This is exercise 3.7.1 from Introduction to Mathematical Statistics 6th edition by Hogg, Mckean, and Craig. The question is: Suppose $Y$ has a $\text{gamma} (\alpha, \beta)$ distribution. Let $X = e^{Y}$. Show that the pdf of $X$ is given by expression (3.7.4). Derive the mean and variance.
I have been able to use the pdf method of transformation to show that the pdf of $X$ is
$$ f(x) = \frac{1}{\Gamma(\alpha) \beta^{\alpha}} x^{-(1 + \beta)/\beta}(\ln x)^{\alpha - 1} $$
for $x> 1$ and $0$ elsewhere, which is expression (3.7.4) mentioned above. I have attempted to derive the mean using the definition of expected value. So we have
\begin{align} E(X) = \int^{\infty}_{1} x f(x)dx &= \int^{\infty}_{1} x \frac{1}{\Gamma(\alpha) \beta^{\alpha}} x^{-(1 + \beta)/\beta}(\ln x)^{\alpha - 1} dx \\ &= \int^{\infty}_{1} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} x^{-1/\beta}(\ln x)^{\alpha - 1} dx. \end{align}
I am not quite sure where to go from here but I thought maybe a change of variables where $u = \ln(x)$ and then $x = e^{u}$ and $dx = xdu$ so that we now have
$$ E(X) = \int^{\infty}_{0} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} u^{\alpha - 1}e^{-u/\beta}xdu = x\int^{\infty}_{0} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} u^{\alpha - 1}e^{-u/\beta}du = x. $$
However, I have a feeling this is not correct, but I'm not sure.
Edit after comment from J.G.
$$ E(X) = \int^{\infty}_{0} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} u^{\alpha - 1}e^{-u/\beta}xdu = \int^{\infty}_{0} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} u^{\alpha - 1}e^{-u(1-\beta)/\beta}du $$
Now I am not sure how to proceed with evaluating this integral. It sort of resembles a gamma distribution.
I think we can evaluate the above integral using the fact that
$$ \int^{\infty}_{0} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} e^{-x/\beta} x^{\alpha - 1} dx = 1 \implies \int^{\infty}_{0} e^{-x/\beta} x^{\alpha - 1} dx = \Gamma(\alpha) \beta^{\alpha}.$$
Therefore,
\begin{align} E(X) = \int^{\infty}_{0} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} u^{\alpha - 1}e^{-u(1-\beta)/\beta}du &= \frac{1}{\Gamma(\alpha) \beta^{\alpha}}\int^{\infty}_{0} u^{\alpha - 1}e^{-u(1-\beta)/\beta}du \\ &= \frac{1}{\Gamma(\alpha) \beta^{\alpha}} \left( \Gamma (\alpha) \left[ \frac{\beta}{1-\beta} \right]^{\alpha} \right) \\ &= \frac{1}{\Gamma(\alpha) \beta^{\alpha}} \left( \frac{\Gamma(\alpha)\beta^\alpha}{(1-\beta)^{\alpha}}\right) \\ &= (1-\beta)^{-\alpha}. \end{align}
I believe that is the correct answer for the $E(X)$, but it would be great to get some confirmation.
