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.

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.
