Originally Posted by

**Laurent** Consider again $\displaystyle X(s)=\int_0^\infty \frac{e^{-st}}{1+t^2}dt$. We have $\displaystyle \sigma_0=0$. And we have a uniform bound: for any $\displaystyle s\in\mathbb{C}$ such that $\displaystyle \Re(s)\geq 0$, we have $\displaystyle \left|\frac{e^{-st}}{1+t^2}\right|=\frac{e^{-\Re(st)}}{1+t^2}\leq \frac{1}{1+t^2}$. Thus, $\displaystyle |X(s)|\leq \int_0^\infty\frac{dt}{1+t^2}<\infty$ is bounded on this half-plane!

**I can't see my mistake; this would mean that this more precise conclusion is again wrong (?)... **

If $\displaystyle X(\alpha_0)=+\infty$, this is however correct, and simple. It suffices to consider limits along the real axis. Indeed, $\displaystyle s\mapsto X(s)$ decreases, therefore either it diverges or it is bounded toward $\displaystyle \sigma_0$. Assume by contradiction that $\displaystyle X(s)\leq M$ for all $\displaystyle s>\sigma_0$. Then, for all $\displaystyle A>0$, for all $\displaystyle s>\sigma_0$, $\displaystyle \int_0^A f(t)e^{-st}dt\leq M$. For a fixed $\displaystyle A$, we can take the limit $\displaystyle s\to\sigma_0$, and get: $\displaystyle \int_0^A f(t)e^{-\sigma_0 t}dt\leq M$. Therefore, $\displaystyle \int_0^\infty f(t) e^{-\sigma_0 t} dt\leq M <\infty$, in contradiction with $\displaystyle X(\alpha_0)=+\infty$.

This is not correct, since we don't know a priori that we can expand in power series at $\displaystyle \sigma_f$ (and a posteriori we can't). In order to reduce to what I did for $\displaystyle \sigma=-1$, you can look at $\displaystyle F(\sigma+1+s)$, i.e. change $\displaystyle f$ into $\displaystyle f(t)e^{-(\sigma+1)t}$. Then the abcissa is -1.

Or, plainly, it corresponds to expanding in power series at $\displaystyle \sigma+1$, where the radius of convergence is 1. Inside the circle of center $\displaystyle \sigma+1$ and radius 1 (and only there), $\displaystyle F(s)=\sum_{k\in\mathbb{N}_{0}}\bigg(\frac{(-1)^{k}}{k!}\int_{0}^{\infty}f(t)t^{k}\mathrm{e}^{-(\sigma_{f}+1)t}\mathrm{d}t\bigg)(s-(\sigma_{f}+1))^{k}$ and you continue the same: I want the singularity on the right of the circle, hence the change of sign, which amounts here to changing $\displaystyle s-(\sigma_{f}+1)$ into the opposite, as you did.