Suppose $(X,Y)$ have joint pdf given by:
$$f_{X,Y}(x,y) = \frac{c}{\left(1+\frac{x^2+y^2}{2}\right)^2}$$
Find $c$ such that $f_{X,Y}$ is a legitimate pdf.
I know that $f_{X,Y}$ has to be positive or 0 given that $c \geq 0$. But I'm stuck on the integration to check what value of $c$ allows the pdf to equal 1.
I've tried setting $x = \arctan(u)$ with different values of $a$ to see if I can get a similar result as mentioned on the Integral of $\frac{1}{(1+x^2)^2}$ but the $y^2$ component in the denominator is making things tricky when evaluating the first double integral with respect to $x$.
Any help or guidance would be greatly appreciated!
