Solve this integration $\int_{-\infty}^{+\infty}(x^2+1)\delta{(x^2-x-6)}\,dx$?

Question

I know that $\delta(g(x))=\sum_i\dfrac{\delta(x-x_i)}{g'(x)}$, but I want it's proof, am not getting how to proof this.

Answer 1

First note that,

$$\delta_a(x) = 1 \qquad \text{if $x= a$ and $\delta_a(x) = 0$ if $x\neq a$}$$ and for any measurable function $f$

$$\int_{\mathbb R}f(x)\delta_a(dx) = f(a).$$

$$\delta_0\equiv \delta.$$ But $$ x^2 -x-6 = 0 \Longleftrightarrow x=3 ~~or ~~x=-2$$

Hence

$$\delta(x^2 -x-6) = 1 \qquad \text{if $ x=3 $ or $x=-2$ and $\delta(x^2 -x-6) = 0 $ if else}$$

Therefore $$\delta(x^2 -x-6) = \delta_3+\delta_{-2}$$

Hence for any measurable function $f$ we have

$$\int_{\mathbb R}f(x)\delta(x^2 -x-6)dx = \int_{\mathbb R}f(x) (\delta_3+\delta_{-2})dx= \int_{\mathbb R}f(x) \delta_3(dx)+\int_{\mathbb R}f(x)\delta_{-2}(dx)= f(3)+f(-2)$$

Thus

$$\int_{\mathbb R}(x^2+1)\delta(x^2 -x-6)dx = 3^2+1+2^2+1 = 15$$

Answer 2

I know the answer, the integration will give

$\int_{-\infty}^{+\infty}(x^2+1)\delta(x^2-x-6)\,dx$

$\implies\displaystyle\delta(x^2-x-6)=\delta\left[(x+2)(x-3)\right]=\frac{\delta(x+2)}{-5}+\frac{\delta(x-3)}{5}$

So, the integration becomes

$\int_{-\infty}^{+\infty}(x^2+1)\left[\frac{\delta(x+2)}{-5}+\frac{\delta(x-3)}{5}\right]\,dx$

$\implies\dfrac{(x^2+1)}{-5}|_{x=-2}+\dfrac{(x^2+1)}{5}|_{x=3}=1$

But what’s my doubt is , How to proof this?

$\implies\displaystyle\delta(x^2-x-6)=\delta\left[(x+2)(x-3)\right]=\frac{\delta(x+2)}{-5}+\frac{\delta(x-3)}{5}$

Help me Guys

EDIT

This is how I tried the proof

$\displaystyle\int_{-\infty}^{+\infty}f(x)\delta[(x-a)(x-b)]\,dx$

But we know that $\displaystyle\boxed{\int_{-\infty}^{+\infty}\delta[(x-a)(x-b)]\,d[(x-a)(x-b)]=1}$

Right?

$\implies\int_{-\infty}^{+\infty}(2x-(a+b))\delta[(x-a)(x-b)]\,dx=1$

So,

$\displaystyle\int_{-\infty}^{+\infty}f(x)\delta[(x-a)(x-b)]\,\frac{d[(x-a)(x-b)]}{2x-(a+b)}$

$\implies\lim_{\epsilon\to 0}\left(\int_{0}^{a+\epsilon}+\int_{a+\epsilon}^{b}\right)f(x)\delta[(x-a)(x-b)]\,\frac{d[(x-a)(x-b)]}{2x-(a+b)}$

$\implies\left(\dfrac{f(a)}{2a-(a+b)}+\dfrac{f(b)}{2b-(a+b)}\right)$

Hence, $\displaystyle\boxed{\delta[g(x)]=\sum_i\frac{\delta(x-x_i)}{g'(x)}}\tag*{}$

REGARDS!

VM