\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 ? homework probability-distributions share | improve this question edited Dec 10 '11 at 22:12 Davide Giraudo 35.1k 3 26 75 asked Dec 10 '11 at