Study the convergence of the improper integral $$\int_{-\infty}^\infty \frac{d^k}{dx^k}\left\{\frac{\sin(\pi(x-t))}{\pi(x-t)}\right\}\Biggl|_{x=n}\ \cdot \ \frac{d^k}{dx^k}\left\{\frac{\sin(\pi(x-t))}{\pi(x-t)}\right\}\Biggl|_{x=m} dt$$ for $n,m,k\in\mathbb N$, and if it is convergent to give an its estimate from above.
My attempt: I have tried to use the relationship $$\frac{d^k}{dx^k}\left\{\frac{\sin\pi x}{\pi x}\right\}=\pi^k\int_0^1 s^k \cos\left(s\pi x+k\frac{\pi}{2}\right)ds$$ but without success.
Any suggestions please?
Maybe this helps. No guarantee about computational errors.
Let $\tilde{f}(\omega )$ be the Fourier transform of $\sin (\pi x)/(\pi x)$. Then
\begin{eqnarray*} \frac{\sin (\pi (x-t))}{\pi (x-t)} &=&\int d\omega \exp [i\omega (x-t)] \tilde{f}(\omega ) \\ \partial _{x}^{k}\frac{\sin (\pi (x-t))}{\pi (x-t)} &=&\int d\omega (i\omega )^{k}\exp [i\omega (x-t)]\tilde{f}(\omega ) \end{eqnarray*} \begin{eqnarray*} &&\int_{-\infty }^{+\infty }dt\int d\omega _{1}(i\omega _{1})^{k}\exp [i\omega _{1}(x_{1}-t)]\tilde{f}(\omega _{1})\int d\omega _{2}(i\omega _{2})^{k}\exp [i\omega _{2}(x_{2}-t)]\tilde{f}(\omega _{2}) \\ &=&\int d\omega _{1}\int d\omega _{2}(i\omega _{1})^{k}(i\omega _{2})^{k}\exp [i(\omega _{1}x_{1}+\omega _{2}x_{2})]2\pi \delta (\omega _{1}+\omega _{2})\tilde{f}(\omega _{1})\tilde{f}(\omega _{2}) \\ &=&2\pi \int d\omega _{1}(i\omega _{1})^{k}(-i\omega _{1})^{k}\exp [i(\omega _{1}(x_{1}-x_{2})]\tilde{f}(\omega _{1})\tilde{f}(-\omega _{1}) \\ &=&2\pi \int d\omega _{1}\omega _{1}^{2k}\exp [i(\omega _{1}(x_{1}-x_{2})] \tilde{f}(\omega _{1})\tilde{f}(-\omega _{1}) \end{eqnarray*}