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How to prove the identity $$\int_{-\infty}^\infty \delta(f(x)) \,s(x) \,dx = \sum_i \frac{s(x_i)}{|f'(x_i)|},$$ where $x_i$ are the zeros of $f$?

I am supposed to use an identity that I've already proved: $$\int_{-\infty}^\infty \delta(bx)\,s(x)\,dx=\frac{s(0)}{|b|}.$$

alexwlchan
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    Try expanding $f$ into a Taylor series about each zero – Simon S Dec 31 '14 at 21:13
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    Since the Dirac delta is not a function, but a distribution, you need to write down what your required statement means. – GEdgar Dec 31 '14 at 21:13
  • @SimonS Could you elaborate slightly? I only know of Taylor series about a single point. –  Dec 31 '14 at 22:39
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    @user41404 The idea is that if you cut up $\mathbb{R}$ into small intervals around each zero and one big chunk where $f$ is bounded away from zero, then the integral on the big chunk is zero and the integral on each little chunk can be estimated by Taylor expanding around the zero in that chunk. This is basically the same as what I wrote. – Ian Jan 01 '15 at 00:11

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Suppose for the moment that $f$ is invertible and $C^1$. Then $f'$ is never zero and you can take $u=f(x)$, $du=f'(x) dx$ to turn the integral into

$$\int_{-\infty}^\infty \delta(u) s \left (f^{-1}(u) \right ) \frac{du}{|f'(u)|} \\ = \frac{s \left ( f^{-1}(0) \right )}{|f'(f^{-1}(0))|}$$

by the definition of $\delta$. If $f$ is only locally invertible, meaning that the set of points where $f'=0$ is discrete, then we can do the same thing: break up into intervals where it is invertible, take $u=f(x)$, and sum up.

If $f$ is not locally invertible but $f'$ is never zero on $A \equiv f^{-1}(\{ 0 \})$, then you need to isolate the region where $f$ is not locally invertible and check that it does not contribute to the integral. Then you can apply the previous argument again.

If $f'$ is zero somewhere on $A$, then both sides of the formula are ill-defined.

This all assumes implicitly that $\delta(f(x))$ is to be defined in such a way that we can make this $u$-substitution in the first place. A better treatment in my opinion would be to make this a definition and then check that it makes sense. Alternately you could proceed through approximate identities. The latter argument would be to prove that if $A_\varepsilon$ is an approximate identity, then $\int_{\mathbb{R}} A_\varepsilon(f(x)) s(x) dx$ converges to the desired quantity as $\varepsilon \to 0^+$. This is all sensibly defined because each $A_\varepsilon$, unlike $\delta$, is actually a function.

Ian
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