Can we calculate the integral $\int_{-a}^a \frac{\sin(\pi x)}{\pi x}e^{2\pi i xt} \, dx$? It is well known, that $\frac{\sin(\pi x)}{\pi x}$ (the sinc function) is the Fourier transform of the characteristic function on $[-1/2,1/2]$. Is there a way to calculate the integral $$\int_{-a}^a \frac{\sin(\pi x)}{\pi x}e^{2\pi i xt} \, dx$$
for $a>0$? That means we calculate a "finite" inverse Fourier transform of the sinc function. 
 A: Hint
For the antiderivative, consider 
$$I=\int \frac{e^{i \pi x}} {\pi x}e^{2\pi i xt} \, dx=\int \frac{e^{i \pi  (2 t+1) x}}{\pi  x}\,dx=\frac{1}{\pi }\text{Ei}(i \pi  (2 t+1) x)$$ where appears the exponential integral function.
A: Denoting by $\mathcal{F}$ the Fourier transform and by $\mathrm{sinc}$ the sinc function, you know three things: (up to normalisation of the Fourier transform)


*

*$\mathcal{F}^{-1}(\mathrm{sinc}) = \chi_{[-1/2,1/2]}$,

*$\mathcal{F}^{-1}(\chi_{[-1/2,1/2]}) = \mathrm{sinc}(- \cdot ) = \mathrm{sinc}$ and hence $\mathcal{F}^{-1}(\chi_{[-a,a]}) = \mathcal{F}^{-1}(\chi_{[-1/2,1/2]}(\cdot/(2a)))=  2a \mathrm{sinc}(2a \cdot)$,

*$\mathcal{F}^{-1}(f  g) = \mathcal{F}^{-1}(f) *  \mathcal{F}^{-1}(g)$, where $*$ denotes the convolution. (This is by the convolution theorem)


So $\int_{-a}^{a} \mathrm{sinc}(x) e^{2\pi i x \cdot} dx = \mathcal{F}^{-1}(\chi_{[-a,a]} \mathrm{sinc}) = 2a \mathrm{sinc}(2 a\cdot) * \chi_{[-1/2,1/2]}$. 
Now we also have that 
$$
2a \mathrm{sinc}(2 a\cdot) * \chi_{[-1/2,1/2]}(y) =  2a\int_{y-1/2}^{y+1/2} \mathrm{sinc}(2 a x) dx = \int_{2a y-a}^{2a y+a} \mathrm{sinc}(x) dx$$
and
$$
\int_{2a y-a}^{2a y+a} \mathrm{sinc}(x) dx = \pi^{-1}\int_{2a\pi y-a\pi}^{2a\pi y+a\pi} \frac{\sin(x)}{x} dx.
$$
So you can write the solution in terms of the antiderivative of $\frac{\sin(x)}{x}$, which is the special function Si. So the answer is
$$
\pi^{-1}(\mathrm{Si}(2a\pi y+a\pi)- \mathrm{Si}(2a\pi y-a\pi)),
$$
which was already said in the comments.
