Distribution of ratio of uniform and exponential random variables This is a homework question, I feel like I'm doing it right, but I can't seem to get the answer to match up.  I have a uniform RV from 2 to 4, and an exponential with mean 4, so $X \sim \text{UNI}(2,4)$, $Y \sim \text{EXP}(4)$.  
I'm looking for the density of $U=\dfrac{Y}{X}$.  So $f(x)=\dfrac{1}{2}$, $f(y)=\dfrac{e^{-y/4}}{4}$, and $f(x,y)=\dfrac{e^{-y/4}}{8}$.  
Now if I substitute $Y=Ux$, and multiply in a Jacobian ($\dfrac{\text{d}}{\text{d}u}ux=x$, this might be a mistake, I'm not completely clear on this), I can get $f(x,u)=\dfrac{xe^{-xu/4}}{8}$.  The marginal density $f(u)$ is found then by integrating out the $x$, which I can do by turning that joint density into a gamma function, where I need to get a $\dfrac{16}{u^2}$ on the bottom, so I end up with $\dfrac{2}{u^2}$ outside of the gamma integral, and after the gamma integrates to 1, f(u)=$\dfrac{2}{u^2}$.  Anyone see a mistake here?
 A: Your answer is clearly wrong since the density does not integrate to $1$ across the positive reals, as Dilip Sarwate.
The way I would approach this would be to take your $f(x,y)=\dfrac{e^{-y/4}}{8}$ and look at 
$$\Pr\left(\frac{Y}{X} \le u\right)=\int_{x=2}^{4} \int_{y=0}^{ux}  \dfrac{e^{-y/4}}{8} dy\, dy= \int_{x=2}^{4}\left(\frac12- \dfrac{e^{-ux/4}}{2}\right) dx $$  $$=1-\frac{2}{u}\left({e}^{-\frac{u}{2}}-{e}^{-{u}}\right),$$ which has the correct limits as $u \to 0$ and $u \to \infty$, and then take the derivative for the density.  
A: You have assumed that $X$ and $Y$ are independent random variables and so I
suppose that the actual problem you are solving is stating this somewhere,
or it is taken from a section titled Independent Random Variables in some book.
For a (continuous) random variable $Y$ with density $f_Y(y)$, the density of
$Z = aY$ is $$f_Z(z) = \frac{1}{|a|}f_Y\left(\frac za\right).$$
Thus, for $x > 0$, given that $X = x$, the conditional density of $U =\frac YX = \frac Yx$ is
$$f_{U\mid X}(u \mid X=x) = xf_Y(xu) = \frac{x e^{-xu/4}}{4}.$$
Now  find the unconditional density of $U$ using
$$f_U(u) = \int_{-\infty}^\infty f_{U\mid X}(u \mid X=x)f_X(x)\,\mathrm dx
= \int_2^4 \frac{x e^{-xu/4}}{4}\times\frac 12 \,\mathrm dx.$$
