set-up
$X_1, \ldots, X_n \sim \mathrm{\ i.i.d.\ } N(\theta, 1).$
The following holds. $$ \sqrt{\frac{2\pi}{n}}\mathbb{E}\left[\frac{1}{\bar{X}}\right] = \int_0^\infty \frac{1}{y}\exp{\left[-\frac{n}{2}(y-\theta)^2\right]}dy - \int_{-\infty}^0 \frac{1}{y}\exp{\left[-\frac{n}{2}(y-\theta)^2\right]}dy $$
Problem
We want to show that RHS first term tends to $\infty$ i.e. $\mathbb E[1/\bar{X}]$ doesn’t exist.
What I know
For any $\epsilon > 0$, RHS first term is $$ \int_0^\infty \frac{1}{y}\exp{\left[-\frac{n}{2}(y-\theta)^2\right]}dy \geq \int_0^\epsilon \frac{1}{y}\exp{\left[-\frac{n}{2}(y-\theta)^2\right]}dy. $$
As $y \to 0$, $$ \exp{\left[-\frac{n}{2}(y-\theta)^2\right]} \to \exp{\left[-\frac{n}{2}\theta^2 \right]} \overset{?}{>}\frac{1}{2} \mathrm{\ for\ all\ } \theta < \epsilon $$ when $\epsilon$ is sufficiently small.
Can you tell me why this inequality holds?
If $Z\sim N(0,1)$ recall that $E(1/Z)$ does not exist. To see this, recall that $Z^2/2$ is gamma distributed and that for $s\geq -1/2$ we have $E((Z^2/2)^s)=\frac{\Gamma(\frac{1}{2}+s)}{\sqrt{\pi}}$ which implies $$E(|Z|^{2s})=2^s\frac{\Gamma(\frac{1}{2}+s)}{\sqrt{\pi}}.$$ Clearly $\lim_{s\to -1/2}E(|Z|^{2s})=E(1/|Z|)=\infty.$ The proof of $E(1/|Z+\theta|)=\infty$ is a little bit more difficult, based on the fact that $(Z+\theta)^2/2$ is a non central xi square distribution. Its Laplace transform is $L(s)=(s+1)^{-1/2}\exp\left(-s\theta^2/2(s+1)\right)$ which leads to the divergent integral$$E(1/|Z+\theta|)=\int_0^{\infty}L(s) s^{-1/2}\frac{ds}{\sqrt{2\pi}}.$$