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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.

violettagold
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  • Please mention the source where you are reading from. In particular, I have never seen a function inside the delta, so this must be defined in the text before it's use – Sarvesh Ravichandran Iyer Feb 12 '20 at 04:48
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    It is known in general that $$\int_{-\infty}^{+\infty},f(x),\delta\big(g(x)\big),\text{d}x=\sum_{t\in g^{-1}(0)},\frac{f(t)}{\big|g'(t)\big|},$$ where $f,g:\mathbb{R}\to\mathbb{R}$ are such that $g$ is continuously differentiable, $f$ is sufficiently well behaved and continuous near the zeros of $g$, and $g'$ does not vanish at each zero of $g$. In other words, $$\delta\big(g(x)\big)=\sum_{t\in g^{-1}(0)},\frac{\delta(x-t)}{\big|g'(t)\big|}.$$ – Batominovski Feb 12 '20 at 05:02
  • This was a homework assignment. Can you please show me how this property work in this question, I can’t get the right answer. What is in this case? – violettagold Feb 12 '20 at 07:09

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Hint:$$\delta(g(x))~=~\sum_{x_0}^{g(x_0)=0}\frac{\delta(x\!-\!x_0)}{|g^{\prime}(x_0)|}~\stackrel{\begin{array}{c}g(x)=x(x-2)\cr g^{\prime}(x)=2x-2 \end{array}}{=}~\frac{1}{2}\delta(x)+\frac{1}{2}\delta(x\!-\!2).$$

Qmechanic
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