Elementary proof of orthogonality of translates of $\text{sinc}$?

Question

Context: The solution I gave here used the theory of tempered distributions; also there were various bits of handwaving that needed to be filled in. I finally found a completely elementary proof of the "Magical Property" - the answer to the question below was the last piece in the puzzle. I got a major chuckle out of the solution - passing it on because I didn't get you guys anything for Christmas... $\newcommand{\sinc}{\text{sinc}}$

Define $\sinc(t)=\sin(t)/t$ as usual.

Question: How can one give an elementary proof that $$\int_{-\infty}^\infty\sinc(t)\sinc(t-n\pi)\,dt=0$$for $n\in\mathbb Z$, $n\ne0$?

Comment: If you note that $\sinc(t)=\frac12\int_{-1}^1e^{ixt}\,dx$ then this is more or less obvious from the Plancherel Theorem. We want a solution much more elementary than that - the answer uses nothing but calculus, and no "hard" calculus either...

Answer 1

We have$\newcommand{\sinc}{\operatorname{sinc}}$ $$ \begin{align*} \int_{-\infty}^\infty\sinc(t)\sinc(t-n\pi)\,dt&=(-1)^n\int_{-\infty}^\infty\frac{\sin^2(t)}{t(t-n\pi)}\,dt\\ &=\frac{(-1)^n}{n\pi}\int_{-\infty}^\infty \sin^2(t)\left(\frac{1}{t-n\pi}-\frac{1}{t}\right)\,dt. \end{align*} $$ Here it is tempting to separate this integral as the difference of two integrals, and observe that a substitution shows those integrals are equal. However, the integrals are separately divergent, so some care is needed. Dropping the constant in front, we can write this latter integral as $$ \begin{align*} \lim_{N\to\infty}\int_{-N}^N \sin^2(t)\left(\frac{1}{t-n\pi}-\frac{1}{t}\right)\,dt&=\lim_{N\to\infty} \int_{-N}^N \frac{\sin^2(t)}{t-n\pi}\,dt - \int_{-N}^N\frac{\sin^2(t)}{t}\,dt\\ &=\lim_{N\to\infty}\left(\int_{-N-n\pi}^{-N}-\int_{N-n\pi}^N\right)\frac{\sin^2(t)}{t}\,dt. \end{align*} $$ The final expression is bounded by $2\pi n/(N-n\pi)$, which goes to $0$ as $N\to\infty$.