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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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    $\delta$ is an even distribution2016-06-11

3 Answers 3

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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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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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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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    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.2012-11-13