Change of variable for integrating a Dirac's delta function

Question

Consider this integral: $$ \int_{-1}^1 \delta(1-x^2)dx \quad . $$ I tried to solve it by doing this change of variable: $$ y = x^2 \\ dy = 2xdx $$ But I have got problems with the limits since both go to $1$, getting a meaningless integral. Any ideas to fix this?

Additionaly I was thinking that it isn't straightforward to extend the method for an integral of the type: $$ \int_{-1}^1 \int_{-1}^1 \delta(1-x^2 - y^2)dxdy \quad . $$

Can you please give me a key to start?

Thank you.

Answer 1

perhaps try splitting the integral, i.e. $$\int_{-1}^1\delta(1-x^2)\,dx=\int_{-1}^0\delta(1-x^2)\,dx+\int_{0}^1\delta(1-x^2)\,dx$$ now deal with the integrals seperately, and the substitution may become meaningful. It didnt work before, since when you substitute over a domian, by setting $y=f(x)$, for some $f$, the function $f$ must be injective.

Answer 2

Let $u=2(1-x)$.

$$ \int_{-1}^1\delta(1-x^2)\,dx=2\int_{0}^1\delta(1-x^2)\,= 2\int_{2}^0\delta\left(u(1-\tfrac{u}{4})\right)(-\tfrac{1}{2})\,du=\int_{0}^2\delta(u)du =\tfrac{1}{2} $$ (using the symmetry of the delta function).