Integral with Dirac and Heaviside function

Question

I want to prove that

$$ \int_{-\infty}^\infty \delta (x) f(\theta (x)) dx = \int_0 ^1 f(y)dy $$

for smooth functions f(y)

where $$\delta (x)$$ is Dirac delta and $$ \theta(x) $$ is the Heaviside function.

I know that the derivative of the Heaviside function is the Dirac function, which seems the only identity that I can use, along with other properties of the Dirac function, but I cannot work the integral.

Can someone help?

Thanks in advance.

Answer 1

I'm not too sure whether it makes sense or not, because $f(\theta(x)) = f(1), \ x > 0, \ = \ f(0), \ x < 0,$ in general will not have finite support and is not smooth at $x = 0$.

The idea of the exercise is probably to use integration by substitution. Since $\delta(x) = \theta(x)'$, you can do a change of variables $y = \theta(x)$, $dy = \theta(x)'dx = \delta(x)dx.$ Then $$\int_{-\infty}^{\infty} \delta(x)f(\theta(x))dx = \int_{\theta(-\infty)}^{\theta(\infty)} f(y)dy = \int_0^1 f(y)dy.$$