# $\lim_{L\rightarrow \infty}\int\limits_{-a^2}^{a^2} \frac{f(x) \sin(Lx)}{\pi x} dx=f(0)$

Why does this work (for all $a>0$). How can you prove the formula
$$\lim_{L\rightarrow \infty}\int\limits_{-a^2}^{a^2} \frac{f(x) \sin(Lx)}{\pi x} dx=f(0)$$

Does it work for all functions f?

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hint: $\ \frac {\sin(Lt)}{\pi t}\to \delta(t)\$ as $L\to\infty$. eq. (9) of link – Raymond Manzoni Aug 26 '12 at 22:05
Why the $a^2$ instead of just $a$? – marty cohen Aug 26 '12 at 22:07
Consult your analysis textbook on "approximations to the identity." – Potato Aug 26 '12 at 22:28
It doesn't hold for all functions. Counterexample $$f(x)=\begin{cases}1\quad x=0\\0\quad x\neq 0\end{cases}$$ – Norbert Aug 26 '12 at 22:43
In many text books they consider $f(x)$ to be continuous. – Mhenni Benghorbal Aug 27 '12 at 1:10