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


$$\begin{array}{ll} \text{maximize} & \displaystyle\int\_0^\infty p(x)x^\alpha \,\mathrm d x\\ \text{subject to} & \displaystyle\int\_0^\infty p(x)\,\mathrm d x = 1\\ & \displaystyle\int\_0^\infty p(x)x \, \mathrm d x = \mu\end{array}$$


I had previously posted this problem on math stack exchange. However, on second thought, I think this may be a more involved problem than I previously thought.




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Here is one Idea I had. Let $M(t)$ be the moment generating function of $p(x)$. We know for integers $n$ that:


$$\frac{d^n}{dt^n}M(t) \bigg|\_{t=0} = \int\_0^\infty p(x)x^n \,\mathrm d x$$. I think I could extend this so:


$$\frac{d^\alpha}{dt^\alpha}M(t) \bigg|\_{t=0} = \int\_0^\infty p(x)x^\alpha \,\mathrm d x$$


Now our problem boils down to this. Find the moment generating function $M(t)$ that has the maximum fractional derivative $\frac{d^\alpha}{dt^\alpha}M(t) |\_{t=0}$ such that $M(0) = 1$ and $M'(0) = \mu$. I do not know how to proceed further. I know little about fractional derivatives.



Maximizing the $\alpha$-moment of a distributution
Let us consider the closely *related* problem: maximize $EX^\alpha$ over all nonnegative random variables (r.v.'s) $X$ with $EX=\mu$. To avoid trivialities, assume that $\mu\in(0,\infty)$. Consider the following cases: 


Case 1: $\alpha>1$. Suppose that $P(X=x)=\mu/x=1-P(X=0)$ for some real $x>\mu$. Then $EX=\mu$, whereas $EX^\alpha=\mu x^{\alpha-1}\to\infty$ as $x\to\infty$. It follows that $\sup\{EX^\alpha\colon X\ge0,\,EX=\mu\}=\infty$. Using an appropriate approximation of the distribution of this discrete r.v. by an absolutely continuous one, we conclude that in this case the supremum in your problem is $\infty$ as well. 


Case 2: $0<\alpha\le1$. Here $x^\alpha$ is concave in $x>0$. So, by Jensen's inequality, 
$\int\_0^\infty p(x)x^\alpha \,\mathrm d x\le (\int\_0^\infty p(x)x \,\mathrm d x)^\alpha=\mu^\alpha$ or, a bit more generally, $EX^\alpha\le(EX)^\alpha=\mu^\alpha$. The latter bound on $EX^\alpha$ is attained when $P(X=\mu)=1$. So, 
$\max\{EX^\alpha\colon X\ge0,\,EX=\mu\}=\mu^\alpha$. In your original problem, this maximum is not attained, since you only allowed absolutely continuous r.v.'s. However, the corresponding supremum is of course also $\infty$. 


Case 3: $\alpha=0$. This case is trivial: here $\int\_0^\infty p(x)x^\alpha \,\mathrm d x=\int\_0^\infty p(x) \,\mathrm d x=1$, given the conditions. 


Case 4: $\alpha<0$. Suppose that $P(X=2\mu)=1/2=1-P(X=0)$. Then $EX=\mu$, whereas $EX^\alpha=\infty$. It follows that $\sup\{EX^\alpha\colon X\ge0,\,EX=\mu\}=\infty$. Using an appropriate approximation of the distribution of this discrete r.v. by an absolutely continuous one, we conclude that in this case the supremum in your problem is $\infty$ as well. 


You certainly don't need fractional derivatives for this problem. 


