How to evaluate products involving the delta function and Cauchy principal value?

Question

Prove that $x\delta(x) = 0$ and $xP(\frac{1}{x})=1$

Here $P$ means the Cauchy principal value.

How can I start this? And if I prove the second, will $xP(\frac{1}{|x|})=1$ also follow?

Definition: $P\left(\frac{1}{|x|},f\right) = \int_{|x|<1}\frac{f(x)-f(0)}{|x|}dx + \int_{|x|>1}\frac{f(x)}{|x|}dx$

Answer 1

We have for any test function $f$,

$$\begin{align} \int_{-\infty}^\infty f(x)(x\delta(x))\,dx&=\int_{-\infty}^\infty (xf(x))\delta(x)\,dx\\\\ &=\left.\left(xf(x)\right)\right|_{x=0}\\\\ &=0 \end{align}$$

Therefore, in the sense of Generalized Functions $x\delta(x)=0$.

Answer 2

You can do that by definition of multiplication a distribution by a $C^\infty$ function.

Take arbitrary test function $\phi$, then write $$\langle x \delta_0,\phi(x)\rangle=\langle \delta_0,x\phi(x)\rangle = x\phi(x)\big|_{x=0} = 0\cdot \phi(0)=0.$$ Since $\phi$ is arbitrary, we conclude that $x\delta_0=0$.

Similarly for $P(1/x)$: we write for arbitrary test function $$\langle xP(1/x),\phi(x)\rangle = \langle P(1/x),x\phi(x)\rangle=\lim_{\epsilon\to 0}\int_{\epsilon}^{\infty} \frac{x\phi(x)+(-x\phi(-x))}{x}dx = \lim_{\epsilon\to 0}\int_{\epsilon}^{\infty} (\phi(x)-\phi(-x))) dx = \lim_{\epsilon\to 0}\int_{\epsilon}^{\infty} \phi(x) dx +\lim_{\epsilon\to 0}\int_{-\infty}^{-\epsilon} \phi(x) dx =\int_{\Bbb R}\phi(x)dx=\langle 1,\phi\rangle . $$