solve integral with residue theorem

Question

I want to show that for positive $a$

$$\int_{-\infty}^{\infty}{\frac{\cos(x)}{x^2+a^2}} dx = \frac{\pi e^{-a}}{a}$$

I'm not even sure how to define a smart contour… I guess it can't be a half circle, since $\cos(z)$ is unbounded for big imaginary parts. If I take a rectangle, then the vertical lines will have no impact in the limit since $\cos(z)$ is bounded there and $\frac{1}{z^2}$ decreases rapidly, but for the "way back" i can't find a good choice since the nominator isn't periodic… :(

As a function on the whole complex plane I choose $f(z)=\frac{e^{iz}}{z^2+a^2}$, so I just have to take the real part of the solution in the end — crank, Dec 28 '14 at 21:20
oh there will be problems for the way back with this choice-.- — crank, Dec 28 '14 at 21:22
Integrating $e^{iz}/(z^2+a^2)$ over the upper semicircle, we have just one pole. — leonbloy, Dec 28 '14 at 21:32
thanks…omg i failed calculating the residue correctly and so i tried to find my mistake^^ — crank, Dec 28 '14 at 21:39

score 1 · Accepted Answer · answered Dec 28 '14 at 21:40

The trick is rather simple, first we evaluate the integral $$\int_{-\infty}^\infty \frac{e^{i x}}{x^2+a^2}dx$$ using half a circle as you suggested, this function has no problem since the $e^{i z}$ is bounded in upper half plane. The residue theorem yields $$\int_{-\infty}^\infty \frac{e^{i x}}{x^2+a^2}dx=2\pi i\frac{e^{-a}}{2ia}=\frac{\pi}{a} e^{-a}$$ Taking real parts of both side give you rhe answer.

solve integral with residue theorem

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