How to solve Dirac Delta function having 2 centres?

Question

$$\int_{-\infty}^\infty (x^2 + 1) \delta(x^2 - 3x + 2) dx =\ ?$$

Which property of Dirac Delta function can be used to solve this type of problems? Or can we call it a Dirac Delta function having two centres?

Answer 1

As already pointed out you should take a look of the properties of the Dirac delta for the general case.

Nonetheless in this case your intuition gives the correct answer:

you can regard $\delta(x^2-3x+2)$ as $\delta(x-1)+\delta(x-2)$, where $1$ and $2$ are the roots of the polynomial $x^2-3x+2$.

The result is then simply achieved, and the original integral becomes:

$$\int_{-\infty}^{+\infty}(x^2+1)\delta(x-2) + \int_{-\infty}^{+\infty}(x^2+1)\delta(x-1) = (2^2+1)+(1^2+1) = 7$$

Answer 2

$$\int_{-\infty}^\infty (x^2 + 1) \delta(x^2 - 3x + 2) dx=\int_{-\infty}^\infty (x^2 + 1) \delta((x-1)(x-2)) dx$$ $$=\int_{2-\epsilon}^{2+\epsilon} (x^2 + 1) \delta((x-1)(x-2)) dx+\int_{1-\epsilon}^{1+\epsilon} (x^2 + 1) \delta((x-1)(x-2)) dx$$ $$=2^2+1 + 1^2+1 = 7$$

Answer 3

Use the property that $$\delta [g(x)]= \sum_i \frac{\delta(x-x_i)}{\left| g'(x_i)\right|}$$ then, $$\delta(x^2-3x+2)=\frac{\delta(x-1)}{1}+\frac{\delta(x-2)}{1}$$ For more information check this post: Dirac Delta Function of a Function