2
$\begingroup$

I am trying to evaluate the integral below (delta function) and not sure if I evaluated correctly? The integral is the following:

$$\int^{100}_{-100}x^3\sin(2x)\delta(x^2-9)dx$$


I have the following $x=\pm 3$, $$\therefore \int^{100}_{-100}x^3\sin(2x)\delta(x^2-9)dx = \int^{0}_{-100}x^3\sin(2x)\delta(x^2-9)dx + \int^{100}_{0}x^3\sin(2x)\delta(x^2-9)dx= \ (-3)^3\sin(2\cdot(-3)) + (3)^3\sin(2\cdot(3)) = \ -27\sin(-6) + 27\sin(6)$$

| cite | improve this question | | | | |
$\endgroup$
2
$\begingroup$

Start from from property of delta function:

$\delta(x^2-3^2)=\frac{1}{2*3}[\delta(x+3)+\delta(x-3)]$

Your function is even, $f(x)=f(-x)$ due to symmetry you can write

$\int_{-100}^{100} x^3 \sin(2x) \delta(x^2-9)dx = 2\int_{0}^{100} x^3 \sin(2x) \delta(x^2-9)dx = 2\frac{1}{2*3}\left(\int_{0}^{100} x^3 \sin(2x)\delta(x+3)dx+\int_{0}^{100} x^3 \sin(2x)\delta(x-3)dx\right)=2\frac{1}{2*3}\left(0+27\sin(6)\right)=9\sin(6)$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ You should say that you use the fomula given by @Ninad Munshi $\endgroup$ – Jean Marie Sep 8 '19 at 23:30
  • $\begingroup$ I see, how did you get that $$\delta(x^2-3^2)=\frac{1}{2*3}[\delta(x+3)+\delta(x-3)]?$$ $\endgroup$ – Robben Sep 9 '19 at 1:03
3
$\begingroup$

Not quite. Delta function has the property that

$$\delta(f(x)) = \sum_i \frac{1}{|f'(x_i)|}\delta(x-x_i)$$

where $x_i$ are the zeros of $f$.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ how do you define $\delta(f(x))$ $\endgroup$ – reuns Sep 8 '19 at 22:37
  • $\begingroup$ @reuns It's a very common object that shows up in physics and engineering. One could quite literally define the distribution the way I did above, by using the "property". It is undefined if the gradient of the function vanishes at the desired points. $\endgroup$ – Ninad Munshi Sep 8 '19 at 22:40
  • $\begingroup$ You should better define it as $\delta(f(x))=\lim_{n \to \infty} \frac{n}2 1_{|f(x)| < 1/n}$ (limit in the sense of distributions, if it converges) for $f\in C^1$ with finitely many simple zeros on $[-r,r]$ from $f(x) = (x-x_i)(f'(x_i)+o(1))$ you get the result $\endgroup$ – reuns Sep 8 '19 at 22:45
  • $\begingroup$ @reuns Is there any reason to? I don't see a problem with the definition $\delta_{f} = \sum_{i} \frac{1}{|\nabla f(x_i)|} \delta_{x_i}$. This is a fairly standard definition I have even seen in Hormander. Not to mention, if a student's instructor is writing statements like $\delta(x^2-9)$, a level of informality is okay. We're trying to speak to the student's level, not make our answers extraordinarily obtuse for non math students. $\endgroup$ – Ninad Munshi Sep 8 '19 at 22:50
  • $\begingroup$ Well you need it if you plan to look and use $1_{|x|<100}\delta(f(x))$ which is what the OP asked. Not even mentioning things like $f(x^2+y^2-1)$. The limit definitions is what you need to show that we can't do a pairing of $\delta$ with any test function and that in the Fourier transform we have more complicated approximations of $\delta$ thus more convergence problems. $\endgroup$ – reuns Sep 8 '19 at 22:59
1
$\begingroup$

You need to do a change of variables to standardise the argument of the $\delta$ near points where it is $0$: if $a>0$ and $0<\varepsilon \ll a$, we have $$ \int_{a-\varepsilon}^{a+\varepsilon} g(x) \delta(x^2-a^2)\, dx = \int_{-(2a-\varepsilon)\varepsilon}^{(2a+\varepsilon)\varepsilon} g(\sqrt{u+a^2})\frac{1}{2\sqrt{u+a^2}} \delta(u) \, du = \frac{g(a)}{2a} $$ putting $x = \sqrt{u+a^2}$, and similarly for the integral over the negative axis.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ thanks for this different evaluation. $\endgroup$ – Robben Sep 9 '19 at 1:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.