Limit involving probability Let $\mu$ be any probability measure on the interval $]0,\infty[$. I think the following limit holds, but I don't manage to prove it:
$$\frac{1}{\alpha}\log\biggl(\int_0^\infty\! x^\alpha d\mu(x)\biggr) \ \xrightarrow[\alpha\to 0+]{}\ \int_0^\infty\! \log x\ d\mu(x)$$
In probabilistic terms it can be rewritten as:
$$\frac{1}{\alpha}\log\mathbb{E}[X^\alpha] \ \xrightarrow[\alpha\to 0+]{}\ \mathbb{E}[\log X]$$
for any positive random variable $X$.
Can you help me to prove it?

We assume that there is $\alpha_0>0$ such that $\int_0^{+\infty}x^{\alpha_0} d\mu(x)$ is finite. Let $I(\alpha):=\frac 1{\alpha}\log\left(\int_0^{+\infty}x^\alpha d\mu(x)\right)$ and $I:=\int_0^{+\infty}\log xd\mu(x)$. 
Since the function $t\mapsto \log t$ is concave, we have $I(\alpha)\geqslant I$ for all $\alpha$. 
Now, use the inequality $\log(1+t)\leqslant t$ and the dominated convergence theorem to show that $\lim_{\alpha\to 0^+}\int_0^{+\infty}\frac{x^\alpha-1}\alpha d\mu(x)=I$. Call $J(\alpha):=\int_0^{+\infty}\frac{x^\alpha-1}\alpha d\mu(x)$. Then 
$$I\leqslant I(\alpha)\leqslant J(\alpha).$$
