How can this Fourier transform be found?

Question

Let $g_a(x)$ be defined by

$$g_a(x) = \dfrac{\sin(ax)}{\pi x},$$

I want to find the Fourier transform of $g_a$, that is,

$$\mathcal{F}(g_a)(\xi)=\int_{-\infty}^{\infty} \dfrac{\sin(ax)}{\pi x}e^{-2\pi ix\xi}dx.$$

I just have no idea on how to proceed. At first, the $x$ in the denominator suggested to use contour integration and the residue theorem. However there's a problem: the pole $x = 0$ lies along the integration path.

I've been thinking for a while but I have no idea on what to do here.

How can I find this Fourier transform rigorously? What has to be done?

Thanks in advance!

Answer 1

Note that $$ \frac{\sin(ax)}{\pi x}=\frac{e^{iax}-e^{-iax}}{2\pi ix}= \frac{1}{2\pi}\int_{-a}^{a}e^{isx}ds $$ Therfore the given function is $\frac{1}{\sqrt{2\pi}}\mathcal{F}^{-1}\chi_{[-a,a]}$, where $\chi_{[-a,a]}$ is the characteristic function of the interval $[-a,a]$. So the Fourier transform of the above is $$ \mathcal{F}\left(\frac{\sin(ax)}{\pi x}\right)=\frac{1}{\sqrt{2\pi}}\chi_{[-a,a]}. $$