Good afternoon! I can't prove $$x \cdot \delta^m(x)=-m\delta^{(m-1)}(x), m=1,2,3....$$ I have found that $\int x \cdot \delta'(x)dx =x \cdot \delta(x)-\int \delta(x)dx$, as a result $ x\cdot\delta'(x)=-\delta(x)$, but for the next derivative it doesn't work. I can't understand when $n$ appears.
Thanks for help!
Be careful! You need to do more than just pairing the distribution on the constant function $1$.
We have \begin{align*} \int_\mathbb{R} f(x)x\delta'(x)=-\int_\mathbb{R}[f(x)x]'\delta(x)=-f(0)=-\int_\mathbb{R} f(x)\delta(x) \end{align*} which we can write as $$ x\delta'+\delta=0.\tag{1} $$ Assuming(*) you have seen the product rule for differentiating distributions, we have \begin{align*} 0&=\delta'+x\delta''+\delta'=x\delta''+2\delta'\tag{$1'$}\\ 0&=\delta''+x\delta^{(3)}+2\delta''=x\delta^{(3)}+3\delta''\tag{$1''$}\\ 0&=\delta^{(3)}+x\delta^{(4)}+3\delta^{(3)}=x\delta^{(4)}+4\delta^{(3)}\tag{$1'''$}\\ &\vdots \end{align*}
(*)If you haven't seen the product rule, you can proceed as follows \begin{align*}\require{cancel} \int_\mathbb{R} f(x)x\delta^{(m)}(x)&=-\int_\mathbb{R} [f(x)x]'\delta^{(m-1)}(x)\\ &=\dots\\ &=(-1)^m\left.\frac{\mathrm{d}^m}{\mathrm{d}x^m}\right\rvert_{x=0}[f(x)x]\\ &=(-1)^m[\cancel{f^{(m)}(0)\cdot 0}+m f^{(m-1)}(0)]\\ &=-m\cdot(-1)^{m-1}f^{(m-1)}(0)\\ &=-m\int_\mathbb{R} f(x)\delta^{(m-1)}(x) \end{align*} for all test function $f$.
Using Fourier transform
First recall that $$ \mathcal{F}\{u'\} = i\xi \, \mathcal{F}\{u\}, \quad \mathcal{F}\{xu\} = i\frac{d}{d\xi} \mathcal{F}\{u\}, \quad \mathcal{F}\{\delta\} = 1. $$
From this follows that $$ \mathcal{F}\{\delta^{(m)}\} = (i\xi)^m \mathcal{F}\{\delta\} = (i\xi)^m. $$
Then, $$ \mathcal{F}\{x\delta^{(m)}\} = i\frac{d}{d\xi} \mathcal{F}\{\delta^{(m)}\} = i\frac{d}{d\xi} (i\xi)^m = i^2 m \, (i\xi)^{m-1} = -m \, \mathcal{F}\{\delta^{(m-1)}\}. $$
Thus, since the Fourier transform is bijective, $$ x\delta^{(m)} = -m \, \delta^{(m-1)}. $$
Using generalized product rule for functions
We are used to the product rule $(fg)' = f'g + fg'.$ For higher order derivatives this generalizes to $$ (fg)^{(m)} = \sum_{k=0}^{m} {m \choose k} f^{(k)} g^{(m-k)}. $$
Applying this to the left hand side of the identity to be shown gives, $$\begin{align} \langle x \delta^{(m)}, \varphi \rangle &= \langle \delta^{(m)}, x \varphi \rangle \\ &= (-1)^m \langle \delta, (x \varphi)^{(m)} \rangle \\ &= (-1)^m \langle \delta, \sum_{k=0}^{m} {m \choose k} x^{(k)} \varphi^{(m-k)} \rangle \\ &= (-1)^m \sum_{k=0}^{m} {m \choose k} \langle \delta, x^{(k)} \varphi^{(m-k)} \rangle \\ &= (-1)^m \sum_{k=0}^{m} {m \choose k} \left. x^{(k)} \varphi^{(m-k)} \right|_{x=0} \\ &= \{ \text{ only the term for $k=1$ will contribute } \} \\ &= (-1)^m \, m \, \varphi^{(m-1)} \\ &= -m \, (-1)^m\varphi^{(m-1)} \\ &= -m \, \langle \delta^{(m-1)}, \varphi \rangle. \\ \end{align}$$
Since this is valid for any test function $\varphi$ we have $x\delta^{(m)} = -m\delta^{(m-1)}.$
I have above used a common notation for distributions acting on a test function. You can think of $\langle u, \varphi \rangle$ as $\int u(x) \, \varphi(x) \, dx.$