# Integration involving dirac delta function

I know that $\delta(0)=1$, and $\delta(x)=0$ otherwise. So for the integral $\int_{-\infty}^{\infty}\delta(6-2x)x^2$, why can't you say that $\delta(6-2x)= \delta(0)$ at $x=3$, and therefore evaluating $x^2$ at $x=3$ you get 9? I know you have to do it by substitution. But why??

And how do you tackle a problem where the limits aren't $-\infty$ to $\infty$ and where there's a quadratic in the delta function? So

$\int_{0}^{\infty} \delta(x^2+x-6)x^2 = 0$?

• It is misleading to say $\delta(0)=1$. You should think of $\delta$ as going to $\infty$ as its argument goes to zero. It's the integral of $\delta$ that is $1$. – Dan Piponi May 11 '16 at 20:39

The delta function of a function $f(x)$ is given by $$\delta(f(x))=\sum_{i=1}^n\frac{\delta(x-x_i)}{|f'(x_i)|}$$ where the $x_i$ are the roots of $f(x)$. So for $f(x)=x^2+x-6=(x+3)(x-2)$ we have $$\delta(f(x))=\frac{\delta(x-2)}{5}+\frac{\delta(x+3)}{5}$$ \begin{align} \int_{0}^{\infty} \delta(x^2+x-6)x^2 \mathrm d x &=\int_{0}^{\infty}\left(\frac{\delta(x-2)}{5}+\frac{\delta(x+3)}{5}\right)x^2\mathrm d x\\ &=\int_{0}^{\infty}\frac{\delta(x-2)}{5}x^2\mathrm d x\\ &=\int_{0}^{\infty}\frac{\delta(x-2)}{5}(2)^2\mathrm d x\\ &=\frac{4}{5}\int_{0}^{\infty}\delta(x-2)\mathrm d x\\ &=\frac{4}{5} \end{align} observing that the point $-3$ is outside of interval of integration and then $\int_{0}^{\infty}\frac{\delta(x+3)}{5}x^2\mathrm d x=0$.

For $g(x)=6-2x$ we have \begin{align} \int_{0}^{\infty} \delta(6-2x)x^2 \mathrm d x =\int_{0}^{\infty}\frac{\delta(x-3)}{2}x^2\mathrm d x =\int_{0}^{\infty}\frac{\delta(x-3)}{2}3^2\mathrm d x =\frac{9}{2}\int_{0}^{\infty}\delta(x-3)\mathrm d x &=\frac{9}{2} \end{align}

• Why do both fractions have a denominator of 5? Why doesn't one have denominator 3, since one of the delta functions has $x_i=3$ and one have denominator 2? – user13948 May 11 '16 at 21:04
• $f'(x)=2x+1$ so we have $f(2)=5$ and $f(-3)=-5$ and then $|f(2)|=|f(-3)|=5$ – alexjo May 11 '16 at 21:06
• See this link on wikipedia – alexjo May 11 '16 at 21:09
• Although this is obvious, it is still notable that you can really only compose the Dirac delta with a function which has simple zeros only. $\delta(x^2)$ is a non-meaningful object. – Ian May 11 '16 at 21:33
• @Ian yes of course and with $f'(x_i)\neq 0$ in an interval of $x_i$. – alexjo May 11 '16 at 21:54

Notice:

$$\delta(6-2x)x^2=\frac{x^2\delta(x-3)}{2}=\frac{9\delta(x-3)}{2}$$

So:

$$\int_{-\infty}^{\infty}\delta(6-2x)x^2\space\text{d}x=\lim_{n\to\infty}\int_{-n}^{n}\frac{9\delta(x-3)}{2}\space\text{d}x=\frac{9}{2}\lim_{n\to\infty}\int_{-n}^{n}\delta(x-3)\space\text{d}x=$$ $$\frac{9}{2}\lim_{n\to\infty}\left[\theta(x-3)\right]_{-n}^{n}=\frac{9}{2}\lim_{n\to\infty}\left(\theta(n-3)-\theta(-n-3)\right)=$$ $$\frac{9}{2}\left(\lim_{n\to\infty}\theta(n-3)-\lim_{n\to\infty}\theta(-n-3)\right)=\frac{9}{2}\left(1-0\right)=\frac{9}{2}$$

• The reason the manipulation in the box is allowed isn't at all obvious to me. Having said that, I've never actually seen the definition of the delta function as given in alexjo's answer, so our introduction to it obviously had bits missing. We were given a physicist's intro, very brief! ;) – user13948 May 11 '16 at 21:32
• @Karacoreable It is "the thing you would get if substitution made sense". – Ian May 11 '16 at 22:01
• @Ian If I did do a substitution, y = 6-2x, then wouldn't I divide by -2 rather than 2? – user13948 May 11 '16 at 22:20
• @Karacoreable Yes, except the limits would reverse; reversing them back would give the desired result. – Ian May 11 '16 at 22:35

Can I give you a stupid one, but it fits with your intuition....

$\delta(x)$ is a curve that is 0 everywhere except in an infinatessimal area where it is infinitely tall and: $\int_a^b \delta(x)\,dx = 1$ if $a<0<b.$

$\delta(2x)$ his half a wide and just as tall, and

$\int_a^b \delta(2x)\,dx = 1/2$ if $a<0<b$

$\int_a^b \delta(6-2x)x^2 \,dx= (1/2) 3^2$ if $a<0<b.$

$\int_0^\infty \delta(x^2 + x -6)x^2\,dx$

$\int_0^\infty \delta((x-2)(x+3))x^2\,dx$

Clearly x = -3 is out of the interval... at x = 2

$\lim_\limits{h\to0}\int_{2-h}^{2+h}\delta((x-2)(x+3))x^2\,dx$

$\lim_\limits{h\to0}\int_{2-h}^{2+h}\delta((x-2)5)4\,dx = 4/5$