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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 $D$.

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    $\begingroup$ 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. $\endgroup$ – Surb Apr 27 at 11:53
  • $\begingroup$ Essentially, you swap two integrals which is only allowed once they are absolutely convergent and that is where it comes in in the proof $\endgroup$ – Stan Tendijck Apr 27 at 12:28

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