It is well known that
$$ \lim_{L\to\infty} \frac{\sin(L x)}{x} = \delta(x) $$
in the sense of distribution. Does anybody know of the error term in the above equation ? I am interested in the leading term $f_1(x)$ in an expansion of the form
$$ \frac{\sin(L x)}{x} = \delta(x) +\sum_{n=1} f_n (x)$$
where each $f_n(x)$ goes to zero faster than $f_m(x)$ ($n>m$) when $L\to \infty$.
It is quite easy to obtain the delta function identity using Fourier transform, as the Fourier transform of the sinc is the rect function (the indicator function of the set $[-L,L]$ (by the way, this is pretty much Ron's answer below). However when I try to obtain a series expansion I can't go beyond a tautology ($f_1(x) = \delta(x) - \sin(L x )/x$).
Well, you can use Parseval's theorem to express the integral
$$\int_{-\infty}^{\infty} dx \, f(x) \frac{\sin{L x}}{\pi x} = \frac{1}{2 \pi} \int_{-L}^L dk \, \hat{f}(k)$$
where
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$ $$f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-i k x}$$ $$f(0) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) $$
Assume $f$ is even. Then we may write
$$\int_{-\infty}^{\infty} dx \, f(x) \frac{\sin{L x}}{\pi x} = f(0) - \frac{2}{2 \pi} \int_L^{\infty} dk \, \hat{f}(k)$$
The error term for large $L$ then depends on the behavior of $\hat{f}$ for large values of $k$. This in turn may depend on the continuity of the derivatives of $f$.