I have often seen the following equality in Physics textbooks. $$\int_{\mathbb{R}}\delta\left(\alpha x\right)f\left(\alpha x\right)|\alpha|dx=\int_{\mathbb{R}}\delta(u)f(u)du$$ or $$\int_{-\infty}^\infty \delta(\alpha x)\,dx =\int_{-\infty}^\infty \delta(u)\,\frac{du}{|\alpha|} =\frac{1}{|\alpha|} .$$ What does this equality mean mathematically, and how can this be made rigorous in the context of Dirac mass? Does this have any bearing on the change of variables in an integration where a function is being integrated with respect to a Dirac mass? for example, in the following integral

$$\int_{\mathbb{R}}f(\alpha x)d\delta_{x_0}(x),$$ does a scaling factor of $|\alpha|$ come out somehow?

Thank you.


1 Answer 1


To understand this properly, I suggest to look at distributions and how operations with distributions are defined in the first place.

A distribution is an object that acts on the space of infinitely differentiable and compactly supported functions in a linear and continuous way (check a textbook or Wikipedia for the precise definition). I.e. a distribution $T$ on a set $\Omega\subset\newcommand{\RR}{\mathbb{R}}\RR^n$assigns to any infinitely differentiable and compactly supported function $\phi$ defined on $\Omega$ a complex number $T(\phi)$. Then one notes that locally integrable function $f$ defined on $\Omega$ induced a distribution $T_f$ via the operation $$T_f(\phi) = \int_\Omega f(x)\phi(x)dx.$$ Now one can try to define operations which one can do to a locally integrable function also for a distribution by analogy. Take, for example, translation (in the case $\Omega=\RR^n$): define the operation $t_y(f)$ defined by $t_y(f)(x) = f(x-y)$. Observe that $$T_{t_y(f)}(\phi) = \int t_y(f)(x)\phi(x)dx = \int f(x-y)\phi(x)dx = \int f(x)\phi(x+y)dx = T_f(t_{-y}(\phi)).$$ I.e. "translating the function $f$ is the same as translation the test function $\phi$ in the opposite direction". Hence, one defines the translation of the distribution $T$ as $$t_y(T)(\phi) := T(t_{-y}\phi).$$ (Work out, that the translation of the Dirac $\delta$ is what you think it should be.) Now you should be able to do the same thing with scaling $s_a(f)(x) = f(ax)$.

  • $\begingroup$ I like this answer a lot! Both rigorous and comprehensible. $\endgroup$
    – Lord_Farin
    Feb 28, 2013 at 7:37
  • $\begingroup$ So, if we define $s_aT(\phi)=|a|^{-n}T(s_{1/a}\phi)$, for Dirac Mass this becomes $s_a\delta_{x_0}=|a|^{-n}\delta_{x_0}$, but what bearing would that have on a possible change of variable formula for the Dirac mass. $\endgroup$
    – user38404
    Feb 28, 2013 at 7:38
  • $\begingroup$ The notation $\displaystyle \int \delta(\alpha x) f(x)\,\mathrm dx$ is not but a shorthand physics uses where $(s_\alpha \delta) (f)$ is intended. It has the pleasant property of being well-behaved WRT intuitive application of change of variables in context of distributions. $\endgroup$
    – Lord_Farin
    Feb 28, 2013 at 7:41
  • $\begingroup$ As for the formula for change of variables: There are formula for that (with quite general measures, see here en.wikipedia.org/wiki/…) but often there is some assumption on absolute continuity involved (somehow, the change of variables is not allowed to "concentrate mass where the measure can't see it). However, what is your motivation to get a formula for the change of variables? In the last example of the OP, just apply the definition of integration w.r.t $\delta_{x_0}$, evaluating the function to be integrated at $x_0$. $\endgroup$
    – Dirk
    Feb 28, 2013 at 7:49
  • $\begingroup$ So, may I think of $s_a\delta$ as a "rescaled measure" that gives a formula for change of variables which is similar to the usual change of variables when integration is done with respect to the Lebesgue measure? $\endgroup$
    – user38404
    Feb 28, 2013 at 8:31

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