<s>[INST] integral of normal distribution how to do this integral:
$$ \mathop{\int\int}_{y+2x>0}  x y \frac1{2\pi\sigma_x\sigma_y}e^{ -\frac{(x-\mu_x)^2}{2\sigma_x^2}}\cdot e^{ -\frac{(y-\mu_y)^2}{2\sigma_y^2}} dx dy$$
Both x and y are normally distributed and mutually independent. I need to calculate the above integral.
 [/INST] For simplicity, I let $\sigma_x=\sigma_y=1$. Then we have 
\begin{align*}
&\mathop{\int\int}_{y+2x>0}\frac{xy}{2\pi}e^{ -\frac{(x-\mu_x)^2}{2}}\cdot e^{-\frac{(y-\mu_y)^2}{2}}dxdy\\
=&\int_{-\infty}^\infty dx\int_{-2x}^\infty\frac{xy}{2\pi}e^{-\frac{(x-\mu_x)^2}{2}}\cdot e^{-\frac{(y-\mu_y)^2}{2}}dy\\
=&\frac{1}{2\pi}\int_{-\infty}^\infty xe^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty ye^{-\frac{(y-\mu_y)^2}{2}}dy\\
=&\frac{1}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty ye^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x\mu_y}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy\\
=&\frac{1}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_y}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x\mu_y}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy
\end{align*}
You can have explicit expression for each term above except for the last one.
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