Trying to find the mean of a Random Variable but I cannot do the integration. Please consider the problem below and my partial solution to it. Is it right so far? I do not know how to perform the integration and I am hoping somebody can point me in the right direction.
Thanks,
Bob
Problem:
The joint pdf of a bivariate r.v. $(X,Y)$ is given by:
\begin{eqnarray*}
f_{XY}(x,y) &=& \frac{1}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) } \\
\end{eqnarray*}
Find the mean of $X$.
Answer:
\begin{eqnarray*}
u_x &=&
 \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x \frac{1}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) } \, dy \, dx \\
\end{eqnarray*}
Now to perform this double integral, I am going to do it in polar coordinates.
\begin{eqnarray*}
u_x &=&
 \int_{0}^{2 \pi} \int_{0}^{\infty}
  r^2 \cos{\theta} \frac{1}{\sqrt{3 \pi } }
   e^{ -\frac{2}{3} ( r^2 - r^2 \cos{\theta} \sin{\theta} ) }
    \, dr \, d\theta \\
u_x &=&
\int_{0}^{2 \pi} \int_{0}^{\infty}
 r^2 \cos{\theta} \frac{1}{\sqrt{3 \pi } }
 e^{ \frac{2}{3}r^2 ( \cos{\theta} \sin{\theta} - 1 ) }
  \, dr \, d\theta \\
\end{eqnarray*}
How do I complete the integration?  

Don't go polar. Complete the square instead: $x^2 - xy + y^2 = \frac{3}{4}x^2 + (y - \frac{x}{2})^2$. The integrand $$\frac{x}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) }$$ becomes
$$\frac{x}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} e^{-\frac{2}{3}(y-\frac{x}{2})^2}$$
Integrate it first over $dy$. Can you continue?
