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Just so there are no misunderstandings let me first ask whether it is true that:

$$ \int_{-\infty}^{\infty}x\delta(x)\mathrm{d}x=0. $$

If that is not true, then I don't know anything about the Dirac delta distribution and I will be off to correct this :)

Otherwise I have this question: Why can we take a delta functional of $x$? As far as I've read, the Dirac delta is a tempered distribution and it must act on Schwartz functions. I am not convinced that this is true for $f(x)=x$.

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  • $\begingroup$ The equation is definitely true. The integral can be viewn as the expected value of a constant random variable $X = 0$. $\endgroup$ – GDumphart Oct 13 '14 at 11:41
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You're right, it is not strictly correct to apply - in general - a distribution with $x \mapsto x$ because it is not a compactly supported function.

But the Dirac is itself a compactly supported distribution, so you can extend its domain to all $\mathcal{C}^\infty$ functions.

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The Dirac delta $\delta$ distribution can be paired with a Schwarz function, but we can make sense of pairing it with more general objects too. For example, for a continuous function $f$, $$\int f(x) \delta(x) \,dx = f(0),$$ which in particular confirms your computation.

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