I don't know how to compute the Fourier tranform of this function:
$f(x) = \frac{\sin \pi a x}{\pi x}$
I know that $\frac{\sin \pi a x}{\pi x} = \frac{e^{i \pi a x} - e^{- i \pi a x}}{2i \pi x}$
Then I plug this function in the formula for the Fourier transform and I get:
$$\int_{\mathbb{R}}\frac{e^{i \pi a x} - e^{- i \pi a x}}{2i \pi x} e^{-i s x} dx =$$
$$= \frac{1}{2 \pi i } \int_{\mathbb{R}} \frac{e^{ix( \pi a - s)} - e^{- i x( a \pi +s)}}{x} dx=$$
$$=\frac{1}{2 \pi i } \int_{\mathbb{R}} \frac{e^{ix( \pi a - s)}}{x} dx - \int_{\mathbb{R}} \frac{e^{- i x( a \pi +s)}}{x} dx$$
What should I do know? We can get rid of $i$ in the exponent by changing variables, but that doesn't help much
We have that: $$ g(t) = \int_{\mathbb{R}}\frac{\sin(\pi x)}{\pi x}e^{-itx}\,dx = \mathbb{1}_{(-\pi,\pi)}(t)+\frac{1}{2}\mathbb{1}_{\{\pi\}\cup\{-\pi\}}(t).\tag{1}$$
To prove such an identity, we may compute the inverse Fourier transform of $\mathbb{1}_{(-\pi,\pi)}$, or notice that:
$$ g(t) = 2\int_{0}^{+\infty}\frac{\sin(\pi x)}{\pi x}\cos(tx)\,dx = \int_{0}^{+\infty}\frac{\sin((t+\pi)x)-\sin((t-\pi)x)}{\pi x}\,dx.\tag{2}$$ However, since: $$\int_{0}^{+\infty}\frac{\sin(ax)}{x}\,dx = \frac{\pi}{2}\operatorname{sign}(a),\tag{3}$$ $(1)$ just follows. With the same steps, we have:
$$ f(t) = \int_{\mathbb{R}}\frac{\sin(\pi a x)}{\pi x}e^{-itx}\,dx = \operatorname{sign}(a)\cdot g\left(\frac{t}{|a|}\right).\tag{4}$$