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I'm trying to show that

$$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}$$

Where $a_{i}$ are the roots of the function $f(x)$. I've tried to proceed by using a dummy function $g(x)$ and carrying out:


Then making the coordinate substitution $u$ = $f(x)$ and integrating over $u$. This seems to be on the right track, but I'm unsure where the absolute value comes in in the denominator, and also why it becomes a sum.

$$\int_{-\infty}^{\infty}\frac{du}{\frac{df}{dx}}\delta(u)g\big(f^{-1}(u)\big) = \frac{g\big(f^{-1}(0)\big)}{\frac{df}{dx}\big(f^{-1}(0)\big)}$$

Can any one shed some light? Wikipedia just states the formula and doesn't actually show where it comes from.

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Split the integral into regions around $a_i$, the zeros of $f$ (as integration of a delta function only gives nonzero results in regions where its arg is zero) $$ \int_{-\infty}^{\infty}\delta\big(f(x)\big)g(x)\,\mathrm{d}x = \sum_{i}\int_{a_i-\epsilon}^{a_i+\epsilon}\delta(f(x))g(x)\,\mathrm{d}x $$ write out the Taylor expansion of $f$ for $x$ near some $a_i$ (ie. different for each term in the summation) $$ f(a_i+x) =f(a_i) + f'(a_i)x + \mathcal{O}(x^2) = f'(a_i)x + \mathcal{O}(x^2) $$ Now, for each term, you can show that the following hold: $$ \int_{-\infty}^\infty\delta(kx)g(x)\,\mathrm{d}x = \frac{1}{|k|}g(0) = \int_{-\infty}^\infty\frac{1}{|k|}\delta(x)g(x)\,\mathrm{d}x $$ (making a transformation $y=kx$, and looking at $k<0,k>0$ separately **Note: the trick is in the limits of integration) and $$ \int_{-\infty}^\infty\delta(x+\mathcal{O}(x^2))g(x)\,\mathrm{d}x = g(0) = \int_{-\infty}^\infty\delta(x)g(x)\,\mathrm{d}x $$ (making use of the fact that we can take an interval around 0 as small as we like)

Combine these with shifting to each of the desired roots, and you can obtain the equality you're looking for.

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In order to give a rigorous proof of this fact, you require a precise definition of the composition of a distribution with a smooth function. This can be found in Schwartz' presentation but requires delicate functional analysis and is not too helpful in computing explicit examples. A simpler and more direct definition was given by Sebastiao e Silva and is as follows. Most distributions $f$ of practical import have finite order, i.e., have the form $D^nF$ where $F$ is continuous and the derivative is in the sense of distributions (for the record, this is also true locally for any distribution). One then defines $f \circ \phi$ to be $$\left(\dfrac 1{\phi'} D\right )^n F\circ \phi.$$ The motivation for this definition and the proof that it is independent of the representation of $f$ (this is the hard part) can be found in the articles of Sebastiao e Silva. A more accessible version is contained in the book "Introduction to the theory of distributions" by Campos Ferreira which is based on lectures of the former.

Using this definition and the fact that the $\delta$-distribution is half of the second derivative of the absolute value function, one can give a rigorous proof of the formula in the query. I can supply the details if so desired.

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The big thing you're overlooking is that you got the coordinate substitution $u = f(x)$ wrong. You assumed:

  • $f(x)$ is an invertible function
  • $\lim_{x \to -\infty} f(x) = -\infty$
  • $\lim_{x \to +\infty} f(x) = +\infty$

If these were true, then your calculation is correct -- and the sum has only one term and the thing inside the absolute value is positive. But when these aren't true your calculation is wrong.

Now that you know your mistake, it should be worth trying to the calculation again, being careful to get the substitution right.

If you need more hints,

  • The absolute value will appear when you properly handle the possibility that $f(x)$ is decreasing
  • The sum will appear when you properly handle non-invertible $f(x)$; e.g. by splitting the domain of integration into regions each of which $f(x)$ is invertible.
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