Marginal distribution of $X$ given $f_{XY}(x,y)=ce^{-x^2-y^2-xy-x}$ I'm supposed to calculate the marginal distribution of $X$ given that the joint distribution is $ce^{-x^2-y^2-xy-x}$ for some $c>0$ and I know that that means $$f_X(x)=c\int_{-\infty}^\infty e^{-(x^2+y^2)-xy-x}dx$$ and I learned in class that you can use polar coordinates to simplify integrals similar to this but I am not really sure if that can be applied to this problem? Any tips on how to proceed would be greatly appreciated, thanks!
 A: One way to integrate is to complete the square in the exponent, firstly for $y$ and then for $x$, to help you find $c$. We want to put the integrand into the form of a Standard Normal pdf so we also introduce a factor of $\frac{1}{2}$:
\begin{align}
1 &= c \int_{x=-\infty}^{\infty} \int_{y=-\infty}^{\infty} \exp\left[-\dfrac{1}{2}\left(\sqrt{2}y + \dfrac{x}{\sqrt{2}}\right)^2 -\dfrac{1}{2}\left(\sqrt{\dfrac{3}{2}}x + \sqrt{\dfrac{2}{3}}\right)^2 + \dfrac{1}{3} \right]dy\;dx. \\
\end{align}
Letting $u = \sqrt{2}y + \dfrac{x}{\sqrt{2}},$ and $v = \sqrt{\dfrac{3}{2}}x + \sqrt{\dfrac{2}{3}}$, we have $dy=\dfrac{du}{\sqrt{2}}$ and $dx=\sqrt{\dfrac{2}{3}}dv,$ so:
\begin{align}
1 &= \dfrac{2\pi e^{1/3}}{\sqrt{3}} c \int_{u=-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}u^2} du \int_{v=-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}v^2} dv \\
&= \dfrac{2\pi e^{1/3} c}{\sqrt{3}} \\
\therefore\quad c &= \dfrac{\sqrt{3}}{2\pi e^{1/3}}.
\end{align}
The same technique, but just integrating over $y$, allows you to then find the marginal pdf $f_X(x)$.
