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?
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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.