integral of normal distribution

Question

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.

Answer 1

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.