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Does the equality below hold?

$$f(x)=\int_{-\infty}^{\infty}f(x')\delta(x-x')dx=\int_{-\infty}^{\infty}f(x')\delta(x'-x)dx$$

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Why this is true? how could you get $f(x)$ be integrating with variable $dx$? – Bombyx mori Nov 13 '12 at 0:44
    
$\delta$ is an even distribution – user1952009 Jun 11 at 0:40
up vote 1 down vote accepted

Well, $$\int_{\mathbb R} f(x)\delta (x - a)dx = {f(a)\int_{- \infty }^\infty {\delta (x - a)dx}} = f(a) = {f(a)\int_{- \infty }^\infty {\delta (a - x)dx}}$$ so YES, if you change your $dx$ to $dx'$.

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for both sides $dx$ by $dx'$? – user46060 Nov 13 '12 at 1:16
    
Yes. You should have: $f(x) = \int {f(x')\delta (x' - x)dx' = \int {f(x')\delta (x - x')dx'}}$. You could also change $a$ (in my answer) to $x'$. It is just notation. Whatever you prefer. – glebovg Nov 13 '12 at 1:37

The correct form should be:

$f(y)=\int_{-\infty}^{\infty}f(x) \delta(x-y)dx = \int_{-\infty}^{\infty}f(x) \delta(y-x)dx$

When $x=y$, we have $\delta(y-y)=\delta(0)$ and this performs a sampling on the $f(x)$ function at the point $x=y$.

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Yes since the argument of the delta distribution equals zero when x=x' in either case, and the result is f(x) for either integral, assuming dx is changed to dx'.

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