# Expectation of log(1/X) when X follows inverse gaussian distribution

Can anybody help me in this question? How to derive the expectation of log(1/X) when X follows inverse Gaussian distribution?

Found out a type of approximation for $\mathbb{E}[\log(X)]$ while $X \sim \mathcal{N}(\mu,\sigma^2)$ in this link (Expected value and variance of log(a)). Can this be extended to the present case to get an approximate solution?

Any help will be appreciated.

• $\log(1/X)=-\log(X)$, did you consider that? – mpiktas Jul 21 '15 at 7:33