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Compute the following integral using residues:

$$\int_0^{\infty}\frac{\cos x}{x^2+a^2} d x $$

$\int_0^{\infty}\frac{\cos x}{x^2+a^2} dx =\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos x}{x^2+a^2} dx =\frac{1}{4}\int_{- \infty}^{\infty}\frac{e^{ix}+e^{-ix}}{x^2+a^2} dx$.

After the substitutions $z=e^{ix}$ and $\cos(x)=\frac{1}{2}(z+\frac{1}{z})$ I get:

$\frac{1}{4}\int_{-\infty}^{\infty}\frac{e^{ix}+e^{-ix}}{x^2+a^2} dx =\frac{1}{4}\int_{-\infty}^{\infty}\frac{z+\frac{1}{z}}{e^{2iz}+a^2}\times{dz}{ie^iz}=\int_{-\infty}^{\infty}\frac{z^2+1}{z(2ie^{3iz}+2a^2ie^{iz})}dz$

In the expression $\frac{z^2+1}{z(2ie^{3iz}+2a^2ie^{iz})}$ I detected only one residue at 0, which is a simple pole. However my computation is giving me the result $\lim_{z\to 0}\frac{z^2+1}{z(2ie^{3iz}+2a^2ie^{iz})}=\frac{1}{2i+2a^2 i}$ which is wrong according with the solution.

Questions:

What am I doing wrong? How should I compute this integral?

Thanks in advance!

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Pedro Gomes
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    Integrate the complex function $f(z)=\frac{e^{iz}}{z^2+a^2}$ over half a circle. (I assume $a\ne 0$ here). At the end use the fact that $\cos x$ is the real part of $e^{ix}$. – Mark Jun 22 '19 at 17:17
  • If you use that substitution then $x^2\to -\ln^2{(z)}$ not $e^{2iz}$ – Peter Foreman Jun 22 '19 at 18:03
  • The integral for $z$ is around the unit circle an infinite amount of times. Also $z=e^{ix}$ leads to $x^2=-(ln(z))^2$. – herb steinberg Jun 22 '19 at 18:14

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