I try to prove that $f(z)= \int_0^{+\infty}e^{zt}t^{-t} dt $ is holomorphic.
What I did: Let $\Gamma $ be a closed path in $\mathbb{C}$
If we apply Fubini's theorem on the measurable function $(t,z) \rightarrow e^{zt}t^{-t}$ we have $\int_{\Gamma} \int_0^{+\infty}e^{zt}t^{-t} dt = \int_0^{+\infty} \int_{\Gamma}e^{zt}t^{-t} dt $.
However, $\mathbb{C} $ is convex and $z \rightarrow e^{zt}t^{-t}$ is holomorphic so after Cauchy's theorem $\int_{\Gamma}e^{zt}t^{-t} dt=0$
We deduce that $\int_{\Gamma} f(z) dz = 0 $, then with Morera's theorem, we show that f is holomorphic.
But how to ensure that $(t,z) \rightarrow e^{zt}t^{-t}$ is integrable to use Fubini's theorem properly?
Many thanks.
The integral is improper only at $\infty$, since $$ \lim_{t\to0^+}e^{tz}t^{-t}=1. $$ Moreover, it is uniformly convergent on compact subsets of $\Bbb C$. If $|z|\le R$ and $t\ge2$ $$ \bigl|\,e^{tz}t^{-t}\,\bigr|\le\Bigl(\frac{e^R}{t}\Bigr)^t\le\frac{e^{2R}}{t^2}. $$