but if you try this on some divergent series (and yes, also on $\zeta(s) $, for $ s \in \mathbb {C}$ except $s=1$) it works fine. There is only some problem when $x$ crosses singularity.
