Evaluate integral using Fourier transform

Question

$\int_{0}^{+\infty }\frac{1}{1+x^{2}}$

using Fourier transform of $$\mathfrak{F}\left \{ e^{-\left | x \right |} \right \}=\frac{2}{1+\omega ^{2}}$$

I am curious how to approach this. I am looking for hints only.

Answer 1

Notice that by evenness $$\int^\infty_0 \frac{d\omega}{1+\omega^2} = \frac 1 2 \int^\infty_{-\infty} \frac{d\omega}{1+\omega^2}.$$

It seems you're using the definition $$\hat f(\omega) = \int^\infty_{-\infty} f(x) e^{-i\omega x} dx$$ meaning that the inversion formula is $$f(x) = \frac 1 {2\pi} \int^\infty_{\infty} \hat f(\omega) e^{i\omega x} d\omega.$$ In your case, using the inversion formula gives $$e^{-\lvert x \rvert} = \frac{1}{2\pi} \int^\infty_{-\infty} \frac{2e^{i\omega x}}{1+\omega^2} d \omega.$$ Can you take it from here?

Answer 2

Hint

Well, $$ \int_{-\infty}^\infty \frac{2}{1+\omega^2}e^{i\omega x}\frac{d\omega}{2\pi}$$ is the inverse Fourier transform of $\frac{2}{1+\omega^2},$ isn't it? And this, evaluated at $x=0$ is pretty closely connected to what you want to calculate.