I want to prove the following relation. For any real numbers $\lambda$ and $\nu$, we have \begin{equation} \int_{-\infty}^\infty {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau= {\operatorname{sinc}}(\lambda-\nu ). \end{equation} This is my proof. $$4\pi^2\int_{-\infty}^\infty {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau=$$ $$= \int_{-\infty}^\infty {2\pi} {\operatorname{sinc}}\big({\tau}-\lambda\big) {2\pi} {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau=$$ $$\int_{-\infty}^\infty {2\pi} {\operatorname{sinc}}\big({\tau}-\lambda\big)\overline{ {2\pi} {\operatorname{sinc}}\big({\tau}-\nu\big)}d\tau=$$ $$\int_{-\infty}^\infty \mathcal F( u_\lambda)(\xi)\overline{\mathcal F( u_\nu)}(\xi)d\xi$$ which becomes, solving the Fourier transform \begin{align} &=2\pi \int_{-\infty}^\infty u_\lambda(\xi)\overline{( u_\nu)}(\xi)d\xi \notag \\ &=2\pi \int_{ -{\pi}}^{{\pi} }e^{i (\lambda-\nu)\xi} d\xi \notag \\ &=4\pi^2 {{\sin \pi\big(\lambda-\nu)}\over{\pi(\lambda-\nu) }}\notag \\ &=4\pi^2 {\operatorname{sinc}}(\lambda-\nu ) \end{align}
This statement can be proved in other ways?
Thank you very much.
Hint: $~I(k)=\displaystyle\int_{-\infty}^\infty\frac{\sin(x-a)}{x-a}\cdot\frac{\sin(x-b)}{x-b}~e^{-k~(x-a)(x-b)}~dx\quad=>\quad I'(k)~=~?$
Of course, this will require some knowledge about Euler's formula and Gaussian integrals.