I am trying to evaluate this integral:
$$\int_{0}^{\infty }\frac{\cos(x)}{1+x^{2}}dx$$
My attempt:
$$\int_0^{\infty}\frac{\cos(x)}{(x+i)(x-i)}dx=1/2 \int_{-\infty}^{\infty} \frac{\cos(x)}{(x+i)(x-i)}dx$$
Where do I go from here? My teacher told me that I need $$\oint \frac{e^{iz}}{(z+i)(z-i)}dz$$ but I don't understand how we go from $$1/2 \int_{-\infty}^{\infty} \frac{\cos(x)}{(x+i)(x-i)}dx$$ to $$\oint \frac{e^{iz}}{(z+i)(z-i)}dz$$
Thanks
Define
$$f(z):=\frac{e^{iz}}{z^2+1}\;,\;\;C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;|z|=R\;,\;\text{Im}(z)\ge 0\}\;,\;\;\Bbb R\ni R>>0$$
The only (simple, BTW) pole of $\,f\,$ inclosed by the above closed curve is $\,z=i\,$ , and
$$\text{Res}(f)_{z=i}\lim_{z\to i}(z-i)f(z)=\frac{e^{-1}}{2i}$$
So by Cauchy's Theorem:
$$2\pi i\,\text{Res}(f)_{z=i}=\frac\pi e=\oint\limits_{C_R}f(z)\,dz=\int\limits_{-R}^R\frac{e^{ix}}{x^2+1}+\int\limits_{\gamma_R}f(z)\,dz$$
But the Evaluation theorem gives (with $\;z=x+iy\,$):
$$\left|\int\limits_{\gamma_R}f(z)\,dz\;\right|\le\max_{|z|=R}\frac{e^{-y}}{R^2-1}R\pi\xrightarrow[R\to\infty]{}0$$
so we get
$$\frac\pi e=\lim_{R\to\infty}\oint\limits_{C_R}f(z)\,dz=\int\limits_{-\infty}^\infty\frac{\cos x+i\sin x}{x^2+1}dx$$
Now just compare real-imaginary parts and using the real one is an even function get the result $\;\displaystyle{\frac\pi{2e}}\,$