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

Question

Is this proof formal enough? I plan on being a theoretical physicist one day, so I want to get into the good habit of being mathematically strict.

My proof:

$u=x$; $du=dx$

$v = \delta (x)$; $dv = -\delta (x)$

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

$$x\delta (x) = 0$$

We now integrate both sides in order to properly use the Kronecker delta function.

$$\int x \delta (x) dx = \int 0dx$$

It is known that the $\int_{-\infty}^{+\infty} f(x)\delta (x)=f(0)$. Thus,

$$0=0$$

---

Answer 1

Let $f$ be any smooth compactly supported function. Then:

$$\int x\delta'(x)f(x)\ dx=\int\delta'(x) xf(x)\ dx=-\int\delta(x)(xf'(x)+f(x))\ dx=-0f'(0)-f(0)=-f(0)$$