Dirac delta function have this property:
\begin{equation}
\delta(f(x))=\textstyle \sum_i\frac{\delta(x-a_i)}{\lvert f^\prime(a_i)\rvert}.
\end{equation}
And its derivation is:
\begin{eqnarray}
\int_{-\infty}^{\infty}g(x)\delta(f(x))&=&\sum_i\int_{a_i-\epsilon}^{a_i+\epsilon}g(x)\delta(f(x)),\qquad f(a_i)=0\cr
&=&\sum_i\int_{f(a_i-\epsilon)}^{f(a_i+\epsilon)}g(f^{-1}(y))\delta(y)\frac{1}{\lvert f^\prime(f^{-1}(y))\rvert}dy,\qquad x=f^{-1}(y)\cr
&=&\sum_i\frac{a_i}{\lvert f^\prime(a_i)\rvert}.\cr
&&\therefore\ \delta(f(x))=\sum_i\frac{\delta(x-a_i)}{\lvert f^\prime(a_i)\rvert}.
\end{eqnarray}
Here are questions:
- How about if $f^\prime(a_i)$ is 0?
- I think $\int_{-\infty}^{\infty}g(x)\delta(f(x))=\sum_i g(a_i)$ because Dirac delta function $\delta(f(x))$ in LHS is zero at $a_i$, and so $\int\delta(x)f(x)\ dx$ is just $f(x)$. What's wrong with this?