I have the next equation:$ \int_{-\infty }^{\infty}e^{x}\delta (x^{2}-2x)dx
$
The solution for this equation is:
$\int_{-\infty }^{\infty}e^{x}\delta (x^{2}-2x)dx= \int_{-\infty }^{\infty}e^{x}[\frac{1}{2}\delta (x)+\frac{1}{2}\delta (x-2)]dx=\frac{1}{2}(1+e^{2}) $
I don’t understand how they split the delta function and got $\frac{1}{2}$ in both of the Deltas. I have tried to look at the properties of the function but haven’t found something similar.
Hint:$$\delta(g(x))~=~\sum_{x_0}^{g(x_0)=0}\frac{\delta(x\!-\!x_0)}{|g^{\prime}(x_0)|}~\stackrel{\begin{array}{c}g(x)=x(x-2)\cr g^{\prime}(x)=2x-2 \end{array}}{=}~\frac{1}{2}\delta(x)+\frac{1}{2}\delta(x\!-\!2).$$