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

[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

 

[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.
Your answer is correct

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

 

[stevendaryl]

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.

 

[stevendaryl]

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.
Your answer is correct

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

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

 

[BvU]

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.

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 !

 

[stevendaryl]

Wikipedia discusses this very integral here:

https://en.wikipedia.org/wiki/Trigonometric_integral#Auxiliary_functions

(although it's not at all obvious to me why \int_{0}^\infty \frac{sin(t)}{t+x} dt = \int_{0}^\infty \frac{exp(-xt)}{1+t^2} dt)

[edit: the integrals are over t, not x]