I am trying to solve the following double integral for a problem that came up during my research:
$$\int_0^{\infty}\int_0^q \frac{1}{q} \Phi\left(\frac{u-\mu l-kQ}{\sigma l}\right)\mathrm du f_L(l)\mathrm dl$$
where $\Phi(x)$ is the cdf of the standard normal distribution and $f_L$ is the pdf of the gamma distribution.
I am able to calculate the inner integral as
$$\int_0^q \frac{1}{q}\Phi\left(\frac{u-\mu l-kQ}{\sigma l}\right)du =\frac{1}{q}\left(q*\Phi\left(\frac{-\mu l+(k+1)Q}{\sigma l}\right)+\sigma l\left(\phi(\frac{-\mu l+(k+1)Q}{\sigma l}\right)-\phi\left(\frac{-\mu l+kQ}{\sigma l}\right)\right)+(nq-\mu l)\left(\Phi\left(\frac{-\mu l+(k+1)Q}{\sigma l}\right)-\Phi\left(\frac{-\mu l+kQ}{\sigma l}\right)\right)$$
All constants are $>0$.
The pdf of the gamma distribution is $f(x) = \frac{\beta^{\alpha} x^{\alpha-1} e^{-x\beta}}{\Gamma(\alpha)} \quad \text{ for } x \geq 0 \text{ and } \alpha, \beta > 0$.
Thank you for your help!
Update: I found the following integral table of the error function (the normal cdf mainly consists of the error function): http://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn1p1_A1b.pdf
Equation 9 in Section 4.2 almost looks like what I am looking for except that I have a fraction as an argument where the variable is in the nominator and the denominator. I played around a bit but with no success but this may be a a way how to solve it.