$\int dx \, \delta(h(x))=\frac{1}{\mid h'(x) \mid}$ integration with delta-function I am confused about the following picture coming from a textbook:  

My question is why the following holds:  
$$\int dx \, \delta(h(x))=\frac{1}{\mid h'(x) \mid}$$  
if you just simply consider there is only one intersection point.  
My naïve derivation is the following:   
$$\left. \int dx \, \delta(h(x))=\lim_{\varepsilon\rightarrow0}x \, \right|_{x^*-\epsilon}^{x^*+\epsilon}$$  if $x^*$ is the only intersection point
 A: Delta functions, and in general distributions, are determined by their action on test functions. And in addition, since "delta function" is not a function, we have to give a meaning to the notation $\delta(h(x))$.
As the first step, here is a heuristics that tells why the reciprocal of the derivative pops up: let $h : \mathbb{R} \to \mathbb{R}$ be a nice bijection with the nice inverse $g$. Then for any test function $\varphi$,
$$ \int_{\mathbb{R}} \delta(h(x)) \varphi(x) \, dx
\stackrel{x = g(y)}{=} \int_{\mathbb{R}} \delta(y)\varphi(g(y)) |g'(y)| \, dy
= \varphi(g(0)) |g'(0)|
= \frac{\varphi(g(0))}{|h'(g(0))|}. $$
Since $x^* := g(0)$ is the unique zero of $h$, this tells that $\delta(h(x)) = \frac{1}{|h'(x^*)|} \delta(x - x^*)$ modulo the fact that we have not defined $\delta \circ h$ and this kind of computation is a priori valid only when $\delta$ is a regular function, which is of course not the case.
So let us give a possible way of defining $\delta\circ h$ and justifying its property. The idea is that $\delta$ can be approximated by regular functions, and the manipulation above may be justified for $\delta_n$ instead. So let $(\delta_n)$ be a sequence of locally integrable functions such that $\delta_n \to \delta$ in the space $\mathcal{D}'(\mathbb{R})$ of distributions. For simplicity, we make an assumption that $\delta_n$ concentrates at $0$ even in pointwise sense, whose precise meaning is that


*

*$ \forall \epsilon > 0$, we have $\displaystyle \lim_{n\to\infty} \sup_{|x|\geq\epsilon} |\delta_n(x)| = 0. $


Although this is a technical assumption for a short proof, it is satisfied for a wide range of examples of $(\delta_n)$ that are commonly used (such as rectangular peak, Gaussian kernel or Dirichlet kernel). So it is general enough for our justification. Also we make the following assumptions on the regularity of $h$:


*

*$h$ is in $C^1(\mathbb{R})$, i.e., $h$ is continuously differentiable.

*The zero set $A = \{x : h(x) = 0\}$ of $h$ is isolated.

*$h'(a) \neq 0$ for each $a \in A$.


Then we can find a disjoint open intervals $\{ U_a : a \in A \}$ such that $h$ restricted to $U_a$ is invertible with the local $C^1$-inverse $g_a : V_a \to U_a$.
Now for each test function $\varphi \in \mathcal{D}(\mathbb{R})$, the function $h$ is non-zero on $K = \operatorname{supp}(\varphi) \setminus \cup_{a\in A} U_a$, which is compact. So there is $\epsilon > 0$ such that $|h| \geq \epsilon$ on $K$. Then
\begin{align*}
\int_{\mathbb{R}} \delta_n(h(x)) \varphi(x) \, dx
&= \sum_{a \in A} \int_{U_a} \delta_n(h(x)) \varphi(x) \, dx + \int_{K} \delta_n(h(x)) \varphi(x) \, dx \\
&= \sum_{a \in A} \int_{V_a} \delta_n(y) \varphi(g_a(y)) |g_a'(y)| \, dy + \mathcal{O}\left( \| \varphi\|_{L^1} \sup_{|x|\geq\epsilon} |\delta_n(x)| \right).
\end{align*}
Since $A$ is isolated and $\varphi$ is compactly supported, there are only finitely many indices $a \in A$ for which $\varphi$ does not vanish on $U_a$. So we can take limit $n\to\infty$ termwise, and as $n\to\infty$ the above integral converges to
$$ \lim_{n\to\infty} \int_{\mathbb{R}} \delta_n(h(x)) \varphi(x) \, dx
= \sum_{a \in A} \varphi(g_a(0)) |g_a'(0)|
= \sum_{a \in A} \frac{\varphi(a)}{|h'(a)|}. $$
This tells that the distributional limit of $\delta_n(h(x))$ is exactly $\sum_{a\in A} \frac{1}{|h'(a)|}\delta(x - a)$.
A: This is intended to be a non-rigorous approach. Someone else can probably give a rigorous proof based on distributions.
I've found it helpful to keep in mind what the delta function does: 
$\delta (x)=0$ for $x \neq 0$ and $\displaystyle \int_a^b \delta (x)dx = 1$  for $a < 0 < b$ (normalization).
To start, $\delta (h(x))$ will vanish except where $h(x) = 0$. Suppose this happens at $x^*$. We can then write
$$\displaystyle \int_{-\infty}^{\infty} \delta (h(x))dx = \int_{x^* - \epsilon}^{x^* + \epsilon} \delta (h(x))dx$$
When $x \approx x^*$ , $h(x) \approx h'(x^*)(x - x^*)$. Using this with the scaling property, $\delta (ax) = \dfrac{1}{|a|} \delta(x)$,
$$\int_{x^* - \epsilon}^{x^* + \epsilon} \delta (h(x))dx = \int_{x^* - \epsilon}^{x^* + \epsilon} \delta (h'(x^*)(x - x^*))dx = \dfrac{1}{|h'(x^*)|} \int_{x^* - \epsilon}^{x^* + \epsilon} \delta (x - x^*)dx = \dfrac{1}{|h'(x^*)|}$$
The last integral evaluates to $1$ from the normalization property.
