Residue Theorem Problem

Question

$$\int_0^\infty \frac{\cos x}{(x^{2} +1)^{2}}dx$$ Can someone help to solve it because I failed.

Answer 1

Hints in Highlights:

Define

$$\;f(z)=\cfrac{e^{zi}}{(z^2+1)}\;,\;C_R:=\left\{ z=x+iy\in\Bbb C\;|\;|z|=R\,,\,\,-R\le x\le R\,,\,\;y\ge0\right\}\,,\,\,\text{so $C_R$ is}:$$

Of course, with $\;1<R\in\Bbb R\;$ and $\;\gamma_R\;$ is the circle's arc.

Observe that within this contour, $\;f\;$ has one unique double pole, namely $\;z=i\;$,

and its residue at it is:

$$\lim_{z\to i}\left[(z-i)^2 f(z)\right]'=\lim_{z\to i}\left[\frac{e^{zi}}{(z+i)^2}\right]'=\lim_{z\to i}\frac{ie^{zi}(z+i)-2e^{zi}}{(z+i)^3}=\frac{-2e^{-1}-2e^{-1}}{(2i)^3}=$$

$$=\frac{-4e^{-1}}{-8i}=\frac1{2ei}$$

so

$$\int_{C_R} f(z)\,dz=\frac{\pi}e$$

Use now Jordan's lemma, or the estimmation lemma or whatever, to show that

$$\lim_{R\to\infty}\int_{\gamma_R}f(z)\,dz=0$$

so we get

$$\frac{\pi}e=\lim_{R\to\infty}\int_{C_R} f(z)\,dz=\lim_{R\to\infty}\left(\int_{-R}^R\frac{e^{xi}}{(x^2+1)^2}dx+\int_{\gamma_R} f(z)\,dz\right)=\int_{-\infty}^\infty\frac{\cos x+i\sin x}{(x^2+1)^2}dx$$

Take the real part (observe the function is even...) and get

$$\int_0^\infty\frac{\cos x}{(x^2+1)^2}dx=\frac\pi{2e}$$

Of course, you need to justify all the steps above...