Normal Ratio Distribution with CDF Method I think I'm missing something glaringly obvious here that's causing problems for me in the entire subject.
I have two independent standard normal random variables, X and Y ~N(0,1), and I need to find the density of U=Y/X.  I start with f(x,y)=$\frac1{2\pi}e^{-x^2/2}e^{-y^2/2}$, then set Y=ux, and take a double integral of that.  I'm leaving out my limits of integration, because my problem is that I know that I have to integrate with $\iint$y f(x,y), but I have no idea where that y (which is needed to integrate $e^{-y^2/2}$) comes from.  Is this a Jacobian?  None of my notes or our textbook mentions any use of a Jacobian in the CDF method.  Is this just impossible with the CDF method?
 A: I'm not 100% sure what approach you are taking.  I would approach this problem by considering the CDF that you want, which is $F(t) = P(X/Y<t)$ for $t>0$ (for now, in order to avoid some issues).  Equivalently, consider $P(X<tY)$ for positive $Y$, and $P(X>tY)$ for negative $Y$.  With 
$f(x,y)=\frac{1}{2\pi}\exp\left(\frac{1}{2}(x^2+y^2)\right)$,
$$P(X<tY) = \int_{0}^{\infty} \int_{-\infty}^{tY} f(x,y) dx dy$$
This can be switched to polar coordinates using the same trick found on the page http://en.wikipedia.org/wiki/Gaussian_integral:
$$P(X<tY) = \int_{\arctan(1/t)}^{\pi} \int_{0}^{\infty} \frac{1}{2\pi}r\mathrm{e}^{-r^2/2} dr d\theta = \frac{1}{2}-\frac{1}{2\pi}\arctan(1/t) $$
The $\arctan$ arrives by noticing that you are integrating $\theta$ from the line $x=ty$ to $\pi$.  For $t>0$, $\arctan(1/t) = \pi/2 - \arctan(t)$.  By symmetry,
$$
  F(t) = P(X<tY|Y>0) + P(X>tY|Y<0) \\ 
  = 1 - \frac{1}{\pi} \arctan(1/t) \\
  = \frac{1}{2}+\frac{1}{\pi}\arctan(t)
$$
which is the CDF of the Cauchy distribution.  You can continue this to $t=0$ by integrating over half of the plane and getting $1/2$.  You can handle $t<0$ by itegrating $\theta$ from $\arctan(1/t)$ to $0$.  The issue is that the range of $\arctan$ is $(-\pi/2,\pi/2)$.  
Finally, differentiate to get the density function $F'(t)=\frac{1}{\pi}\frac{1}{1+t^2}$.
