Show that $\int_{-\infty}^{\infty} \frac{\sin(2\pi(x-a))}{\pi(x-a)}\frac{\sin(2\pi(x-b))}{\pi(x-b)}dx = \frac{\sin2\pi(a-b)}{\pi(a-b)}$ I want to show that for any $a,b \in \mathbb{R}$ we get $\int_{-\infty}^{\infty} \frac{\sin(2\pi(x-a))}{\pi(x-a)}\frac{\sin(2\pi(x-b))}{\pi(x-b)}dx = \frac{\sin2\pi(a-b)}{\pi(a-b)}$. A hint for this exercise is to use Parseval's identity.
Attempt: For $f(x)=e^{isx}\chi_A(x)$, where $A=[-t,t]$ we get the Fourier transform:
$
\hat{f(\gamma)} = \int_{-t}^te^{iy(s-2\pi  \gamma)}dy = [\frac{e^{iy(s-2\pi  \gamma)}}{i(s-2\pi  \gamma)}]^t_{-t} = \frac{2\sin[t(s-2\pi\gamma)]}{s-2\pi\gamma}
$
In particular for $s=2\pi a$, $\gamma=b$ and $t=1$ we get
$
\hat{f(\gamma)} = \frac{\sin 2\pi(a-b)}{\pi(a-b)}
$
Then if we consider Parsevals's identity we get that:
$
\int_{-\infty}^\infty \big|\frac{\sin 2\pi(a-b)}{\pi(a-b)}\big|^2db = \int_{-\infty}^\infty \big| e^{i2\pi a x}\chi_{[-1,1]}\big|^2 dx = \int_{-1}^1 \big| e^{i2\pi a x}\big|^2 dx = 2
$
But I didn't see how to proceed to solve the problem.
 A: Let $f(x)=2\text{sinc}(2\pi x) $. Note that $f$ is even, so
$$\int_{-\infty}^\infty f(x-a)f(x-b)dx= \int_{-\infty}^\infty f(x-(a-b))f(x)dx= \int_{-\infty}^\infty f(a-b-x)f(x)dx=(f*f)(a-b) $$
Then, we calculate
$$\widehat{(f*f)}(\xi)=\hat{f}(\xi)^2=\chi_{[-1,1]}(\xi)^2=\chi_{[-1,1]}(\xi)$$
Hence,
$$(f*f)(a-b)=\widehat{\chi_{[-1,1]}}(a-b)=f(a-b)$$
A: EDIT: I had to modify my answer because partial fraction decomposition of the integrand resulted in the difference of two divergent integrals.

Assume that $a \ne b$, and let
$$I(r) = \int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(x-a) \right)}{\pi(x-a)} \frac{\sin\left(2 \pi(x-b) \right)}{\pi(x-b)} \, \mathrm dx, $$ where $r$ is a nonegative parameter that is not equal to $1$.
Using  partial fraction decomposition and the trig identity $$\sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha)\sin(\beta),$$ we get
$ \begin{align} I (r)   &= \small \frac{1}{\pi^{2}(a-b)} \left( \int_{-\infty}^{\infty}\frac{\sin \left(2 \pi r(x-a) \right)\sin\left(2 \pi(x-b) \right)}{x-a} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(x-a) \right)\sin\left(2 \pi(x-b) \right)}{x-b} \, \mathrm dx\right) \\ &= \small \frac{1}{\pi^{2}(a-b)} \left(\int_{-\infty}^{\infty}\frac{\sin \left(2 \pi ry \right)\sin\left(2 \pi(y+a-b) \right)}{y} \, \mathrm dy - \int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(z+b-a) \right)\sin\left(2 \pi z \right)}{z} \, \mathrm dz\right) \\ &= \small\frac{1}{\pi^{2}(a-b)} \int_{-\infty}^{\infty} \frac{\sin(2 \pi ry) \sin(2 \pi y) \cos\left(2 \pi(a-b\right))+ \sin(2 \pi ry)\cos(2 \pi y)\sin\left(2 \pi(a-b) \right)}{y} \, \mathrm dy \\ & \small- \frac{1}{\pi^{2}(a-b)}\int_{-\infty}^{\infty} \frac{\sin(2 \pi rz) \sin(2 \pi z) \cos\left(2 \pi r(a-b\right))- \cos(2 \pi rz)\sin(2 \pi z)\sin\left(2 \pi r(a-b) \right)}{z} \, \mathrm dz\\ &= \small \frac{ \sin \left(2 \pi(a-b) \right)}{\pi^{2}(a-b)} \int_{-\infty}^{\infty} \frac{\sin(2 \pi ry) \cos(2 \pi y)}{y} \, \mathrm dy + \frac{ \sin \left(2 \pi r (a-b) \right)}{\pi^{2}(a-b)}\int_{-\infty}^{\infty} \frac{\sin(2 \pi z) \cos(2 \pi rz)}{z} \, \mathrm dz. \end{align}$
If $\alpha > \beta \ge 0$, we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin( \alpha x) \cos(\beta x)}{x} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin\left((\alpha-\beta)x \right) + \sin\left((\alpha+\beta)x \right)}{x} \, \mathrm dx \\  &= \frac{\pi}{2} \left(\operatorname{sgn}(\alpha-\beta) + \operatorname{sgn}(\alpha+\beta) \right) \\ &= \pi. \end{align} $$
And if $ \beta > \alpha \ge 0$, we have $$\int_{-\infty}^{\infty} \frac{\sin( \alpha x) \cos(\beta x)}{x} \, \mathrm dx =0. $$
Therefore, $$\int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(x-a) \right)}{\pi(x-a)} \frac{\sin\left(2 \pi(x-b) \right)}{\pi(x-b)} \, \mathrm dx = \begin{cases} \frac{ \sin \left(2 \pi r (a-b) \right)}{\pi (a-b)}    &\text{ if } 0 \le r < 1\\ \frac{ \sin \left(2 \pi(a-b) \right)}{\pi (a-b)}  &\text{ if } r >1 \end{cases}$$
The case $r=1$ should follow by continuity.
