# Evaluating integral delta function

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

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

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

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

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 $$f$$.

• how do you define $\delta(f(x))$ – reuns Sep 8 '19 at 22:37
• @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. – Ninad Munshi Sep 8 '19 at 22:40
• 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 – reuns Sep 8 '19 at 22:45
• @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. – Ninad Munshi Sep 8 '19 at 22:50
• 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. – reuns Sep 8 '19 at 22:59

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.

• thanks for this different evaluation. – Robben Sep 9 '19 at 1:04