Show that if X and Y are independent N(0, 1)-distributed random variables, then X/Y ∈ C(0, 1). 
Show that if $X$ and $Y$ are independent $N(0, 1)$-distributed random variables, then $X/Y ∈ C(0, 1)$. C is cauchy distribution..

First trial:
1) Transformation
Let $Y=V,U=X/V,=>X=UV$;$|J|=v$
$f(u,v)=f_x(uv)f_y(v)|J|=1/2\pi ve^{-\frac{(uv)^2}{2}}e^{-\frac{v^2}2}$
To get $f_u(u) =\int_{-∞}^{+∞}1/2\pi ve^{-\frac{(uv)^2}{2}}e^{-\frac{v^2}2}=0$
Second trial: 2) CDF
$F(U\le u)=F(X \le uY)=\int_{-∞}^{+∞}F(X \le uy)f_y(y)dy=1/2\pi\int_{-∞}^{+∞} (\int_{-∞}^{uy}e^{-\frac{(uy)^2}{2}}du)e^{-\frac{y^2}2}dy=?$
Can you help me?
 A: I carry on your second trial: $\frac{X}Y\leq u $
Now we discriminate between two cases $Y>0$ and $Y<0$. Multiplying the equation above by $Y$.
$P(\frac{X}Y\leq u)=(X\leq uY| Y>0)+P(X\geq uY| Y<0)$
$=\int_{0}^{\infty}\int_{0}^{uy} f(x,y) \,dx \, dy +\int_{-\infty}^{0}\int_{uy}^{0} \,dx \, dy$
Then we differentiate w.r.t. u to get the pdf. This can be done by using the Leibniz rule.
$p(u)=\int_{0}^{\infty} y\cdot  f(uy,y) \, dy- \int_{-\infty}^{0} y\cdot  f(uy,y) \, dy$
$=2\int_{0}^{\infty} y\cdot  f(uy,y) \, dy$
$=2\cdot\frac{1}{2\pi\sigma_x\sigma_y}\int_{0}^{\infty} y\cdot \exp\left( -\left(\frac{y^2}{2\sigma_x^2}+\frac{u^2y^2}{2\sigma_y^2}\right)\right) \, dy$
$=\frac{1}{\pi\sigma_x\sigma_y}\int_{0}^{\infty} y\cdot \exp\left( -y^2\left(\frac{1}{2\sigma_x^2}+\frac{u^2}{2\sigma_y^2}\right)\right) \, dy$
Let $c=\frac{1}{2\sigma_x^2}+\frac{u^2}{2\sigma_y^2}$. The integral becomes equal to
$\int_{0}^{\infty} y\cdot \exp\left( -cy^2\right) \, dy$, which is $\frac1{2c}$. Thus we have
$p(u)=\frac{1}{\pi\sigma_x\sigma_y}\cdot \frac{1}{2\cdot(\frac{1}{ 2\sigma_x^2}+\frac{u^2}{2\sigma_y^2})}=\frac{1}{\pi\sigma_x\sigma_y}\cdot \frac{1}{(\frac{1}{ \sigma_x^2}+\frac{u^2}{2\sigma_y^2})}$
$=\boxed{\large{\frac1{\pi}\frac{\frac{\sigma_x}{\sigma_y}}{u^2+\left(\frac{\sigma_y}{\sigma_y}\right)^2}}}$
This is the pdf of the  cauchy distribution with $\gamma=\frac{\sigma_x}{\sigma_y}$ and $x_0=0$.
