Let $X,Y$ be Exponential random variables with parameter alpha=lambda and beta=mu. Assume $t=u$.
Appendix: Evaluation of the integral: \begin{align} f_{X/Y}(u) = {} & \alpha\beta \int_0^\infty e^{-(\alpha u+\beta)y} y \, dy \\[8pt] = {} & \frac{\alpha\beta}{(\alpha u+\beta)^2} \int_0^\infty e^{-(\alpha u+\beta)y} (\alpha u+\beta) y \big( (\alpha u + \beta) \, dy \big) \\[8pt] = {} & \frac{\alpha\beta}{(\alpha u+\beta)^2} \int_0^\infty e^{-v} v \,dv \\[8pt] = {} & \frac{\alpha\beta}{(\alpha u+\beta)^2}. \end{align}
$$\mathbb E\left[\frac{X}{Y}\right]=\int uf_{\frac{X}{Y}}(u)du$$
What are the limits of integration on $u$?
Someone told me that the limits of integration are from zero to infinity.
I don't believe them and I think instead, the limits of integration are from zero to $\frac{x}{y}$ I understand the expected value yields an real number and using x/y as the upper bound wouldn't give me an integral convergent to some real number but that's not a great explanation. .
because of the bounds on the integral where $t=u$
For $t>0$ we have $$\begin{align} \mathbb P\left(\frac XY>t\right) &= \iint_{\{(x,y)\in\mathbb R^2\ :\ 0\leqslant ty\leqslant x\}} \lambda\mu e^{-\lambda x}e^{-\mu y}\end{align}$$
I'm looking at the bounds on $ty$ which gives me the bounds for $t$ or $u$.