Integral of Dirac delta function evaluated at a differential function

Question

I am having some troubles in evaluate the following: \begin{align} \int_{a}^{b} \delta \left( f \left(x \right)\right) f'\left( x\right) dx, \end{align} where $f: \left[a,b \right] \to \mathbb{R}$ is differentiable, $f\left( a\right)\neq 0, f\left(b \right)\neq0$ and if $f\left( x\right) = 0$, then $f'\left( x\right) \neq 0$. How can I proceed? Thank you!

Answer 1

using the substitution $y=f(x)$ you get: $$I=\int_{f(a)}^{f(b)}\delta(y)dy$$ and this would be $I=1$ if $0\in[f(a),f(b)]$ or $I=0$ if $0\notin[f(a),f(b)]$

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To answer your comment, remember that: $$\int_a^b\delta(x)dx=1,\,0\in[a,b]$$ since the function $\delta$ has one peak for every zero for the function inside of it. So for a function $f(x)$ with $n$ roots $i=[1,n],\,\alpha_i\in[a,b]$ we can say that there will be $n$ delta peaks within the region $[a,b]$ and so: $$\int_a^b\delta(f(x))dx=\int_a^b\sum_{i=1}^n\delta(x-\alpha_i)dx=n$$

As for the second part, notice that if we use the same substitution as in the original question we get: $$\int_{f(a)}^{f(b)}\delta(y)\frac{|f'(x)|}{f'(x)}dy=\int_{f(a)}^{f(b)}\delta(y)\operatorname{sgn}(y')dy$$ Now think about what this represents