Simplification ideas for an integration Does anyone see any shortcuts for calculating the following integral analytically?
$$
-\int_{-\infty}^{+\infty} \frac{1}{(\pi x)^2}\sin{(2\pi x)}\sin{(2\pi n x)} e^{2\pi i k x} dx \tag{1}
$$


*

*Where $n\in \mathbb{N}.$ 

*Alternate writing of the two sin terms: $\sin{(2\pi x)}\sin{(2\pi n x)} = -\frac{1}{4}(e^{-2\pi i x}-e^{2\pi i x})(e^{-2\pi i n x}-e^{2\pi i n x})$, but it does not seem to change matters too much when substituted in (1).


You are looking for the Fourier transform of a product, that is the convolution between the Fourier transform of $\frac{\sin(2\pi x)}{\pi x}$ and the Fourier transform of $\frac{\sin(2\pi m x)}{\pi x}$. That is quite easy to compute, since the Fourier transform of $\frac{\sin x}{x}$ is just the indicator function of a symmetric interval with respect to the origin, multiplied by a constant. It follows that our integral, as a function of $k$, is compact-supported and its graph is a trapezoid.
