I am trying to use Sohozki's theorem to find the support of the given singular generalized function:
$$\bigg(\mathscr{P}\frac{\cos{kx}}{x} ,\phi\bigg)=P.V.\int_{-\infty}^\infty \frac{\cos{kx}}{x}\phi(x)\text{d}x$$
However, after re-writing it to
$$\bigg(\mathscr{P}\frac{\cos{kx}}{x} ,\phi\bigg)=P.V.\int_{-\infty}^\infty \frac{e^{-ikx}}{2x}\phi(x)\text{d}x+P.V.\int_{-\infty}^\infty\frac{e^{ikx}}{2x}\phi(x)\text{d}x$$
I get
$$\bigg(\mathscr{P}\frac{\cos{kx}}{x} ,\phi\bigg)=\int_{-\infty}^{-\epsilon} \frac{e^{-ikx}}{2x}\phi(x)\text{d}x+\int_{\epsilon}^\infty\frac{e^{-ikx}}{2x}\phi(x)\text{d}x +\int_{-\infty}^{-\epsilon} \frac{e^{ikx}}{2x}\phi(x)\text{d}x+\int_{\epsilon}^\infty\frac{e^{ikx}}{2x}\phi(x)\text{d}x$$
However, from here I am stuck. I have the fact that:
$$P.V.\int \frac{1}{x}\text{d}x=0,$$ and also that
$$\phi(x) - \phi(0) = \phi'(\theta x)x, \theta \in [0, 1],$$
where $\phi$ is the test function (basic function) and it is a locally summable and $\in C^\infty$. Also that $\phi_k\to\phi$ as $k\to\infty$. However, the exponential terms make it difficult to reach a conclusion for this integral, and to find its value by Sohozki's theorem.
Any suggestions are welcome!
Thanks