Identity in distributional sense, principal value $1/x$

Question

How to prove that $$ T_{n}(\phi) =\lim_{r \to 0} \int_{r \le |x|} \frac{\cos(nx)}{x}\phi(x)\,\mathrm dx = \cos(nx)\ \operatorname {p.v.}\frac{1}{x}, $$ where $\phi$ belongs to $D(R)$ ($\phi$ is a test function).

Unfortunately, I don't know what to do with $\cos(nx)$ in the integral.

I know that $$ \cos(nx)\ \operatorname {p.v.}\frac{1}{x}= \cos(nx) \lim_{\epsilon \to 0} \int_{\epsilon \le |x|} \frac{\phi(x)}{x}\,\mathrm dx, $$ where $\phi$ belongs to $D(R)$.

I corrected my post.

Thanks for your help.

Answer 1

The definition of $fT$ for $f\in C^\infty$ and $T$ being a distribution is $$ (fT)(\phi) := T(f\phi). $$

What does this expand to when you have $f(x)=\cos nx$ and $T=\operatorname{pv}\frac1x$ defined by the following? $$ T(\phi) := \lim_{\epsilon \to 0} \int_{\epsilon \le |x|} \frac{\phi(x)}{x}\,\mathrm dx $$

Answer 2

I think you just applied the definition wrong. Recap- if $T\in \mathcal D'$ $f\in C^\infty,$ then $fT\in\mathcal D'$ is defined to be $$ \langle fT,g\rangle := \langle T,fg\rangle \quad ( g\in \mathcal D)$$ In the case $f(x)=\cos(nx)$, $T\phi =\operatorname{pv}\frac1x\phi = \lim_{r\to0}\int_{|x|>r} \frac{\phi(x)}x dx,$ we see that $$ \langle f\operatorname{pv}\frac1x,g\rangle = \langle \operatorname{pv}\frac1x,fg\rangle = \lim_{r\to0}\int_{|x|>r} \frac{\cos(nx)g(x)}x dx$$ and we see the product appears inside the integral.