The Dirac Delta function for a fraction

Question

I was studying quantum theory and came to know that the unrealistic function representing the position of a particle can be represented in the form of an infinitely peaked Gaussian or simply the Dirac delta function $\delta(x-x_0)$ where $x_0$ is the definite position with zero uncertainty. The Dirac function can also be written as $$\delta(x-x_0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik(x-x_0)}dk$$ or simply $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik(x)}dk\tag{1}$$ Now in solving a problem where it was asked to find the momentum distribution of the given position wave function given as $$\psi(x)=Ae^{\frac{i}{\hslash}p_0x}$$ where $A$ is a real positive constant. While solving for the momentum wave function $a(p)$ using Fourier transform I get, $$a(p)=\frac{A}{\sqrt{2\pi\hslash}}\int_{-\infty}^{\infty}e^{ikx}dx$$ where the $k={(p_0-p)}/{\hslash}$ . Now just by using the equation $(1)$, I get, $$a(p)=\sqrt{\frac{2\pi}{\hslash}}A\delta(k)\\\implies a(p)=\sqrt{\frac{2\pi}{\hslash}}A\delta\left(\frac{p_0-p}{\hslash}\right)\tag{2}$$ This was simply my answer, but the book shows a step forward which I am unable to understand how they did this. It shows the final answer as $$a(p)=\sqrt{{2\pi\hslash}}A\delta(p-p_0)\tag{3}$$ which is obviously the expected one as the position wave function $\psi(x)$ given is the basis eigenfunction for momentum, so we can expect to get $a(p)$ as in equation $(3)$ so that we can find the momentum of the particle to be $p_0$ with zero uncertainty. So my question at last is that how to operate such a Dirac function for which the argument is a fraction(as in equation $(2))$?

Answer 1

$\delta(ax) = \frac{1}{|a|} \delta(x)$