While there is no trouble finding the formula for functions with distinct simple roots

$$\delta(f(x)) = \sum_{i:f(x_i)=0} \frac{\delta(x-x_i)}{|f'(x_i)|} \tag{1}$$

e.g. on Wikipedia, I cannot find anything on functions with roots of higher multiplicity. Thus my question, how to treat that?

  • $\begingroup$ Simply use $$\int_{-\infty}^\infty \delta(f(x))\varphi(x)dx \overset{def}=\lim_{\epsilon \to 0^+}\int_{-\infty}^\infty \frac{1_{|f(x)| < \epsilon}}{2 \epsilon}\varphi(x)dx $$ Note $\delta(f(x))$ is then well-defined as a distribution iff the limit exists and is continuous in $\varphi\in C^\infty_c(\mathbb{R})$, which is not the case when $f(x) \ne \mathcal{O}(x-a)$ at every $a, f(a) = 0$ $\endgroup$
    – reuns
    Commented Dec 12, 2017 at 11:20
  • $\begingroup$ @reuns Indeed, so for $f(x)$ with critical points $\delta(f(x))$ is not a "pure" distribution, but rather a "weak" or "critical" one which limits the test functions $\phi$ to those vanishing at the critical points of $f(x)$. And if my answer is not horribly wrong it is actually sufficient that the $\phi$ have simple roots at $f$'s critical points, and that's just a countable set... This is probably related to the problems of function inversion at critical points $\endgroup$ Commented Dec 12, 2017 at 13:02
  • $\begingroup$ If $f$ has only one zero at $0$ and $f'(0) = 0, f''(0) = 2C \ne 0$ then $1_{|f(x)| < \epsilon} \approx 1_{|C x^2| < \epsilon}$ so that $$\lim_{\epsilon \to 0^+}\int_{-\infty}^\infty \frac{1_{|f(x)| < \epsilon}}{2 \epsilon}\varphi(x)dx=\lim_{\epsilon \to 0^+}\int_{-(\epsilon/C)^{1/2}}^{(\epsilon/C)^{1/2}} \frac{1_{|f(x)| < \epsilon}}{2 \epsilon}\varphi(x)dx = C^{-1/2} \lim_{x \to 0} \varphi(x) x^{-3/2}$$ (if the limit exists) $\endgroup$
    – reuns
    Commented Dec 12, 2017 at 14:32
  • $\begingroup$ Possible duplicate: math.stackexchange.com/q/2481114/11127 $\endgroup$
    – Qmechanic
    Commented Dec 12, 2017 at 19:07
  • $\begingroup$ @reuns I'm afraid I cannot entirely follow your derivation, could you please elaborate in an answer? $\endgroup$ Commented Dec 17, 2017 at 19:48

1 Answer 1


For each root consider the local Taylor series, truncated to the first non-vanishing term:

Let $x_i$ denote the distinct roots of $f(x)$ with a multiplicity of $n_i$ respectively. Then

$$\delta(f(x)) = \sum_i \delta\Bigg(\frac{f^{n_i)}(x_i)}{n_i!}(x-x_i)^{n_i} + \mathcal O\Big((x-x_i)^{n_i+1}\Big)\Bigg).$$

Since $\delta$ vanishes almost everywhere, the higher order terms can be dropped, and thus only $\delta(ax^n)$ needs to be determined. Change of variables yields

$$\begin{align*} \delta(ax^n)\,dx &= \delta((\sqrt[n]ax)^n)\frac{d(\sqrt[n]ax)}{|\sqrt[n]a|} \\ &= \delta((\sqrt[n]ax)^n)\frac{d(\sqrt[n]ax)^n}{|\sqrt[n]a\cdot n(\sqrt[n]ax)^{n-1}|} \qquad\Bigg|\quad ax^n \to x \\ &= \delta(x)\frac{dx}{|n\sqrt[n]ax^{\frac{n-1}n}|} \\ &= \frac{\delta(x)}{|n\sqrt[n]{ax^{n-1}}|}\,dx. \end{align*}$$

Putting it all together we then get

$$\boxed{\delta(f(x)) = \sum_i\frac{\delta(x-x_i)}{\Bigg|n_i\sqrt[n_i]{\frac{f^{(n_i)}(x_i)}{n_i!}(x-x_i)^{n_i-1}}\Bigg|}}$$

For $n_i=1$ this is consistent with $(1)$, for $n_i>1$ the singularity remains and apparently can only be cancelled when the $\delta$-distribution is applied to a function with a root of at least order $1-\frac1{n_i}$.

It might be interesting to consider $n_i\to\infty$, but it's getting late...


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