I am having the following expression. This is the PDF of Nakagami-Lognormal Distribution. I want to express in terms of Gauss-Hermite abscissas and weights. How can I do it?
$$f_Z(z)=\int_0^{\infty}\frac{2m^mz^{2m-1}\exp(-\frac{mz^2}{\Omega})}{\Gamma(m)\Omega^m}\frac{1}{\sqrt{2\pi}\lambda\Omega}\exp\left(-\frac{(\ln \Omega-\mu)^2}{2\lambda^2}\right)d\Omega$$
So your integrand is (for fixed values of everything except $\Omega$):
$$c_1 \exp(-c_2/\Omega) \Omega^{-m-1} \exp(-c_3 (\ln(\Omega)-\mu)^2).$$
To get a Gaussian-like thing, change variables to $u=\ln(\Omega)$, so $du=d \Omega/\Omega$ giving
$$c_1 \exp(-c_2/e^u) e^{-mu} \exp(-c_3 (u-\mu)^2)$$
where the new limits are $-\infty$ to $\infty$. Now merge the linear part into the quadratic and complete the square:
$$c_1 \exp(-c_2/e^u) \exp(-c_3(u-\mu-m/(2c_3))^2) c_4$$
Then change variables to $w=\sqrt{c_3} (u-\mu-m/(2c_3))$. Then your integrand is a bounded smooth function times $\exp(-w^2)$ and the domain of integration is the whole line. So this form of the problem is amenable to Gauss-Hermite quadrature.