Why is $\pm \int_{-\infty}^{\infty} f(y/k)\delta (y)\frac{dy}{k} = \pm \frac{1}{k} f(0) $? From Griffiths, Introduction to Electrodynamics, pg. 48:

In the image above, the author makes the statement in the title of this question. But shouldn't it be
$$\pm \int_{-\infty}^{\infty} f(y/k)\delta (y)\frac{dy}{k} = \pm \frac{1}{k^2} f(0) \text{ ?}$$
Why can't you say that $f(y/k) = \frac{1}{k} f(y)$
 A: As MichaelHardy notes, it is not true that $f(y/k) = f(y)/k$, in general.
Let $h(x) = x/k$ and notice that $h(0) = 0$. Then, when we change variables $kx = y$, we have that $k\,dx = dy$. Hence, when we make the substitution in the integral, we have the composition $f\circ h$ in the numerator:
$$
\int_{-\infty}^\infty f(x)\delta(kx)\,dx = \int_{-\infty}^\infty \frac{f(y/k)}{k}\delta(y)\,dy = \int_{-\infty}^\infty \frac{(f\circ h)(y)}{k}\delta(y)\,dy=  \frac{1}{\vert k\vert}(f\circ h)(0) = \frac{1}{\vert k\vert}f(0)
$$
since $\int_{-\infty}^\infty g(t)\delta(t)\,dt = g(0)$, by property of the delta function. We have $\pm 1/k$ as a result of following through with our change of variables and noticing the cases where $k \lt 0$ and $k \gt 0$. I think you should see that we don't get $\pm 1/k^2$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Indeed, there are too many unnecessary steps. It's sufficient with
  $\ds{\delta\pars{kx} =
{\delta\pars{x} \over \verts{\dd\pars{k\xi}/\dd\xi}_{\xi = 0}} = {\delta\pars{x} \over \verts{k}}}$

\begin{align}
\int_{-\infty}^{\infty}\mrm{f}\pars{x}\delta\pars{kx}\,\dd x & =
\int_{-\infty}^{\infty}\mrm{f}\pars{x}\,{\delta\pars{x} \over \verts{k}}\,\dd x
= {1 \over \verts{k}}\,\mrm{f}\pars{0}
\end{align}
