I need to compute the following integral: \begin{equation} \lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{x+i\epsilon}. \end{equation} Applying the Sokhotski-Plemelj theorem, we have \begin{equation} \lim_{\epsilon\to 0}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{x+i\epsilon} = \mathcal{P}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{x}-i\pi. \end{equation} I am wondering how the Cauchy principal value is defined in this case when we have a singularity at finite value as well as at infinity. My book tells me that the Cauchy principal value equals zero, but I do not see why.
I have already tried the usual definition, but it is not applicable because both integrals are divergent: \begin{equation} \mathcal{P}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{x} = \lim_{\epsilon\to 0}\left(\int_{-\infty}^{-\epsilon}\mathrm{d}x\,\frac{1}{x}+\int_{\epsilon}^{\infty}\mathrm{d}x\,\frac{1}{x}\right) \end{equation}