# What is $E[1/X]$ when $X$ is a standard normal random variable?

It's long time ago that I took the calculus class, so I dare to ask. If $X\sim N(0,1)$, what is $\mathbb{E}(1/X)$? $$\mathbb{E}(1/X) = \int_{-\infty}^\infty \frac1x \cdot \frac1{\sqrt{2\pi}} \exp\left(-\frac{x^2}2\right) dx.$$ Can I just claim $\mathbb{E}(1/X) = 0$ as $\frac1x \exp\left(-\frac{x^2}{2}\right)$ is an odd function even when it is not bounded?

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Isn't it an odd function? – Thomas Andrews Dec 10 '11 at 20:18
The integral does not exist. There is no good reason to take existence in the PV sense as relevant. – André Nicolas Dec 10 '11 at 20:38
The is rather similar to asking for the expected value of the Cauchy distribution, which also gives 0 if we adopt the Cauchy PV; but that is not consider relevant/appropiate. In particular, if we use the Lebesgue integral (the sane thing to do in probability) it's clear that the integral does not exist. – leonbloy Dec 10 '11 at 21:27

$$\int_0^\infty \frac1x e^{-x^2/2}\;dx \ge \int_0^1 \frac1x e^{-1}\;dx = \infty,$$ and similarly $$\int_{-\infty}^0 \frac1x e^{-x^2/2}\;dx = -\infty.$$
The Cauchy principal value is $0$ because it's an odd function, i.e. $$\lim_{a\to0+}\left(\int_{-\infty}^{-a}+\int_a^\infty\right) \frac1x e^{-x^2/2} \; dx = 0.$$