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Let $X$ be a standard normal random variable. What $\alpha$ minimizes $E|X|^{\alpha}$?

Numerically, $\alpha$ turns out to be equal to $\sqrt{3}/2-\varepsilon$ where $\varepsilon$ is of the order $10^{-7}$.

This problem is equivalent to solving $\psi(\beta)+\ln 2 = 0$ where $\beta=(\alpha+1)/2$ and $\psi$ is the digamma function.

Is there any particular reason $\alpha$ is so close to this algebraic number, or is it just a "random coincidence"?

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    $\begingroup$ More precisely, $\alpha \approx .86602523656318915183$, or continued fraction $[0, 1, 6, 2, 6, 2, 6, 7, 1, 2, 32, 1, 29, 2, 1, 1, 1, 38]$ as opposed to $\sqrt{3}/2 \approx .86602540378443864675$ or continued fraction $[0,1,6,2,6,2,6,2,\ldots]$. I suspect a coincidence. Another nice approximation (based on the large element $32$ in the continued fraction) is $27453/31700$, with error less than $3 \times 10^{-11}$. $\endgroup$
    – Robert Israel
    Commented Jun 28, 2018 at 23:41

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