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I'm working a proof and I've been able to show the following fact:

$$-\int_{-\infty}^{\infty} x \frac{d}{dx} \delta(x)dx=\int_{-\infty}^{\infty}\delta(x)dx$$

I am trying to conclude with the following statement:

$$x\frac{d}{dx}\delta(x)=-\delta(x).$$

I know this is not true in general for arbitrary functions (i.e. two functions are not necessarily the same because their integrals are equal on some given interval), but I believe it is true in this case. Is there a way to justify it?

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2 Answers 2

The $\delta$-function is not a function in the ordinary sense but rather a distribution. The statement that two distributions $d_1(x)$ and $d_2(x)$ are the same is equivalent to $\int d_1(x) f(x)dx = \int d_2(x) f(x) dx$ for all $f(x)$ taken from an appropriate space of functions (e.g., Schwartz space for tempered distributions).

You want to show that $$\int x \left(\frac{d}{dx} \delta(x)\right) f(x) dx= - \int \delta(x) f(x) dx;$$ in fact $$\int x \left(\frac{d}{dx} \delta(x)\right) f(x) dx= - \int \delta(x) f(x) dx -\int \delta(x) x f'(x) dx $$ as can be shown by partial integration (where the boundary terms vanish due to the fact that $f(x) \to 0$ at $|x|\to\infty$ sufficiently fast).

The last term vanishes because $$\int \delta(x) x f'(x) dx = x f'(x) \Big|_{x=0} =0. $$

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hint: $-\int_{-\infty}^{\infty} x \frac{d}{dx} \delta(x)dx=\int_{-\infty}^{\infty}\delta(x)dx$then$$-\int_{-\infty}^{\infty} x \frac{d}{dx} \delta(x)dx-\int_{-\infty}^{\infty}\delta(x)dx=0$$ $$-\int_{-\infty}^{\infty}[ x \frac{d}{dx} \delta(x)-\delta(x)]dx=0$$ easily conclude by properties of integral

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$\int_{-1}^1 dx(x-x^3)=0$ so $x=x^3$, "easily conclude by properties of integral" –  Jonathan May 12 '13 at 15:52

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