The expected value of $1/X$ for $X\sim\mathcal{N}(\mu,\sigma^{2})$ is formally written as \begin{equation} \mathbb{E}[X^{-1}]=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}\frac{1}{x}\exp\left(-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right)\,\mathrm{d}x, \end{equation} which obviously does not converge. I am interested in the Cauchy principal value of this integral, i.e. \begin{equation} \mathbb{E}[X^{-1}]=\frac{1}{\sqrt{2\pi}\sigma}\lim_{\epsilon\nearrow 0}\left(\int_{-\infty}^{-\epsilon}+\int_\epsilon^\infty\right)\frac{1}{x}\exp\left(-\frac{(x-\mu)^{2}}{2\sigma^{2}}\right)\,\mathrm{d}x. \end{equation}
I would have thought that this has been worked out a long time ago but could not find a solution anywhere. Does anyone have a source detailing a solution?