# Evaluating an improper integral that involves $\exp(-|x|)$

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.

• It's not clear if it helps, but if you break up the integral above as you suggested and rearrange, you get$$\frac{1}{2\pi i}\int_0^\infty \frac{2xe^{-x}}{x^2-z^2}dx$$ – Thomas Andrews Aug 8 '13 at 15:16
• Try proving that $f$ is holomorphic without computing it. Hints: Fubini and Morera. – mrf Aug 8 '13 at 15:17
• @mrf Could you please tell me which fact named after Fubini you are talking about? – Pteromys Aug 8 '13 at 15:24

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 $$\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}h=\int_{-\infty}^{+\infty}\frac{e^{-|x|}}{(x-z_0)^2}\mathrm dx.$$
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 $(e^{-z}E(-z)- e^{z}E(z))/(2\pi i).$
• I follow the title of the question "Evaluating an improper integral that involves $\exp(−|x|)$" – user64494 Aug 8 '13 at 16:53