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I am trying to find a formal proof for the following property of Dirac delta function:

for any function $f$ : $$\int_{-\infty}^{+\infty} \delta(x)f(x)dx=f(0),$$ where $\delta$ is Dirac delta generalized function.

Sorry if the above $f$ must have some properties but I think it shouldn't. That is what I want the whole text and the proof.

Is there anyone who knows?

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  • $\begingroup$ This is more or less the definition of $\delta$ ($f$ should be continuous on a neighborhood of $0$ for it to make sense). $\endgroup$ – mrf Jan 22 '16 at 13:41
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    $\begingroup$ You need the theory of distributions to understand why that property holds. You cannot use real calculus, because $\delta(x)$ is not a function. $\endgroup$ – Von Neumann Jan 22 '16 at 15:38
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Indeed, as you were told, there is no proof of the formula you're asking, this is a definition. However, here is the way why the expression on the left is used in so many contexts.

Consider a sequence of function $$ \delta_\epsilon(x)=\begin{cases}\frac{1}{2\epsilon},&-\epsilon<x<\epsilon,\\ 0,&\mbox{otherwise}. \end{cases} $$

It is reasonable to expect that $$ \delta(x)=\lim_{\epsilon\to 0}\delta_\epsilon(x). $$

Now consider the integral and assume that we are free to change the order of operations: $$ \int_{\mathbb R}\delta(x)f(x)dx=\lim_{\epsilon\to 0}\int_{\mathbb R}\delta_{\epsilon}(x)f(x)dx=\lim_{\epsilon\to 0}\int_{-\epsilon}^{\epsilon}\frac{1}{2\epsilon} f(x)dx=\lim_{\epsilon\to0}2\epsilon\cdot \frac{1}{2\epsilon} f(\xi), $$ where due to the mean value theorem $\xi\in(-\epsilon,\epsilon)$.

Hence we can conclude that $$ \lim_{\epsilon\to 0}f(\xi)=f(0), $$ which gives you a "proof" of the original formula.

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