TITLE: Abscissa of Convergence for the Laplace Transform of $f(t)=e^t \sin(e^t)$
QUESTION [1 upvotes]: I am trying to solve the following question: 
Show that the abscissa of convergence for the function $f(t)=e^t \sin(e^t)$ is zero, i.e the unique number $\sigma$ such that the integral $\int_0^\infty e^{-zt}f(t) \, dt$ (the laplace transform of $f$) is converges if $\operatorname{Re}z> \sigma$ and diverges if $\operatorname{Re}z< \sigma$ is $0$. 
I thought that $\sigma=1$ after some comparison but my text book says that it is 0. Any help would be great.

REPLY [1 votes]: You are right that the straightforward comparison 
$$|e^{t-zt} \sin(e^t)|\le e^{t(1-\operatorname{Re}z)}$$
only yields a conclusive result for $\operatorname{Re}z>1$. But let's  integrate by parts: 
$$
\int_1^A e^{t-zt} \sin(e^t)\,dt = -e^{-zt} \cos(e^t)\bigg|_1^A 
- z\int_1^A e^{-zt} \cos(e^t)\,du   
$$
Now use $|e^{-zt} \cos(e^t)|\le   e^{-t \operatorname{Re}z }$.
As $A\to\infty$, both terms have finite limits provided $\operatorname{Re}z>0$.
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