I'm trying to give sense to the (self-made) statement:
The analytic continuation of $\int_0^\infty e^t\;t^{s-1}\;dt$ is holomorphic in $s=0$.
At first sight this could seem completely unreasonable, because the integral does not converge for any $s\in\mathbb{C}$. But there is an heuristic argument that suggests the result.
Let $e^t_n:= \sum_{i=0}^n \frac{t^i}{i!}$ denote the $n$-th truncated Taylor series of the exponential. Then $\int_0^\infty e^t_n\;t^{s-1}\;dt$ still is not convergent for any $s\in\mathbb{C}$, but:
\begin{align*} \int_0^\infty e^t_n\;t^{s-1}\;dt= & \sum_{i=0}^n \frac{1}{i!}\int_0^\infty t^{i+s-1}\;dt\\ =& \sum_{i=0}^n \frac{1}{i!}\left(\int_0^1t^{i+s-1}\;dt +\int_1^\infty t^{i+s-1}\;dt\right)\\ =& \sum_{i=0}^n \frac{1}{i!}\left(\frac{1}{s+i}-\frac{1}{s+i}\right) =0, \quad \text{for any }n \text{ in }\mathbb{N}. \end{align*}
Where in the last step we used the analytic continuation $\int_0^1 t^s\; dt= \frac{1}{s+1}$ (resp. $\int_1^\infty t^s\; dt= -\frac{1}{s+1}$) valid for $\Re(s)>-1$ (resp. $\Re(s)<-1$).
This argument, which is a standard one for the Mellin transform, fails for the whole exponential because we cannot exchange the integral with the infinite and divergent series of polynomials defining the exponential.
Any idea on how to tackle this problem? Or I'm trying to do something completely out of the point?
Thank you in advance!