This will be a quick post on handling functions inside Dirac-delta function. Let us start simple and figure out the constant scaling of the argument first, i.e., \(\delta(\alpha \, x)\). This will certainly be proportional to \(\delta(x)\) since the zero is still at \(x=0\). But it will pick up an overall factor since Dirac delta function is defined as a density, i.e., it integrates to \(1\). Let’s see what \(\delta(\alpha \, x)\) does under the integral:
\[\begin{eqnarray} \lim_{R\to \infty}\int^R_{-R}dx \delta(\alpha \, x) = \lim_{R\to \infty}\int^{R/\alpha}_{-R/\alpha}d\left(\frac{y}{\alpha}\right) \delta( y) = \frac{1}{\alpha}\lim_{R\to \infty}\int^{R/\alpha}_{-R/\alpha}dy \delta( y)=\begin{cases} \frac{1}{\alpha}, & \text{if $\alpha>0$}\\ -\frac{1}{\alpha}, & \text{if $\alpha<0$} \end{cases}=\frac{1}{|\alpha|} \tag{1}, \end{eqnarray}\] where the negative sign appears since we have to flip the integral limits when \(\alpha<0\).
Now Consider a function \(f(x)\) which has its zeros at points \(x_i\). Now we just stick this inside \(\delta()\), and it will have its peaks at \(x=x_i\):
\[\begin{eqnarray} \delta\left(f(x)\right)&=&\sum_i c_i \delta(x-x_i) \tag{2}, \end{eqnarray}\] and the goal is to find the \(c_i\)’s. We expand \(f(x)\) around \(x_i\) as \(f(x)=(x-x_i)f'(x_i)\) when \(x\) is in the vicinity if \(x_i\). This gives \[\begin{eqnarray} \delta\left(f(x)\right)_{x\sim x_i}&=&\delta\left((x-x_i) f'(x_i)\right)=\frac{1}{|f'(x_i)|}\delta\left(x-x_i \right), \tag{3} \end{eqnarray}\] wher we used Eq. (1). Putting it all together, we get: \[\begin{eqnarray} \delta\left(f(x)\right)&=&\sum_i \frac{1}{|f'(x_i)|}\delta\left(x-x_i \right), \tag{4} \end{eqnarray}\] which expresses the Dirac-delta of a function as a series of basic Dirac-delta functions.