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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 ($f_1(x) = \delta(x) - \sin(L x )/x$).

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I think the question you should ask is, what is the error in assuming that $$\int_{-\infty}^{\infty} dx \, f(x) \frac{\sin{L x}}{x} \approx f(0)$$ – Ron Gordon Aug 9 '13 at 16:05
@RonGordon Absolutely. Isn't it in fact the same thing? – lcv Aug 9 '13 at 16:33
Sort of. With integrals, though, you do not need to ask difficult questions about distributions. – Ron Gordon Aug 9 '13 at 16:34

Well, you can use Parseval's theorem to express the integral

$$\int_{-\infty}^{\infty} dx \, f(x) \frac{\sin{L x}}{\pi x} = \frac{1}{2 \pi} \int_{-L}^L dk \, \hat{f}(k)$$


$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$ $$f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-i k x}$$ $$f(0) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) $$

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 $f$.

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Thanks (why should I avoid posting "thanks"?), but this is exactly were I got stuck. Is it possible to go further? – lcv Aug 9 '13 at 17:00
The error term is now the integral of $\theta(k-L)$ times $\hat{f}(k)$ (and some constants). Fourier transforming back the theta you get a tautology. One should make some expansion before this point. PS I can't assume $f$ to be even. – lcv Aug 9 '13 at 17:10
@lcv: what can you assume about $f$? – Ron Gordon Aug 9 '13 at 17:35
@lcv: by the way, my last equation is a very general statement about the error. Note that this error depends on $f$; I know of no better statement of the error term for a general $f$. – Ron Gordon Aug 9 '13 at 17:42
@lcv: of course. But my point is, what more do you want without a further specification of $f$? – Ron Gordon Aug 9 '13 at 18:14

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