# A Why can't I use contour integration for this integral?

1. Dec 16, 2016

### ShayanJ

Consider the integral $\displaystyle \int_{-\infty}^\infty \frac{e^{-|x|}}{1+x^2}dx$. I should be able to use contour integration to solve it because it vanishes faster than $\frac 1 x$ in the limit $x \to \infty$ in the upper half plane. It has two poles at i and -i. If I use a semicircle in the upper half of the plane, I should only consider the residue at i which is:

$\displaystyle \lim_{z\to i} \frac{e^{-|z|}}{i+z}=\frac{e^{-|i|}}{2i}=\frac{1}{2ie}$

So the integral is equal to $\frac{2\pi i}{2ie}=\frac \pi e \approx 1.15$. But wolframalpha.com gives an expression involving Ci(x) and Si(x) that approximates to 1.24. What is wrong here?

Thanks

2. Dec 16, 2016

### BvU

Looks to me Wolfram is doing some numerical integration. When I do a primitive numerical integration with excel I get about the same answers they do.

(How did you get this Ci and Si answer from wolfie ?)

3. Dec 16, 2016

### stevendaryl

Staff Emeritus
I don't know off the bat how to derive the right answer, but contour integration by using residues only works for analytic functions, and the use of an absolute value, $|x|$, makes your integrand non-analytic.

4. Dec 16, 2016

### stevendaryl

Staff Emeritus
That link is for the integrand $\dfrac{e^{itz}}{1+z^2}$, not $\dfrac{e^{-|z|}}{1+z^2}$

5. Dec 16, 2016

### BvU

Yeah, I missed that completely. though to be smart and use $t=i$

(For the numerics looked at $[0,\infty>$)

Changed your avatar, didn't you ? Less colorful but more cultural !

6. Dec 16, 2016

### stevendaryl

Staff Emeritus
Last edited: Dec 16, 2016