Show that $f(z)=\int_0^\infty {e^{zt}\over t+1}dt$ with $\Re (z)<0$ is holomorphic

Let $$f:D:=\{z\in\mathbb{C}:\Re(z)<0\}\to\mathbb{C}$$ defined by $$z\mapsto\int_0^\infty{e^{zt}\over t+1}dt$$. Show that $$f$$ is holomorphic.

In my solution I'm not using that the domain of $$f$$ satisfies $$\Re(z)<0$$ which is make me doubt in this solution. My question is where is the mistake in the solution and what is the right way? Thanks.

Attampt:

Let $$\Gamma$$ be a boundary of rectangle $$M\subset D$$. $$\int_\Gamma f(z)dz=\int_\Gamma\int_0^\infty{e^{tz}\over t+1}dtdz \\ =\int_0^\infty{1\over t+1}(\int_\Gamma e^{tz}dz)dt=\int_0^\infty{1\over t+1}\cdot 0\cdot dt=0$$ By Morera's theorem $$f$$ is analytic in $$D$$.

• Your solution is good as far as $\Re(z)<0$. Indeed, this insure you that your function is well defined. Indeed, if $\Re(z)\geq 0$, $\int_0^\infty \frac{e^{zt}}{t+1}dt$ doesn't converge. – Surb Apr 27 at 11:53
• Essentially, you swap two integrals which is only allowed once they are absolutely convergent and that is where it comes in in the proof – Stan Tendijck Apr 27 at 12:28