From Kanwal's Generalized Functions pg. 49:
Question: How does the author move from
$$ ∫_{-∞}^∞ δ[f(x)]ϕ(x)dx = ∫_{-∞}^∞ δ(y)\frac{ϕ[x(y)]}{f'(x)}dy $$
via Integration by Substitution at the start of the second equation? Specifically how does $f'(x)$ end up in the denominator (this makes no sense to me since $x$ is now interpreted as some fixed constant?)? It seems to me that we should actually have
$$ ∫_{-∞}^∞ δ[f(x)]ϕ(x)dx = ∫_{-∞}^∞ δ(y)ϕ[x(y)]f'(y)dy $$
How am I mistaken?
If your question is only about the change of variable you can read it backwards. We want to give meaning to $\delta(f(x))$, start with $$ \int_{\mathbb{I}} \delta(f(x)) \, \phi(f(x)) \, f'(x) dx \; , $$ where $\mathbb{I}$ is some neighbourhood of the point $x_1$ where $f(x_1)=0$. And now change variables. Let $y=f(x)$, $dy = f'(x) dx$ , then $$ \int_{\mathbb{I}} \delta(f(x)) \, \phi(f(x)) \, f'(x) dx \; = \; \int_{f(\mathbb{I})} \delta(y) \, \phi(y) \, dx \; = \; \phi(0) . $$ This is independent of the neighbourhood $\mathbb{I}$. In this region $f'(x)$ is nonzero by assumption. The distribution $\delta(f(x))$ must constraint $x=x_1$ and cancel the factor $ f'(x)$, thus the only possibility is $$ \delta(f(x)) \; = \; \frac{ \delta(x-x_1) }{ f'(x_1) } \; . $$
Thus \begin{align} \int_{\mathbb{I}} \delta(f(x)) \, \phi(f(x)) \, f'(x) dx \; & = \; \int_{\mathbb{I}} \delta(x-x_1) \, \phi(f(x)) \, \frac{f'(x)}{f'(x_1)} \\ &= \; \phi(f(x_1)) \, \frac{f'(x_1)}{f'(x_1)} \; = \; \phi(0) \end{align}
A simple example
Calculate the integral (for $x_0>0$) $$ \int_{0}^{\infty} \delta(x^2 - x_0^2) \, \phi(x) dx \; . $$ From one side we could use $y=x^2-x_0^2$, $dy = 2x dx$. Let $x(y) = \sqrt{y+x_0^2}$ where $x(0) = x_0$, then $$ \int_{0}^{\infty} \delta(x^2 - x_0^2) \, \phi(x) dx \; = \; \int_{-x_0^2}^{\infty} \delta(y) \frac{\phi[x(y)]}{2x(y)} dy \; = \; \frac{\phi(x_0)}{2x_0} \; . $$ On the other hand, we could use the expression for the delta function. Let $f(x)=x^2-x_0^2$. In this region, $f$ has only one zero (that happens at $x=x_0$) and $f'(x)=2x > 0$. Thus $$ \delta(x^2-x_0^2) = \frac{\delta(x-x_0)}{2x_0} \; , \text{ for } x> 0 \; . $$ Evidently, this gives the same answer for the integral.
Note that the statement of the book can be generalised to incorporate negative values to the derivative and more than one point where $f(x)=0$. In this example, for the general case where $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=x^2-a^2$ the Dirac delta is $$ (\delta \circ f)(x) \; = \; \delta(x^2-a^2) \; = \; \frac{1}{2\vert a \vert } \, \Big( \,\delta(x-a) + \delta(x+a) \, \Big) \; .$$