# Can $\int_{0}^{\infty}\frac {\cos{x}}{(1 + x^2)} dx$ be evaluated without complex analysis?

Can the integral $\int_{0}^{\infty}\frac {\cos{x}}{(1 + x^2)} dx$ be evaluated by differentiation under integral or any other method without involving complex analysis?

I tried using the function $\cos{x}\exp(-m(1+x^2))/(1 + x^2)$ Then used $\cos{x} =\frac {\exp(ix)+\exp(-ix)} {2}$ and applied formula for Gaussian integral. But then I cannot seem to integrate the differentiated (with respect to $m$) function $\frac{\exp\left(-m-\frac{1}{4m}\right)}{\sqrt{m}}\,$ from $0$ to $\infty$.

• See Tunk-Fey's answer here: math.stackexchange.com/questions/756817/create-a-huge-problem/… – Robert Z Mar 7 '18 at 12:41
• I was stuck with the integral $\int_{m=0}^\infty \frac{\exp\left(-m-\frac{1}{4m}\right)}{\sqrt{m}}\,dm$, while Tunk-Fey solved a similar integral ($\int_{y=0}^\infty \frac{\exp\left(-4y-\frac{1}{y}\right)}{\sqrt{y}}\,dy$), so now I now how to evaluate these. Thanks. – Archisman Panigrahi Mar 7 '18 at 14:35

## 1 Answer

If you have the Fourier transform on ${\mathbb R}$ at your disposal you can start with the elementary integral $$\int_0^\infty e^{-t}\,\cos(\omega t)\>dt={1\over 1+\omega^2}\ ,$$and extend $t\mapsto e^{-t}$ to an even function on ${\mathbb R}$.