To prove a function is analytic on the upper half plane.

Question

The map $z\mapsto \int_0^\infty e^{itz}dt$ is well defined on the upper half plane $\mathbb{H}=\{z\in\mathbb{C}: \Im{z}>0\}.$ Is it also analytic on $\mathbb{H}?$

I tried to prove using Morera's theorem but failed.

Answer 1

$\int_0^{\infty} \frac {e^{it(z+h)}-e^{itz}} h \to \int_0^{\infty} ite^{itz} dt$ by an application of DCT: $|e^{it(z+h)}-e^{itz}| =e^{-t\Im z} |e^{ith}-1|$ and $|e^{ith}-1| \leq e^{t|h|-1} $. Note that for $t \neq 0$ and $|h|<\frac 1 {|t|}$, $\frac { e^{t|h|-1}} {|h|} $ is bounded and $e^{-t\Im z}$ is intergable on $(0,\infty)$. The argument for $t=0$ is similar.
[$\frac {e^{x}-1 }x$ is bounded on any bounded subset of $\mathbb R \setminus \{0\}$]