Mean and Variance of a Random Variable with a Given PDF Suppose that I have a random variable $X$ with the following PDF:
$$f_X(x)=\frac{(ln(x))^{\alpha-1}}{\Gamma(\alpha)\beta^\alpha x^{1+\frac{1}{\beta}}}$$ for $1<x<\infty$, $\alpha>0$, and $\beta>0$.
Suppose I want to calculate the mean and variance of this distribution.  But I don't know how to compute the integral:
$$\int_{1}^{\infty}\frac{(ln(x))^{\alpha-1}}{\Gamma(\alpha)\beta^\alpha x^{\frac{1}{\beta}}}dx$$
for the first moment.  Similarly, I don't know how to compute the integral for the second moment needed for the variance either. Can this definite integral be evaluated in terms of $\alpha$ and $\beta$?  If so, how?  If not, how else would you calculate the mean and variance?
 A: hint simply change variables to $u=\ln(x)$ and what you get should be related to the integral representation for the Gamma function.
A: We can compute the $k$-th moment this way:$$\int_{1}^{\infty}x^k\frac{(ln(x))^{\alpha-1}}{\Gamma(\alpha)\beta^\alpha x^{\frac{1}{\beta}+1}}dx = \frac{1}{\Gamma(\alpha)\beta^{\alpha}}\int_{0}^{\infty} \frac{e^{uk}u^{\alpha-1}}{e^{\frac {u}{\beta}}}du = \frac{1}{\Gamma(\alpha)\beta^{\alpha}(\frac{1}{\beta}-k)^{\alpha}}\int_{0}^{\infty} t^{\alpha-1}e^{-t}dt = \frac{\Gamma(\alpha)}{\Gamma(\alpha)\beta^{\alpha}(\frac{1}{\beta}-k)^{\alpha}} = (1-k\beta)^{-\alpha}.$$
Where we used these two transformations: $u=ln(x)$ and $t= u(\frac 1 {\beta}-k)$.
A: Even if do not know the substitution as mentioned by the answer, you can make use of the fact that the integral of the density is equal to $1$ if it is a valid pdf, i.e.  $\alpha, \beta > 0$ . Therefore, 
$$\begin{align} E[X^k] &= 
\int_1^{\infty} x^k \frac {\ln(x)^{\alpha-1}} {\Gamma(\alpha)\beta^{\alpha}x^{\frac {1} {\beta}+1}}dx \\
&= \int_1^{\infty} \frac {\ln(x)^{\alpha-1}} {\Gamma(\alpha)\beta^{\alpha}x^{\frac {1} {\beta}-k+1}}dx \\
&= \frac {\left(\frac {\beta} {1-k\beta}\right)^{\alpha} } {\beta^{\alpha}}\int_1^{\infty} \frac {\ln(x)^{\alpha-1}} {\Gamma(\alpha)\left(\frac {\beta} {1-k\beta}\right)^{\alpha}x^{\left(\frac {\beta} {1-k\beta}\right)^{-1} +1}}dx \\
\end{align}$$
Note that the integrand is a valid pdf with new parmeters $\displaystyle \alpha' = \alpha, \beta' = \frac {\beta} {1 - k\beta}$ if $\alpha',\beta' > 0$ And this is equivalent to $\displaystyle k < \frac {1} {\beta}$. In such case, the integral is equal to $1$, and the remaining constants can be simplified to the answer $(1 - k\beta)^{-\alpha}$
