Product of Two Random Variables that are both independent $U[0,1] $ distributed "Suppose that $X$ and $Y$ are independent $U[0,1]$-random variables. Find the probability density function of the product $V = XY$"
I started this problem by stating $'X ~ U(0,1)$ and $Y ~ U(0,1)`$ and since $V = XY$ then $Y = V/X$ 
I used the integral from $0$ to $1$ $∫f_x(x)f_y(v/x)(1/|x|)dx$
I have $f_x(x) = 1$ and $f_y(v/x)$ also equal $1$.
After plugging in for each of the values, I got the pdf to come out to be just $f_V(v) = ∫(1/|x|)dx$ from $0$ to $1$ which would simplify down to $0$ after integrating from $0$ to $1$ since they $X$ and $Y$ are both uniform from $0$ to $1$.
Where would I stop the answer at for this question? Is the answer just the integral that $f_V(v)$ is equal to? Or am I doing it incorrectly to were it should not simplify down to $0$?
 A: First note that $0<X<1$ and $0<Y<1$ imply that $0<V<1$.
The pdf of $X$ is
$$f_X(x) = \begin{cases}1 & \text{if }0 < x < 1 \\
0 & \text{otherwise}\end{cases}$$
and the pdf of $Y$ is
$$f_Y(y) = \begin{cases}1 & \text{if }0 < y < 1 \\
0 & \text{otherwise}\end{cases}$$
To apply your formula
$$\int f_X(x) f_Y(v/x)(1/|x|)\ dx$$
we need to work out what $f_Y(v/x)$ is. We have
$$f_Y(v/x) = \begin{cases}1 & \text{if }0 < v/x < 1 \\
0 & \text{otherwise}\end{cases}$$
The inequality $0 < v/x < 1$ is equivalent to $x > v$ because $v$ and $x$ are positive. We also must have $v > 0$ and $x < 1$. Therefore we can write
$$f_Y(v/x) = \begin{cases}1 & \text{if }0 < v < x < 1 \\
0 & \text{otherwise}\end{cases}$$
It follows that the product $f_X(x)f_Y(v/x)$ is $1$ if $0 < v < x < 1$, and $0$ otherwise. Therefore we need to integrate over the interval where $x$ is between $v$ and $1$ (note that $|x| = x$ when $x$ is positive, so we can drop the absolute values):
$$\begin{aligned}f_V(v) &= \begin{cases}
\int_v^1 (1/x)\ dx & \text{if }0 < v < 1 \\
0 & \text{otherwise}\end{cases}\\
&= \begin{cases}-\ln(v)&\text{if }0 < v < 1\\
0 & \text{otherwise}\end{cases}\end{aligned}$$
An interesting feature of this pdf $f_V(v)$ is that it grows toward infinity as $v$ approaches zero from the right. Despite this, you can check that $f_V$ still integrates to $1$.
