Find the PDF of $\max(R_1,R_2)/\min(R_1,R_2)$ for $(R_1,R_2)$ i.i.d. standard normal

Question

$R_1, R_2$ are two independent continuous random variables, they all satisfied in the following normally distributed equation. $$f_1(x)=f_2(x)=\frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2}$$ The problem is to find $R_3=\frac{max(R_1, R_2)}{min(R_1, R_2)}$. So here is how I do this: $$F_3(z)=P\{ \frac{max(R_1, R_2)}{min(R_1, R_2)}≤z \}$$ $$F_3(z)=P\{ \frac{R_1}{R_2}≤z;R_1>R_2 \} + P\{ \frac{R_2}{R_1}≤z;R_1<R_2 \}$$ $$F_3(z)=P\{ R_1≤zR_2; R_1>R_2 \} + P\{ \frac{R_2}{z}≤R_1; R_1<R_2 \}$$ $$f_3(z)=\frac{d}{dz}F_3(z)=P\{R_2\cdot f_{R_1}(zR_2)\} + P\{\frac{R_2}{z^2}\cdot f_{R_1}(\frac{R_2}{z})\}$$ And finally, I stuck here, this is the integral formula I derived, but this is probably wrong, please help me. $$\int_0^\infty x \cdot \frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2} \cdot \frac{1}{\sqrt{2\pi}}e^\frac{-z^2x^2}{2} + \frac{x}{z^2} \cdot \frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2} \cdot \frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2z^2}dx $$