Calculate $\int_{-T}^T {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau$. Let $\lambda$ and $\nu$ be real numbers. Then, 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 bounded integration interval? That is:
\begin{equation}
\int_{-T}^T {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau
\end{equation}
for $T>0$. Is possible to solve in closed form the previous integral?
Thank you very much.
 A: Since, after 17 hours, you still did not receive any answer, I give you below what a CAS obtained
$$I=\int {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau$$ $$2I=\cos (\lambda -\nu ) (\log (\lambda -\tau )-\text{Ci}(2 \tau -2 \lambda ))-\sin
   (\lambda -\nu ) \text{Si}(2 \lambda -2 \tau )$$ from which you can compute any given definite integral. In particular, if $$J=\int_{-T}^T {\operatorname{sinc}}\big({\tau}-\lambda\big) {\operatorname{sinc}}\big({\tau}-\nu\big)d\tau$$ then $$2J=\sin (\lambda -\nu ) (\text{Si}(2 (T-\lambda ))+\text{Si}(2 (T+\lambda )))-$$ $$\cos
   (\lambda -\nu ) \left(\text{Ci}(-2 (T-\lambda ))-\text{Ci}(-2 (T+\lambda ))+\log
   \left(\frac{T+\lambda }{T-\lambda }\right)\right)$$
Edit
If we consider the limit of $J$ when $T$ tends to infinity, the complete result is $$\frac{\pi  \left(\sin (\lambda -\nu )-i \cos (\lambda -\nu ) \left(\left\lfloor
   \frac{\arg (\lambda )}{2 \pi }\right\rfloor -\left\lfloor \frac{\arg (\nu )}{2 \pi
   }\right\rfloor \right)\right)}{\lambda -\nu }$$ the real part of which being $\pi\operatorname{sinc}(\lambda-\nu)$ as Wolfram Alpha gave. The direct integration between infinite bounds gave the same result.
