0

It is known from results about differentiation under integral sign that $\int_0^\infty e^{-zt} f(t) dt$ is holomorphic on the right half-plane given any integrable function $f.$ See Show that integral is analytic for a proof.

Now I see in an exercise the converse: Let $f:[0,\infty] \to \mathbb R$ be bounded and locally integrable. Suppose that there exist a analytic function $F(z)$ defined on an open set containing $\Re z \geq 0$ such that $F(z)=\int_0^\infty e^{-zt} f(t) dt$ for $\Re z >0.$ Then $\int_0^\infty f(t) dt$ converges.

To prove this, I try to use the fact that $\int_0^\infty |f(t)| e^{-\Re zt} dt$ is finite. By monotone convergence theorem, the integral converges to $\int_0^\infty |f(t)| dt$ as $z \to 0^+.$ So what we need to show is that the limit does not blow up, for $f$ to be uniformly integrable.

I find that there are some ambiguity as to how we interpret this integral. $\int_0^\infty$ can be either mean the lebesgue integral $\int_{\mathbb R}$ itself or the improper integral $\lim_{T \to \infty} \int_{[0,T]},$ and the convergence of these two are of course not equivalent (which means the final result we are going to prove is not entirely clear). And the improper integral is harder to work with, with more tricky differentiation and things like that.

So, what is the interpretation of $\int_0^\infty$ and how can I prove the result?

Ma Joad
  • 7,696
  • It is not true that $\int_0^\infty |f(t)|dt$ converges, try with $f(t)=\sin(e^t)$ then $F(z)$ is entire. It is a Tauberian theorem, not obvious at all. – reuns Apr 03 '21 at 19:58
  • @reuns So it should be interpreted as an improper integral (with the process of taking a limit)? – Ma Joad Apr 03 '21 at 20:42
  • Yes, but you will hardly find anything harder than such Tauberian theorem in the context of Laplace transforms, you should do some research, follow a textbook and see what you need. Clearly not an "exercice". The much more trivial case is when $f\ge 0$, then $F$ has a singularity at $\sigma$ the abscissa of convergence of $\int_0^\infty e^{-zt} f(t) dt$. – reuns Apr 03 '21 at 20:48

0 Answers0