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Given $\alpha$ and constant $\mu$,

$$\begin{array}{ll} \text{maximize} & \displaystyle\int_0^\infty p(x)x^\alpha \,\mathrm d x\\ \text{subject to} & \displaystyle\int_0^\infty p(x)\,\mathrm d x = 1\\ & \displaystyle\int_0^\infty p(x)x \, \mathrm d x = \mu\end{array}$$

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  • $\begingroup$ What do you mean by optimize? Do you want one $p$ for all $\alpha$, or is $\alpha$ given? $\endgroup$
    – anderstood
    Nov 6, 2017 at 21:18
  • $\begingroup$ For a given value of $\alpha$, what maximizes the integral. $\endgroup$
    – Halbort
    Nov 6, 2017 at 21:21
  • $\begingroup$ Is $\alpha$ integer or, could it take on arbitrary real values? In what range? $\endgroup$
    – Brian Borchers
    Nov 6, 2017 at 21:36
  • $\begingroup$ Arbitrary real value $\endgroup$
    – Halbort
    Nov 6, 2017 at 22:04
  • $\begingroup$ Have you heard of Langrange-Multipliers? $\endgroup$
    – F. Conrad
    Nov 8, 2017 at 19:16

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