0
$\begingroup$

I have to prove that Dirac delta function $\delta$ is equal to this limit:

$$ \delta(x) = \lim_{a\to \infty}\frac{sin(ax)}{\pi x}.$$

We can say, show that the result of the right side of equality is defined as Dirac function:

$$ \delta(x) = \begin{cases} +\infty, & x=0, \\ 0, & x\neq0. \end{cases} $$

Thank you for help.

$\endgroup$
  • $\begingroup$ I'm confused: $\frac{\sin(ax)}{ \pi x }\leq \frac{1}{\pi x} < 1$ for every $x > 1/\pi$ $\endgroup$ – AnalysisStudent0414 Feb 13 '15 at 13:06
  • 1
    $\begingroup$ There's something wrong with this statement, as the limit does not exist for most $x$ (and is not even defined at $x=0$). In any case, I think you meant $0$ if $x \neq 0$ and $+\infty$ otherwise, although I still find this definition quite problematic: what is the difference between $\delta$ and $2\delta$? $\endgroup$ – Pedro M. Feb 13 '15 at 13:09
  • $\begingroup$ I am confused too, when I used l'Hopital rule for limits, the limit change to $ \lim_{a\to \infty}\frac{cos(ax)}{\pi}.$ I think, this limit is undefined, because cosinus oscillates between -1 and 1 in infinity. Is it possible to prove? $\endgroup$ – MrPitivier Feb 13 '15 at 13:19
  • 5
    $\begingroup$ @PedroM.: The statement is correct. It needs to be interpreted as a generalized limit, i.e. $\lim_{a\rightarrow \infty}\int_{-\infty}^{\infty}\frac{\sin(ax)}{\pi x}\phi(x)dx=\phi(0)$ for some test function $\phi(x)$. $\endgroup$ – Matt L. Feb 13 '15 at 13:21
  • $\begingroup$ @MrPitivier You can't use L'Hôpital, as the numerator does not approach infinity. $\endgroup$ – Pedro M. Feb 13 '15 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.