Solving $\int_{-\infty}^\infty f(\tau) {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau$. We know that, for any real numbers $\lambda$ and $\nu$, it has
\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}
Its proof involves Fourier transform, in this way:
$$4\pi^2\int_{-\infty}^\infty {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau=$$
$$= \int_{-\infty}^\infty {2\pi} {\operatorname{sinc}}\big({\tau}-\lambda\big) {2\pi} {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau=$$
$$\int_{-\infty}^\infty {2\pi} {\operatorname{sinc}}\big({\tau}-\lambda\big)\overline{ {2\pi} {\operatorname{sinc}}\big({\tau}-\nu\big)}d\tau=$$
$$\int_{-\infty}^\infty \mathcal F( u_\lambda)(\xi)\overline{\mathcal F( u_\nu)}(\xi)d\xi$$
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}
What happens if we have a function that multiplies two sinc functions? In this way:
\begin{equation}
\int_{-\infty}^\infty f(\tau) {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau
\end{equation}
Do you think that there are assumptions on $f$ ($\neq$ constant C) such that you can solve in closed form the previous integral?
Thank you very much.
 A: We can prove that
$$\int_{-\infty}^\infty f(\tau) {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}{F(\omega_1+\omega_2)e^{j(\omega_1\lambda+\omega_2\nu)}d\omega_1d\omega_2}\qquad(1)$$
where $F(\omega)$ is the Fourier transform of $f(t)$.
Now for $f(t)=c$ we have $F(\omega)=2\pi c\delta(\omega)$ and
$$\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}{F(\omega_1+\omega_2)e^{j(\omega_1\lambda+\omega_2\nu)}d\omega_1d\omega_2}=\frac{c}{2\pi}\int_{-\pi}^{\pi}{e^{j\omega_2(\nu-\lambda)}d\omega_2}=c\frac{\sin(\pi(\nu-\lambda))}{\pi(\nu-\lambda)}$$
Based on (1) we can calculate the integral for nonconstant functions. For example we can obtain closed form solutions if $$f(t)=\cos(\omega_0t)$$ In this case we have that $$F(\omega)=\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]$$
If we calculate the integral then we arrive at the interesting formula 
$$\int_{-\infty}^{\infty}{\cos(\omega_0\tau){\operatorname{sinc}}(\tau-\lambda){\operatorname{sinc}}(\tau-\nu)d\tau }=\Bigg\{\array{0 , \qquad\qquad\qquad\qquad\qquad\quad\textrm{ if } \omega_0\geq 2\pi\\ \frac{1}{2\pi(\nu-\lambda)}\{\sin[\pi(\nu-\lambda)+\omega_0\lambda ]+\sin[\pi(\nu-\lambda)-\omega_0\nu ]\}, \:\textrm{ if }\omega_0\in[0,2\pi]}$$
