How to compute $X+Y$ and $X/Y$ for a density function? I need to solve the problem:

I tried the following:
1) Compute $f(x)$ and $f(y)$:
$$f(x)=e^{-x} ( (-e)^{-\infty} - (-e^{0}))=e^{-x}.$$
2) $$f(u)=\int_{-\infty}^{+\infty}f(x)f(u-x)\,dx=\int_{0}^{\infty}f(x)f(u-x)\,dx\\
=\int_{0}^{\infty}e^{-x}e^{-(u-x)}\,dx=e^{-u}*(\infty-0),$$ which really blocked me in solving it. How can you solve this problem, because I'm sure my approach is very wrong, Thank you!
 A: Because the joint density factors as a product of the marginal densities we know that $X$ and $Y$ are independent, and therefore we may use the formula 
$$
f_{X+Y}(u)=\int_{-\infty}^\infty f_X(u-z)f_Y(z)\,\mathrm dz.\tag{1}
$$
Now $f_Y(z)=e^{-z}\mathbf{1}_{\{z\geq 0\}}$ and $f_X(u-z)=e^{z-u}\mathbf{1}_{\{z\leq u\}}$ and hence
$$
f_X(u-z)f_Y(z)=e^{-z}e^{z-u}\mathbf{1}_{\{u\geq 0,\,z\leq u\}}.
$$
Thus $(1)$ becomes
$$
f_{X+Y}(u)=\int_0^u e^{-u}\,\mathrm du=ue^u,\quad u\geq 0.
$$

To find the density of $V$ we can use the formula
$$
f_V(v)=\int_{-\infty}^\infty |y|f_X(vy)f_Y(y)\,\mathrm dy.
$$
This again uses the independence assumption. To find the joint density of $(U,V)$ you can use the theorem of transformation of densities, with $H:\mathbb{R}\times\mathbb{R}\setminus \{0\}\to \mathbb{R}^2$ given by $$H(x,y)=(x+y,x/y).$$
A: The joint probability distribution function $F_{U,V}(u,v)$ is $0$ if $u < 0$ or $v < 0$.
For $u, v > 0$,
$$\begin{align}
F_{U,V}(u,v) &= P\{U \leq u, V \leq v\}\\
&= P\left\{X+Y \leq u, \frac{X}{Y} \leq v\right\}\\
&= \int_A f_{X,y}(x,y) \,\mathrm dx\,\mathrm dy
\end{align}$$
where $A$ is a triangular region in the first quadrant with vertices $(0,0), (0,u)$, and $\left(\frac{uv}{1+v},\frac{u}{1+v}\right)$. So, you can compute the value of the integral
and then find 
$\displaystyle f_{U,V}(u,v) = \frac{\partial^2}{\partial u \partial v}F_{U,V}(u,v)$
or differentiate the integral using Leibniz's rule to get the joint density
directly.  Since $u$ and $v$ occur only in the limits, no integration will need
to be done, and the result will be the same
as is obtained by the transformation of densities method pointed out to you
by Stefan Hansen.
