Finding $P(X>d)$ when $X\mid \Lambda$ is Pareto and $\Lambda$ is Gamma distributed Supposedly $X$ have a Pareto distribution with parameters $\Lambda$ and $\theta$. Let $\Lambda$ have a Gamma distribution with parameters $\alpha$ and $1$ (i.e., scale parameter $= 1$). Calculate (rounding your answers to five decimal places) the unconditional probability that $X > d$.
I obtained the pdf to be:

How do I calculate for the unconditional probability of $X > d$ ?
 A: To get the cumulative Probability $P(X>d)$, we have to integrate out with respect to $x$, i.e.,
\begin{align}
P(X>d)&=\int_d^\infty f_X(x)dx\\
&=\int_d^\infty\frac{1}{\Gamma[\alpha]}\int_0^\infty\frac{\theta^\lambda\lambda^\alpha e^{-\lambda}}{(x+\theta)^{\lambda+1}}d\lambda dx\\
\end{align}
The issue here is the integral in $\lambda$ which may not be solvable in a closed form. Integration by parts may also not help due to $\alpha$. So, one way to attack this problem is to series-expand the exponential expression $\phi^\lambda=(\tfrac{\theta}{e})^\lambda$, i.e.,
$$
\phi^{\lambda}=\sum_{i=0}^\infty\frac{\lambda^i\log^i{\phi}}{i!}
$$
and then solve the above integral for every term $i$ individually and then sum up, i.e.
\begin{align}
P(X>d)&=\frac{1}{\Gamma[\alpha]}\int_d^\infty \sum_i h_i(x)dx
\end{align}
with $h_i$ satisfying
\begin{align}
h_i(x)&=\underbrace{\frac{\log^i(\phi)}{i!}}_{=:A_i}\int_0^\infty\frac{\lambda^{\alpha+i} }{(x+\theta)^{\lambda+1}}d\lambda\\
&=-A_i\frac{\log(\theta+x)^{-\alpha-2}}{\theta+x}\Gamma[\alpha+2,\lambda\log(\theta+x)]\bigg.\bigg|_{\lambda=0}^{\lambda=\infty}\\
&=A_i\frac{\log(\theta+x)^{-\alpha-2}}{\theta+x}\Gamma[\alpha+2]
\end{align}
where the above integral can be looked up from Wolfram Alpha!. Hence, in total we have
\begin{align}
P(X>d)&=\frac{\Gamma[\alpha+2]}{\Gamma[\alpha]}\sum_{i=0}^\infty A_i\underbrace{\int_d^\infty \frac{1}{(x+\theta)\log(x+\theta)^{\alpha+2}}dx}_{=:g(\alpha,\theta,d)}\\
&=g(\alpha,\theta,d)\frac{\Gamma[\alpha+2]}{\Gamma[\alpha]}\sum_{i=0}^\infty A_i
\end{align}
The right-hand side integral $g(\alpha,\theta)$ is elementary and satisfies
\begin{align}
g(\alpha,\theta,d)&=-\frac{1}{\alpha+1}\frac{1}{\log(x+\theta)^{\alpha+1}}\biggl.\biggr|_d^\infty\\
&=\frac{1}{\alpha+1}\frac{1}{\log(d+\theta)^{\alpha+1}}
\end{align}
Thus, finally we get
\begin{align}
P(X>d)&=g(\alpha,\theta,d)\frac{\Gamma[\alpha+2]}{\Gamma[\alpha]}\sum_{i=0}^\infty A_i\\
&=g(\alpha,\theta,d)\frac{\Gamma[\alpha+2]}{\Gamma[\alpha]}\underbrace{\sum_{i=0}^\infty \frac{\log^i(\phi)}{i!}}_{=\phi}\\
&=g(\alpha,\theta,d)\phi\frac{\Gamma[\alpha+2]}{\Gamma[\alpha]}\\
&=\frac{\Gamma[\alpha+2]}{(\alpha+1)\Gamma[\alpha]}\frac{\overbrace{\phi}^{=\frac{\theta}{e}}}{\log(d+\theta)^{\alpha+1}}\\
&=\frac{\theta\Gamma[\alpha+2]}{e(\alpha+1)\Gamma[\alpha]\log(d+\theta)^{\alpha+1}}
\end{align}
I feel, we have solved this problem in a closed way and no need for approximation or rounding. 
