what is the distribution of a uniform r.v. divided by an exponential r.v.? Say $U=\frac{X}{Y}$. X and Y are independent with each other. X is a Uniform distribution r.v. $X\sim \mathcal{U}(0,1)$. Y is an exponential distribution r.v., $Y\sim\mathcal{Exp}(\lambda)$, whose pdf is $f_Y(y)=\lambda e^{-\lambda y}$. What is the distribution of U? I tried to tackle this problem by caluculating the CDF, as follows:
$F_U(U)=Pr(\frac{X}{Y}\leq U)=Pr(Y\geq \frac{X}{U})=\int_{0}^{1}\int_{\frac{X}{U}}^{\infty}\lambda e^{-\lambda y}dy dx=\frac{U}{\lambda}(1-e^{\frac{\lambda}{U}})$. But this is clearly not right. Because when $u\rightarrow \infty$, $F_{U}(U)\rightarrow \infty$ instead of 1. I want to know what is the mistake that i have made.
 A: Remark: The question has changed. The solution below is for $X$ and $Y$ as described in the question as first posed, where we were looking at $\frac{Y}{X}$ with $Y$ uniform and $X$ exponential. We will not edit that solution, since the general idea can be used to deal with variants.
Solution to original problem:
We give the random variable another name, like $W$, so that $U$ will not do double duty.  We find the cumulative distribution function $F(w)$ of $W$, and then differentiate to find the density. So for fixed $w\gt 0$, we find $\Pr(W\le w)$, that is, $\Pr(Y\le wX)$.
We will assume, as we are probably expected to assume, that $X$ and $Y$ are independent. The problem should really have specified this. 
There is a typo in the density function of $X$. It is $\lambda e^{-\lambda x}$ for $x\gt 0$.
Draw the line $y=wx$. We want the probability that $(X,Y)$ ends up below that line.  It is easier to find the probability of the complement, which is the probability that $(X,Y)$ ends up in the triangle with corners $(0,0)$, $(1/w,1)$, and $(0,1)$. Integrate. We get
$$1-F(w)=\int_{x=0}^{1/w}\int_{y=wx}^1 \lambda e^{-\lambda x}\,dy\,dx.$$
The integration with respect to $y$ is trivial, and the second integration is not too bad. And we do not even have to integrate. If all we want is the density  we can differentiate under the integral sign.
A: I have tested this result. My result as well as that of andre are both correct. We need to user the L'hopital's law as follows
$\lim_{w\rightarrow\infty}\frac{1-e^{\lambda/w}}{\lambda/w}=1$
