How to deal with the dirac delta function with integrand from $0$ to $\infty$? For this identity of the delta function:
$$
\int_{-\infty}^\infty f(x)\delta(g(x)) = \sum_{n = 1}^N\int_{-\infty}^\infty dx f(x)\frac{\delta(x-x_n)}{|g'(x_n)|}
$$
Where $x_n$'s are zeros of $g(x)$. Does it still hold if the integrand goes from $0$ to $\infty$ instead?
Thanks!
 A: Edit: I tried to make my answer a bit more precise after some discussion in the comments.
I want to add the following elements to the discussion.

*

*The formula you wrote above makes sense only for regular enough (let's assume $g$ is a $C^1$ function) and suitable enough functions $g$ (for instance, the zeroes of $g$ must be isolated and $g'(z_0)\neq 0$ for any $z_0$ such that $g(z_0)=0$; the second condition actually implies the first one). In this hypothesis, the formula you wrote above works (in the sense that any of the two quantities is well-defined as soon as the other one is well-defined, and in that case they coincide; so for instance, the sum on the right hand side must be finite, or if it's an infinite series it must converge absolutely; or the series could actually diverge, so in that case the integral is $+\infty$ or $-\infty$ if your series diverges independently on the rearrangement of the summands). If your function $g$ is suitable enough, you can in fact integrate the above expression from zero to infinity, but you miss all the zeroes of $g$ that lie in the interval $(-\infty,0]$. Even if the zeroes of $g$ and the function $f$ are even with respect to zero, you need to multiply by $2$ to make the formula work because you need to count each zero twice.


*What I just wrote only works if $0$ is not a zero of $g$, i.e., $g(0)\neq 0$. If $g(0)=0$, then you really have to think on what you are doing. In this case, you should specify whether you are integrating on the interval $(0,\infty)$ or $[0,\infty)$ to decide whether to include $0$ in the sum on the right-hand side. So, in short, if you just integrate from $0$ to infinity, you have to be careful in seeing where your zeroes of $g$ lie.


*In response to the comments of @Mark Viola: given a Radon measure $\mu$ and a smooth (I think bicontinuous is enough) change of variables $\tau$, it is possible to define the pull-back measure $\mu\circ\tau$ such that
$$ \mu\circ\tau(A)=\mu(\tau(A)), $$
and it is possible to integrate a function with respect to that measure. You can even do things more generally without assuming that your function $\tau$ is invertible (check the first answer of this question). So even if the Dirac delta is not a function (in the mathematical sense), it is still a Radon measure, and you can compose measures with functions under suitable assumptions. And you can integrate continues functions $f$ against a Radon measure (assuming the integral is well defined, so either $f$ is sign-definite, or either the positive or the negative part of $f$ have finite integral); more generally, you can integrate Borel-measurable functions. So no, that integral is not an abuse of notation, it is actually a proper integral in the measure-theoretic sense (I am hiding a lot of details under the carpet, but I think the idea is clear).
Edit.


*A comment on the answer by C4P. Unless I am mistaken, it always makes sense the integral
$$ \int_I fd\delta(x), $$
where $I$ is any interval. You just need to make sure that $I$ is a Borel set (any interval is measurable if we consider Radon measures) and $f$ is a measurable function (continuous functions are Borel-measurable). And clearly, you also need to assume that $f$ is sign-definite, or that the integral of the positive part and the integral of the negative part of $f$ are not both $+\infty$, but this comes for free for the Dirac delta. In the case of the Dirac delta, though, you need to be careful in saying whether your interval is open or closed, essentially because the set $\{0\}$ has non-zero measure.

A: The identity you have given relates to the properties of the delta function, which is a distribution that is often used in the theory of generalized functions. The delta function is defined as
$$\delta(x) = \begin{cases} \infty & x = 0 \\ 0 & x \neq 0 \end{cases}.$$
It is not a function in the usual sense, but rather a distribution that can be used to represent point masses and other singularities.
The identity you have given is known as the "sifting property" of the delta function, and it states that for any smooth function $f(x)$, you have
$$\int_{-\infty}^\infty f(x)\delta(g(x)) = \sum_{n = 1}^N\int_{-\infty}^\infty dx f(x)\frac{\delta(x-x_n)}{|g'(x_n)|},$$
where the $x_n$'s are the zeros of the function $g(x)$. This identity holds as long as the integral on the left-hand side is well-defined.
If you change the limits of the integral on the left-hand side from $-\infty \to \infty$ to $0 \to \infty$, then the identity may not hold in general, since the integral may not be well-defined.
The delta function is not a function in the usual sense, and it is not always possible to integrate it over the whole real line.
If you have specific functions $f(x)$ and $g(x)$ in mind and you want to know whether the identity holds for these functions, you will need to carefully consider the behavior of the functions and the integral to determine whether the integral is well-defined.
