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I am trying to prove that the function $f:\mathbb C\setminus\mathbb R\rightarrow\mathbb C$ defined by $$ f(z) := \frac{1}{2\pi i}\int_{-\infty}^\infty\frac{\exp(-|x|)}{x-z}dx $$ is holomorphic.

I tried to solve it by evaluating the integral. Since |x| introduces non-analicity, I tried dividing the integral into the intervals $(-\infty, 0]$ and $[0,\infty)$. I thought that I could calculate these integrals by using residue calculus, but I have never evaluated this kind of integrals.

I would appreciate if you could provide a clue (not necessarily a complete answer).

Edit: I corrected the problem statement.

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  • $\begingroup$ It's not clear if it helps, but if you break up the integral above as you suggested and rearrange, you get$$\frac{1}{2\pi i}\int_0^\infty \frac{2xe^{-x}}{x^2-z^2}dx$$ $\endgroup$ – Thomas Andrews Aug 8 '13 at 15:16
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    $\begingroup$ Try proving that $f$ is holomorphic without computing it. Hints: Fubini and Morera. $\endgroup$ – mrf Aug 8 '13 at 15:17
  • $\begingroup$ @mrf Could you please tell me which fact named after Fubini you are talking about? $\endgroup$ – Pteromys Aug 8 '13 at 15:24
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As mrf suggest, one can use Fubini's theorem and Morera's one (in order to give a simple condition to check, which is more in the spirit of contour integrals).

An alternative way is to fix $z_0\in\mathbb C\setminus\mathbb R$, and a $\delta$ such that $B(z_0,2\delta)\subset\mathbb C\setminus\mathbb R$, say for each element $z$of this ball, we have $d(z,\mathbb C\setminus\mathbb R)\geqslant r>0$. Then we can prove that $$\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}h=\int_{-\infty}^{+\infty}\frac{e^{-|x|}}{(x-z_0)^2}\mathrm dx.$$

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The integral under consideration can be reduced to the exponential integral in such a way. By the change $x-z=t$ we obtain $$ \int_0^\infty \frac {e^{-x}} {x-z} \,dx =\int_{-z}^\infty \frac {e^{-t-z}}{t}\, dt= e^{-z}E_1(-z),\, |Arg(z)|<\pi . $$ Similarly, by the change $x-z=-s$,
$$ \int_{-\infty}^0 \frac {e^{x}} {x-z} \,dx =\int_\infty^z \frac {e^{-s+z}}{-s}\, (-ds)=- e^{z}E_1(z),\, |Arg(z)|<\pi . $$ Therefore, the integral under consideration equals $(e^{-z}E_1(-z)- e^{z}E_1(z))/(2\pi i).$ It remains to consider the case $\Re z =z$. In this case we have to take the Cauchy principal value of the integral, obtaining $(e^{-z}E(-z)- e^{z}E(z))/(2\pi i).$

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  • $\begingroup$ Maple produces the same answer. $\endgroup$ – user64494 Aug 8 '13 at 16:36
  • $\begingroup$ I follow the title of the question "Evaluating an improper integral that involves $\exp(−|x|)$" $\endgroup$ – user64494 Aug 8 '13 at 16:53
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The integral under consideration is an integral of Cauchy type. Up to the well-known properties of such integrals, it defines an analytic function in the upper complex half-plane and in the lower complex halp-plane.

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