TITLE: Elementary proof of orthogonality of translates of $\text{sinc}$?
QUESTION [5 upvotes]: 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...

REPLY [2 votes]: 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 $N\to\infty$.