Show that $\delta(at)=\left(\frac{1}{|a|}\right)\delta(t)$.

Question

The unit impulse function $\delta(t)$ is defined in terms of the integral $$\int_{-\infty}^\infty x(t)\delta(t)dt=x(0)$$ where $x(t)$ is any test function that is continuous at $t=0$. Show that $\delta(at)=\left(\dfrac{1}{|a|}\right)\delta(t)$.

Answer 1

$$\int x(t)\delta(at)dt=$$ $$\int x(\frac{t'}{a})\delta(t')\frac{1}{a} dt'=$$ $$\frac{1}{a}\int x(t'/a)\delta(t')dt'=$$ $$\frac{1}{a}x(0/a)=\frac{1}{a}x(0)=$$ $$\frac{1}{a}\int x(t)\delta(t)dt$$

You can get the absolute value of $|a|$ with using that $x(-0)=x(0)$ and $\delta(t)=\delta(-t)$ so the sign of $a$ does not matter.