integration - Integrating a function over a random effect (Normal distribution)

I would like to integrate a function with a random effect.
The function is :

$G(t; \beta) = exp(- \lambda t^\gamma \exp(\beta Z))$,

$\beta$ being the random effect taken from a normal distribution of mean 0 and variance $\sigma$.
The solution would be to make a double integration, first on $t$ and then over the distribution of the random effect.

But, I don't know ho to do in practice ! I am ready to use Monte-Carlo integration, I just need to calculate the area under the curve.

Any suggestion ?

Thanks a lot !

Answer

Define $K:=\lambda\exp\beta Z$ so $G=\exp -Kt^\gamma$ and $\int_0^\infty Gdt=\frac{1}{\gamma}\Gamma\left(\frac{1}{\gamma}\right)K^{-1/\gamma}$. Next integrate out the $\beta$ dependence, viz. $$\int_{\Bbb R}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{\beta^2}{2\sigma^2}-\frac{Z}{\gamma}\beta\right)\cdot\frac{1}{\gamma}\Gamma\left(\frac{1}{\gamma}\right)\lambda^{-1/\gamma}d\beta=\frac{1}{\gamma}\Gamma\left(\frac{1}{\gamma}\right)\lambda^{-1/\gamma}\exp\frac{(Z\sigma)^2}{2\gamma^2}.$$