Fourier transform of $\frac{2\sin(x)}{1+x^2}$

Question

I'm trying to calculate FT of $$f(x)=\frac{2\sin(x)}{1+x^2}$$ First of all we have $$ \int_{-\infty}^{+\infty}\left|\frac{2\sin(x)}{1+x^2}\right| \, dx\leq 2\int_{-\infty}^{+\infty}\frac{1}{1+x^2} \, dx=2\pi $$ thus $\mathcal{F}(f)$ is well defined. Now let's compute $\mathcal{F}(f)(\xi)$: $$\mathcal{F}(f)(\xi)=2\int_{-\infty}^{+\infty}\frac{\sin(x)e^{-2\pi i x\xi}}{1+x^2} \, dx = \left[\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\right]=\\=-i\int_{-\infty}^{+\infty} \frac{e^{-2\pi ix\xi + ix}-e^{-2\pi ix\xi -ix}}{1+x^2}\, dx$$

Now I'm not sure how to proceed further. Any help?

Answer 1

Hint: Define the upper semi-circle with center in origin and redius $R$ as the contour and when $z=i$ is a pole of $$\int_{C}\frac{e^{-2\pi ix\xi + ix}-e^{-2\pi ix\xi -ix}}{1+x^2}dx$$ in $C$, find the residue, then you can simplify the integral.

Answer 2

This can be solved using the convolution theorem.

You should first verify for yourself that

$$\mathcal{F}^{-1}\left\{2\pi e^{-|2\pi\xi|}\right\} = \dfrac{2}{1+x^2}$$

and that

$$\mathcal{F}\left\{\sin(x)\right\} = \dfrac{i}{2}\delta\left(\xi + \dfrac1{2\pi}\right)-\dfrac{i}{2}\delta\left(\xi - \dfrac1{2\pi}\right)$$

By the convolution theorem

$$\begin{align}\mathcal{F}\left\{\dfrac{2}{1+x^2} \cdot \sin(x)\right\} &= \mathcal{F}\left\{\dfrac{2}{1+x^2} \right\} * \mathcal{F}\left\{\sin(x)\right\} \\ \\ &= 2\pi e^{-|2\pi\xi|} \space * \space \left[\dfrac{i}{2}\delta\left(\xi + \dfrac1{2\pi}\right)-\dfrac{i}{2}\delta\left(\xi - \dfrac1{2\pi}\right)\right]\\ \\ &= i\pi \left( e^{-\left| 2\pi\left(\xi + \frac1{2\pi}\right)\right|}-e^{-\left| 2\pi\left(\xi - \frac1{2\pi}\right)\right|}\right)\\ \\ &= i\pi \left( e^{-\left| 2\pi\xi +1 \right|}-e^{-\left| 2\pi\xi-1\right|}\right)\\ \end{align}$$