$X\sim \Gamma(p,a)$,find the two dimensional moment generating function of (X, ln X) $X\sim \Gamma(p,a)$,find the two dimensional moment generating function of ($X, \ln X$)
What I have done is 
$$M_{(X,lnX)}(t_{1},t_{2}) = E(e^{t_{1}x+t_{2}lnx}) = E(e^{t_{1}x}x^{t_{2}}) = \int\int e^{t_{1}x}x^{t_{2}}f_{X,lnX}(x_{1},x_{2})dx_{1}dx_{2}$$
I need help to find the joint distribution of $X$ and $\ln X$.
 A: The gamma distribution is an exponential family and $(x, \log x)$ are the expectation parameters, so just write it in exponential family form and you have the density function of $(x, \log x)$.
$$\begin{align}
f_X(x) &= e^{\eta^\intercal T(x) - g(\eta)} \\
\implies \int_{x\in\mathbb R} f_X(x) &= 1 \\
t &= [t_1 t_2]^\intercal \\ 
\implies \mathbb E(e^{t_1 x + t_2 \log x}) &= \mathbb E(e^{t^\intercal T(x)}) \\
&= \int_{x \in \mathbb R}e^{(\eta + t)^\intercal T(x) - g(\eta)} \\
&= \int_{x \in \mathbb R}e^{(\eta + t)^\intercal T(x) - g(\eta +t) + g(\eta+t) - g(\eta)} \\
&= e^{g(\eta+t) - g(\eta)}
\end{align}$$
which gives a general formula for the moment generating function of the sufficient statistics.
For the gamma distribution, the log-normalizer $g(\eta) = \log \Gamma(\eta_2 + 1) - (\eta_2 + 1)\log\left(-\eta_1\right)$
A: The bivariate random variable $(X,\ln X)$ does not have a joint density
and so you cannot use a double integral as you have indicated in your question.  Instead,
the law of the unconscious statistician gives
$$M_{(X,\ln X)}(t_{1},t_{2})=E[e^{t_1X+t_2\ln X}]
=E[e^{t_1X}X^{t_2}]
=\int_0^\infty e^{t_1x}x^{t_2} \cdot \frac{a(ax)^{p-1} e^{-ax}}{\Gamma(p)}
\, \mathrm dx$$
which, after suitable change of variables, should give you something of
the form $\frac{\Gamma(t_2+p)}{\Gamma(p)}$ times some other stuff
involving $a$ and $t_1$ and $t_2$.
