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1)Prove that the dirac delta function property: $$ x\delta'(x)=-\delta(x)$$

2)and : $$\int_{-\infty}^\infty \delta'(x)f(x)dx=-f'(0) \ $$

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By symmetric, do you actually mean that $\delta(-x) = \delta(x)$? –  Alex Zorn Feb 21 '13 at 16:56
    
I still remember we almost had a revolution in class when our professor introduced the derivative of the delta function. –  Ross Millikan Feb 21 '13 at 17:07

2 Answers 2

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There are two ways to prove these properties: the naive way, and the rigorous way. The naive way is simple: the derivative of a distribution is defined through integration by parts: $$ T^\prime[f]=\int T^\prime f=-\int T f^\prime $$ So for the first one,

$$ \int \delta^\prime(x) (xf(x))dx=-\int\delta(x)(xf(x))^\prime dx=-\int\delta(x)(f(x)+xf^\prime(x))dx=-f(0) $$ (you should justify that $xf(x)$ is still a suitable test function, and verify that the last integration is indeed $-f(0)$. )

For the second one,

$$ \int \delta^\prime(x)=-\int\delta(x)f^\prime(x)dx=-f^\prime(0) $$

The rigorous method requires us to take a "delta-sequence" $\delta_n(x)$, then use "normal" integration by parts and take the limit $n\rightarrow\infty$.

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Hint: both of these can be proved by integrating by parts. Integrate the $\delta'(x)dx$ and differentiate the rest.

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