Quote Originally Posted by brenorocha View Post
<br />
\int_0^\infty \frac{\cos x}{x^2+a^2}\,dx<br />
Let a>0.

Define x=au, the integral becomes to \frac{1}<br />
{a}\int_0^\infty {\frac{{\cos au}}<br />
{{u^2 + 1}}\,du} .

Let f(a) = \int_0^\infty {\frac{{\cos au}}<br />
{{u^2 + 1}}\,du} \implies f'(a) = - \int_0^\infty {\frac{{u\sin au}}<br />
{{u^2 + 1}}\,du} .

Now

\int_0^\infty {e^{ - \varphi u} \cos \varphi \,d\varphi } = \text{Re} \int_0^\infty {e^{ - (u - i)\varphi } \,d\varphi } = \text{Re} \,\frac{1}<br />
{{u - i}} = \frac{u}<br />
{{u^2 + 1}}.

So

\int_0^\infty {\frac{{u\sin au}}<br />
{{u^2 + 1}}\,du} = \int_0^\infty {\int_0^\infty {e^{ - \varphi u} \sin (au)\cos \varphi \,d\varphi } \,du} .

Reverse integration order, so we have to find

\int_0^\infty {e^{ - \varphi u} \sin (au)\,du} = \text{Im} \int_0^\infty {e^{ - (\varphi - ai)u} \,du} = \text{Im} \,\frac{1}<br />
{{\varphi - ai}} = \frac{a}<br />
{{\varphi ^2 + a^2 }}.

From here \int_0^\infty {\frac{{u\sin au}}<br />
{{u^2 + 1}}\,du} =\int_0^\infty {\frac{{a\cos \varphi }}<br />
{{\varphi ^2 + a^2 }}\,d\varphi } .

One more substitution defined by \varphi=a\tau,

\int_0^\infty {\frac{{a\cos \varphi }}<br />
{{\varphi ^2 + a^2 }}\,d\varphi } = \int_0^\infty {\frac{{\cos a\tau }}<br />
{{\tau ^2 + 1}}\,d\tau } = f(a).

Then we have the following differential equation f'(a)=-f(a), which is separable and easy to solve, so from here f(a)=ke^{-a}, where k is a constant.

From the initial condition f(0)=\frac\pi2, we have k=\frac\pi2\implies f(a)=\frac{\pi e^{-a}}2.

Since f(a)=f(-a)\implies f(a)=\frac{\pi e^{-|a|}}2.

And finally \int_0^\infty \frac{\cos x}{x^2+a^2}\,dx=\frac{\pi e^{-|a|}}{2a}\,\blacksquare