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Specifically I am working on a problem of the form: $$ \int_0^{\infty} \varphi(x) \delta( x^2 - \alpha^2) dx. $$

I know that when composed with a smooth function $g(x)$ we have: $\delta(g(x))= \displaystyle{\sum_i} \frac{\delta(x-x_i)}{|g'(x_i)|}$, we may then proceed to compute: $$ \int_0^{\infty} \varphi(x) \frac{\delta( x - \alpha)}{2|\alpha|} dx + \int_0^{\infty} \varphi(x) \frac{\delta( x + \alpha)}{2|-\alpha|} dx.$$

It is at this point that I refused to take $$ \int_a^{b} \varphi(x) \delta(x-x_0) dx = \begin{cases} \varphi(x_0) & x_0 \in (a,b) \\ 0 & x_0 \notin (a,b)\\ \end{cases} $$

at face value. In another answer I read it was noted that giving: $ \quad \delta[\varphi]= \displaystyle{\int_{-\infty}^{\infty}} \varphi(x)\delta(x)dx, \ $ is a convenient notation rather than a formal definition of the $ \delta$-distribution with respect to the integral, which makes complete sense to me. My question is this then, how do we make sense of the expression which I am dealing with? I have never taken a class on functional analysis, but I am familiar with the concepts and have a good idea of the algebraic nature of the space of test functions.

Mike_or Mite
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  • Hello! I don't quite understand your problem. You have arrived, it appears, to the correct answer. What exactly doesn't seem right to you? And what is weird about the interval $[0, \infty)$? – Daigaku no Baku Feb 05 '25 at 22:40
  • @DaigakunoBaku I guess "weird" is overly subjective, I meant more to say on an interval not $ (-\infty, \infty) $. As to the point of my question, I included the problem I was working on merely to give context. My question really is, no definition/construction of the dirac delta that I've encountered has given meaning to the expression which I was asked to evaluate, so how might one rigorously understand such an expression? – Mike_or Mite Feb 06 '25 at 01:30
  • What you can do is perform a transformation of the integration variable $x$ to $y = x^2$. Then the Dirac delta function becomes $\delta(y-\alpha ^2)$. Hence the integration becomes a standard one. – M. Wind Feb 06 '25 at 03:00
  • Do you think this answer helps? https://math.stackexchange.com/a/4857780 (referencing a general characterization of $u(g(x))$ where $u$ is a distribution, including $\delta(g(x))$ and $\delta(x - x_0)$ as special cases). – echinodermata Feb 06 '25 at 05:24
  • If we attemp to assign a meaning to the object $\int_0^\infty \phi(x)\delta(x^2-\alpha^2),dx$ as the distribution $\langle \delta \circ f, H\phi\rangle$, then we encounter an immediate problem for $\alpha=0$. Namely, the function $H\phi$ is discontinuous at the origin. Hence, $H\phi$ is not a suitable test function and the object is not defined as a distribution for $\alpha=0$. – Mark Viola Feb 07 '25 at 17:52
  • @MarkViola After having read the answers you linked (which are superb by the way thank you) it seems to me that this sort of expression is only meaningful when we characterize the Dirac Delta as a measure? I hope I'm not missing something glaring. – Mike_or Mite Feb 25 '25 at 18:26
  • @Mike_orMite Well, even if one characterizes $\delta$ as a measure, one still needs to be careful with the lower limit. Is the interval $(0,\infty)$ or is it $[0,\infty)$. – Mark Viola Feb 25 '25 at 18:29

1 Answers1

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Your problem can be easily circumvented by recovering the whole real line as a domain of integration with the help of the indicator function, as it follows : $$ \begin{align} \int_a^b \phi(x)\delta(x-x_0) \,\mathrm{d}x &= \int_{-\infty}^\infty \phi(x)\mathbb{1}_{(a,b)}(x)\delta(x-x_0) \,\mathrm{d}x \\ &= \phi(x_0)\mathbb{1}_{(a,b)}(x_0) \\ &= \begin{cases} \phi(x_0) & \text{if } x_0 \in (a,b) \\ 0 & \text{otherwise} \end{cases} \end{align} $$ As a side note, let's highlight that the final result when the point $x_0$ lies on the boundary of the domain of integration, i.e. in the case of $a = x_0$ or $b = x_0$, is undefined a priori, because it depends on the convention used for the value of the Heaviside function at the origin $-$ i.e. $H(0) \in \{0,\frac{1}{2},1\}$. Indeed, the latter function being the antiderivative of the Dirac delta, integration by parts leads to $$ \begin{align} \int_a^b \phi(x)\delta(x-x_0) \,\mathrm{d}x &= \int_a^b \phi(x)H'(x-x_0) \,\mathrm{d}x \\ &= [\phi(x)H(x-x_0)]_a^b - \int_a^b \phi'(x)H(x-x_0) \,\mathrm{d}x \\ &= \phi(b)H(b-x_0) - \phi(a)H(a-x_0) - \int_{\max(a,x_0)}^{\min(b,x_0)} \phi'(x) \,\mathrm{d}x \\ &= \begin{cases} \phi(x_0) & \text{if } x_0 \in (a,b) \\ \phi(x_0)H(0) & \text{if } x_0 = a \text{ or } x_0 = b \\ 0 & \text{otherwise} \end{cases} \end{align} $$

Abezhiko
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