Let X and Y be random variables with a joint probability density function (joint PDF) given by
$ f_{X,Y}(x,y) \quad=\quad \begin{cases} \frac{c}{1+x^2+y^2} & \text{ if } x^2+y^2<1\,, \\ 0 & \text{ otherwise,} \end{cases} $
where the positive constant c is determined by the requirement that $ f_{X,Y} $ is a PDF.
What is the correct formula for the marginal PDF of X?
I think I have to start off by integrating $ \frac{c}{1+x^2+y^2} $ with respect to y. Which gives me
$ \int \frac{c}{1+x^2+y^2} dy = \frac{c*\arctan{\frac{y}{\sqrt{1+x^2}}}}{\sqrt{1+x^2}} $
But don't know how to continue now.
I'd also like to know if there's any software out there where I can compute this kind of symbolic stuff.
Thanks
I think I finally managed to recall.
[1] I think you made a mistake, you should integrate w.r.t. $x$ not $y$.
That will give you the margin PDF of $Y$.
[2] Make this equation:
$$\int_{-1}^{1} \left(\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{c}{1+x^2+y^2} \,dx\right)\,dy = 1$$
Solving this equation should give you $c$.
[3] Also, the inner integral gives you the marginal distribution of $Y$.
$$f(y) = \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{c}{1+x^2+y^2} \,dx $$
So just solve this as an indefinite integral treating $y$ as constant (similar to what you did but reversed), but then compute the definite integral too (i.e. apply the boundaries). That will give you a function of just $y$, your marginal PDF.