Simple property of Dirac's $\delta$-function. I'm on Page 63 of R. Shankar's "Principles of Quantum Mechanics".
I'm trying to do Exercise 1.10.1 by proving that $\displaystyle{\delta(ax) = \frac{\delta(x)}{|a|}}$, where $a \in \mathbb  R \backslash\{0\}$.
We know that $\displaystyle{\int_{-\infty}^{+\infty}\delta(t)~\mathrm dt = 1}$.
Making the substitution $t=ax$, where $a>0$ gives
$$ \int_{-\infty}^{+\infty} \delta(ax)~\mathrm dx = \frac{1}{a}$$
Making the substitution $t=-ax$, where $a<0$ gives
$$\int_{-\infty}^{+\infty}\delta(-ax)~\mathrm dx = -\frac{1}{a}$$
Since $\delta(t) \equiv \delta(-t)$ we can re-write this as 
$$\int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = -\frac{1}{a}$$
Hence:
$$\int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = 
\left\{ \begin{array}{ccc} 1/a & : & a>0 \\ -1/a &:& a < 0\end{array}\right.$$
$$\implies \ \ \int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = \frac{1}{|a|} $$
I can see why this might suggest the result, since we can conclude that
$$\int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = \int_{-\infty}^{+\infty}\frac{\delta(x)}{|a|}~\mathrm dx$$
However, I can't see why it proves that $\displaystyle{\delta(ax) = \frac{\delta(x)}{|a|}}$. After all, there are many different functions whose integrals over $\mathbb R$ are all equal.
 A: Dirac delta is not a true function. Functions in the integral sign can be understood as differentials $f(x)\,dx$, which describe a small quantity of size $f(x)\,dx$ sitting on a small interval $[x,x+dx]$. For the Dirac delta, however, the 'differential' $\delta(x)\,dx$ describes the unit mass concentrated near $x = 0$. No true function can cope with this situation.
So, if $\delta$ is not a function then how can we deal with them? You can understand them by examining how they react to various test functions. It is like you wear an eye-mask and try to understand an object by touching it. If you carefully touch the object in every possible way you can think of, then you may identify the shape of it. The same analogy works here. If you have an object $?(x)$ over the space, then you can test it by evaluating the integral
$$ \int \varphi(x) \, ?(x) \, dx $$
with every sensible choices of function $\varphi$. This allows you to pinpoint what $?(x)$ is. For instance, $\delta(x)$ is an object which is characterized by the following property
$$ \int \varphi(x) \delta(x) \, dx = \varphi(0)$$
for all nice functions $\varphi$. For your question, notice that
$$ \int \varphi(x) \delta(ax) \, dx
\stackrel{y = ax}{=} \int \varphi\left(\frac{y}{a}\right) \delta(y) \, \frac{dy}{|a|}
= \frac{1}{|a|}\varphi(0). $$
We have tested how $?(x) = \delta(ax)$ behaves when it is acted upon test functions $\varphi$ and now see that the result is just proportional to what $\delta$ yields. This does indeed tell you that $\delta(ax) = \frac{1}{|a|}\delta(x)$.
A: You could make an even stronger argument:
For all $A \subset \mathbb{R},$ we have $$\int_A \delta(ax)dx = \int_A \frac{\delta(x)}{|a|} dx$$
Now you can use this type of result.
