How to find the expectation of $\log x$？ Let $X_1,...,X_n$ be a random sample. The pdf is $f(x\mid\theta)=\theta x^{\theta-1},0<x<1,\theta>0$.
I want to know $\mathbf{E}(\log x)$.
$$\int_0^1 \theta x^{\theta-1} \log x \;dx$$
I don't know how to solve this integral.
 A: Substitute $u = \log x, x = e^u, du = \frac{dx}{x}$ to get
$$ \int_{-\infty}^{0} \theta e^{\theta u} u \;du.$$
The goal is to write this in terms of the Gamma function:
$$\Gamma(s) = \int_0^\infty t^{s - 1}e^{-t} \; dt. $$
So you want to set $t = - \theta u$ above. Alternatively, you can integrate by parts (differentiating $u$ and integrating $e^{\theta u}$).
Integrating by parts is likely simpler, but the Gamma function shows up often enough in probability that it's worth keeping in mind as well.
A: 
I don't know how to solve this integral.

by parts
$$\underbrace{\left[x^{\theta}\log x\right]_0^1}_{=0}-\int_0^1 x^{\theta-1}dx$$
you find $E(\log x)=-\frac{1}{\theta}$
A: This integral can be solved by the use of integration by parts,
\begin{equation*}
\int uvdx=u\int vdx-\int \frac{du}{dx}\left(\int vdx\right) dx
\end{equation*}
\begin{gather*}
\int x^{n}\ln xdx\\
Taking\ \ln x\ as\ u\ and\ x^{n} \ as\ v,\\
\int x^{n}\ln xdx\\
=\ln x\cdotp \frac{x^{n+1}}{n+1} -\int \frac{1}{x} \cdotp \frac{x^{n+1}}{n+1}\\
=\ln x\cdotp \frac{x^{n+1}}{n+1} -\frac{1}{n+1}\int x^{n} dx\\
=\ln x\cdotp \frac{x^{n+1}}{n+1} -\frac{x^{n}}{( n+1)^{2}}\\
\int ^{1}_{0} x^{n}\ln xdx=\frac{-1}{( n+1)^{2}}
\end{gather*}
For your integral,
\begin{equation*}
n=\theta -1
\end{equation*}
Can you complete the rest?
