I have two questions regarding the sinc function in the weak limit, where it can be used as a nascent delta function. The definition:

$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}dx\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}} \phi(x)=\phi(x_{0})$

is said to be valid for any smooth function $\phi(x)$ with compact support. Does that mean that the following is not valid :

$ \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}dx\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}}x=x_{0} $

Moreover, if we expand the sine function, we get:

$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\phi(x)\sum_{n=0}^{\infty}\frac{(-1)^n(x-x_{0})^{2n}}{(2n+1)!(\varepsilon)^{2n+1}}dx =$ $\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} \int_{-\infty}^{\infty}(x-x_{0})^{2n}\phi(x) dx =\phi(x_{0}) $

Is it legit to perform the integration term by term?

  • $\begingroup$ in the second formula $\frac{x}{x-x_0}$ is close to $1$ for $x$ large, so your question becomes related to how to define $\int_M^\infty\sin((x-x_0)/\epsilon)$ $\endgroup$
    – Mircea
    Feb 2, 2012 at 14:29

1 Answer 1


Take $x_0=0$, then the integral $$ \int_{-\infty}^\infty \sin\left(\frac{x}{\varepsilon}\right)dx $$ diverges for every $\varepsilon>0$. You can try a principal value for this. There is a whole branch of analysis on "singular integral operators".

Next try $x_0=1$, even the principal value diverges by oscillation: $$ \int_{-t}^t \sin\left(\frac{x-1}{\varepsilon}\right)\;\frac{x}{x-1}\;dx = \cos\left(\frac{1 + t}{\varepsilon}\right) \varepsilon + \text{Si} \left(\frac{1 + t}{\varepsilon}\right) - \cos\left(\frac{-1 + t}{\varepsilon}\right) \varepsilon + \text{Si} \left(\frac{-1 + t}{\varepsilon}\right) $$

Here, $\varepsilon=1/10$:



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