Expected value $E(X_1X_2)$ for bivariate normal distribution

Given a bivariate normal distribution $(X_1,X_2)$ with parameters $\mu_1,\mu_2,\sigma_1,\sigma_2,\rho$ where $\rho\neq 0$.

I want to find $E(X_1X_2)$. Now $E(X_1X_2):=\int_{-\infty}^\infty \!\int_{-\infty}^\infty \! x_1x_2f(x_1,x_2) \, \mathrm{d}x_2\mathrm{d}x_1$ where $f(x_1,x_2)$ is the bivariate normal density function. Is there a short way to evaluating this integral?

• Is integration required? Or can you use another method? – grand_chat Apr 17 '18 at 22:00
• Is there another method? – Sid Caroline Apr 17 '18 at 22:51

If integration is not required, you could obtain $E(X_1X_2)$ by plugging into the formulae: $$\operatorname{Cov}(X_1, X_2)=E(X_1X_2)-E(X_1)E(X_2)\tag1$$ and $$\rho = \frac{\operatorname{Cov}(X_1,X_2)}{\sqrt{\operatorname{Var}(X_1)\operatorname{Var}(X_2)}}\tag2$$