# Closed form for $\mathbb{E}[\ln (1-p)]$, for $p \sim Beta(\alpha, \beta)$

We know that if $p \sim Beta(\alpha, \beta)$, then $$\mathbb{E}[\ln p] = \psi(\alpha) - \psi(\alpha + \beta)$$ where $\psi(.)$ is the Digamma function. Is there an easy form for $\mathbb{E}[\ln (1-p)]$?

Denote $$1-p = q$$ By the symmetry of the beta distribution, $$q \sim \text{Beta}(\beta, \alpha)$$ Using the identity in your question, we have $$\mathbb{E}[\ln (1-p)]=\mathbb{E}[\ln q]=\psi(\beta)-\psi(\alpha+\beta)$$