Finding Independent exponential random variables

Question

If $X_1$ and $X_2$ are Independent exponential random variables with respective parameters $\lambda_1$ and $\lambda_2$, find the distribution of $Z = \frac{X_1}{X_2}$

My approach would be to make F(Z) = double integral not 100% sure though what the boundary are

Answer 1

\begin{align} 1-F_Z(z):=P(Z> z) &= P(X_1 > zX_2)\\ &= \int_0^\infty P(X_1 > zx_2) p_{X_2}(x_2) \mathop{dx_2}\\ &= \int_0^\infty e^{-\lambda_1 zx_2} \lambda_2 e^{-\lambda_2 x_2} \mathop{dx_2}\\ &= \frac{\lambda_2}{\lambda_1z+\lambda_2}. \end{align}

$$p_Z(z) = \frac{d}{dz} F_Z(z) = \frac{d}{dz} \frac{1}{1+\lambda_2/(\lambda_1 z)} = \frac{1}{(1+\lambda_2/(\lambda_1 z))^2}\cdot \frac{\lambda_2}{\lambda_1 z^2} = \frac{\lambda_2/\lambda_1}{(z+\lambda_2/\lambda_1)^2}$$