Compute expected of a log transformation Let $X$ be a random variable with probability density function $f(x, \theta) = \frac{\theta}{x^{\theta + 1}}$ where $x>1$ and $\theta > 0$.
How can i compute $E[\log X ]$??
I did this way: $$ Y = X \implies X = e^Y$$
Then:
$$E[\log X] = E[Y] $$
so:
$$\theta \int_{0}^{\infty} e^y \frac{1}{e^{\theta + 1}} dy $$
Am i wrong?
 A: The result you want to apply here is the following:

If $X$ is a continuous random variable with density $f$, then for a
  real-valued function $g$, $$ \Bbb
 E\bigl[g(X)\bigr]=\int_{-\infty}^\infty g(x) f(x)\,dx. $$


For your random variable $X$ and for the function $g(x)=\log x$ (assumed to be base $e$), we have, using the above formula,
$$
\Bbb E\bigl[\log(X)\bigr] = \int_1^\infty (\log x) {\theta\over x^{\theta+1}}\,dx 
$$

To compute  $\int (\log x) {\theta\over x^{\theta+1}}\,dx$, we may use integration by parts with $u=\log x$ and $dv= {\theta\over x^{\theta+1}}\,dx $:
$$\eqalign{
\int (\underbrace{\log x\vphantom{1\over3}}_u) \underbrace{{\theta\over x^{\theta+1}}\,dx }_{dv}
&= (\underbrace{\log x\vphantom{1\over3}}_u) \underbrace{{(-x^{-\theta})\vphantom{1\over3}}  }_{v} -\int
  \underbrace{{-x^{-\theta}\vphantom{1\over3}}  }_{v}\,\underbrace{{1\over x}\,dx\vphantom{1\over3}}_{du}\cr
&={-\log x\over x^\theta}+\int x^{-\theta-1}\,dx\cr
&={-\log x\over x^\theta}-{x^{-\theta}\over\theta}+C.\cr
}
$$
Thus, using $\theta>0$,
$$
\Bbb E\bigl[\log(X)\bigr] = \int_1^\infty (\log x) {\theta\over x^{\theta+1}}\,d\theta 
=\Bigl(
{-\log x\over x^\theta}-{x^{-\theta}\over\theta}\Bigr)\biggr|_1^\infty=0-\Bigl(0-{1\over\theta}\Bigr)={1\over\theta}.
$$
A: One should mention that this is an example of a Pareto distribution, named after the Italian economist who used them to model distribution of incomes and of wealth, and they are alleged to model many phenomena (e.g., it has been alleged that 20% of books in a library account for 80% of circulation; you can make it an exercise to show that that corresponds to $\theta=\log_4 5$).
$$
\int_1^\infty (\log x) \frac{\theta}{x^{\theta+1}}\,dx =\int_1^\infty (\log x) \frac{\theta}{x^\theta}\frac{dx}{x} = \int_0^\infty y \frac{\theta}{e^{y\theta}} \,dy = \int_0^\infty y e^{-\theta y}\,(\theta\,dy)
$$
$$
= \frac 1 \theta \int_0^\infty (\theta y) e^{-\theta y} \, (\theta\,dy) = \frac 1 \theta \int_0^\infty u e^{-u}\,du = \frac 1 \theta.
$$
The last step elides an integration by parts:
$$
\int u e^{-u}\,du = \int u \, dv = uv - \int v\,du = ue^{-u} - \int -e^{-u}\,du = ue^{-u}+e^{-u}+\text{constant}.
$$
As $u\to\infty$, L'Hopital's rule shows that this approaches $0$.  When $u=0$, this equals $1$.
A: Another way is to explicitly obtain the distribution of $Y=g(X)=log(X)$; $X = e^Y$  ($X>1$, $Y>0$)
$$f_Y(y) = \frac{f_X(x)}{|g'(x)|}  \biggr|_{x=g{^-1(y)}}= x \frac{\theta}{x^{\theta+1}}\biggr|_{x=e^y} = \theta \; e^{- \theta \, y}$$
which you might recognize as an exponential density, with mean $\frac{1}{\theta}$
