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).

 A: Let's consider the rescaled version of your integral 
$$
I_n(k)=\frac{-2}{\pi }\int_{-\infty}^{\infty} \frac{1}{y^2}\sin(2y)\sin(ny)e^{i y k}dy \quad (1)
$$
differentiating two times w.r.t $k$ gives us 
$$
I_n''(k)=\frac{2}{\pi }\int_{-\infty}^{\infty} \sin(2y)\sin(ny)e^{i y k}dy \quad (2)
$$
Observe that this integral doesn't converge in the usual sense, so we have to be a little bit careful here. I won't go into the details , but i can assure you that everything is well defined if one thinks about all the operations in the sense of (tempered) distributions.
using the facts that
$$
\mathcal{FT}(\cos(a x))= 2 \pi(\delta(k+a)+\delta(k-a))\\
\delta(\gamma x)=\frac{1}{|\gamma|}\delta(x)\\
2\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)-\cos(\alpha+\beta)
$$
we might calculate (2) as 
$$
I_n''(k)= \delta(k+n-2)+\delta(k-n+2)-\delta(k+n+2)-\delta(k-n-2))
$$
Noe for integrating back w.r.t to k we need the following (distributional) identites
$$
\int\delta(x)dx=\Theta(x)+C\\
\int\Theta(x)dx= x \Theta(x)+C
$$
we get
$$
I_n'(k)= \Theta(k+n-2)+\Theta(k-n+2)-\Theta(k+n+2)-\Theta(k-n-2))+C_1
$$
and 
$$
I_n(k)= (k+n-2)\Theta(k+n-2)+(k-n+2)\Theta(k-n+2)\\-(k+n+2)\Theta(k+n+2)-(k-n-2)\Theta(k-n-2))\\+C_1k+C_2
$$
Now all what is left is to fix the constants $C_1,C_2$ 
we can do that by observing 1.) that the $I_0(k)=0$  and that 2.) $I_n(\infty)$ should also vanish (this is a consequence of the Riemann-Lebesgue lemma). Therefore $C_1=C_2=0$ and we have

$$
I_n(k)= (k+n-2)\Theta(k+n-2)+(k-n+2)\Theta(k-n+2)\\-(k+n+2)\Theta(k+n+2)-(k-n-2)\Theta(k-n-2))
$$

which may be simplified by exploiting the properties of the Heaviside function.
A: 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.
