Density function of product Suppose $X$ and $Y$ are continuous random variables with joint density
$$f(x,y)=x+y,\quad 0<x,y<1$$
I am trying to find the density of $XY$. 
I am having trouble applying the formula
$$f_V(v)=\int _{-\infty}^\infty \:f\left(x,\frac{v}{x}\right)\:\frac{1}{x}\:dx$$
As the integral diverges for
$$\int _{0}^1 \:\left(x+\frac{v}{x}\right)\:\frac{1}{x}\:dx$$
and any other bounds ive tried for that matter. Any help appreciated
 A: $$f(x,y) =x+y, 0<x<y<1$$
You are being asked the following.  Only then it makes sense.
Find the joint density function of W = XY and U = X
$$W = XY$$
$$U = X$$
Inverse Transformation is then
$$X = U$$
$$Y = \frac{W}{U}$$
The ranges get transformed from $0<x<y<1 => 0<wz<z<1 => 0<u<1; 0<w<u$
Now take find the Jacobian:
$$D = \begin{bmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial w}\\\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial w}\end{bmatrix}$$
$$D = \begin{bmatrix}1 & 0\\-\frac{w}{u^2} & \frac{1}{u}\end{bmatrix}$$
Now take the determinant of the Jacobian
$$|D| = det(D) = \frac{1}{u}$$
$$f_{U,W}(u,w) = f_{X,Y}(u,\frac{w}{u})\cdot \frac{1}{u}$$
$$ f_{U,W}(u,w) = \left(u+\frac{w}{u}\right)\cdot \frac{1}{u}$$
$$ f_{U,W}(u,w) = \left(1+\frac{w}{u^2}\right); 0<u<1, 0<w<u$$
Sanity Check
$$  \int_{0}^{1}\int_{0}^{u} \left(1+\frac{w}{u^2}\right) du = 1$$
This is what is being asked.
A: The standard approach is to find CDF of $Z=XY$:
$$F_Z(z)= P(XY<z)$$
The region of interest is $0<z<1$, and for this region (draw a picture)
$$P(XY<z)=\int_0^zdx\left(\int_0^1dy(x+y)\right)+\int_z^1dx\left(\int_0^{z/x}dy(x+y)\right)$$
and after taking the integrals
$$f_Z(z)=\frac{dF_Z(z)}{dz}$$
A: Note that $XY < v$ gives us $Y < \tfrac{v}{X}$ and we already know $Y < 1$, hence $Y<\min\{1,\tfrac{v}{X}\}$. It follows that we have to consider two intervals for $X$: [0,t] where $t/X \geq 1$ and $[t,1]$ where $t/X\leq 1$, so the CDF is given by:
\begin{align}
P(XY < v) &= \int_0^1\int_0^{\min\{1,v/x\}} f(x,y)\;dy\;dx\\
&= \int_0^v \int_0^1 f(x,y)\;dy\;dx + \int_v^1\int_0^{v/x}f(x,y)\;dy\;dx\\
&= \int_0^v \left[\,xy + \tfrac12 y^2\,\right]_0^1\;dx + \int_v^1 \left[\,xy + \tfrac12 y^2\,\right]_0^{v/x}\;dx\\
&= \tfrac12\int_0^v (2x+1)\;dx + \int_v^1 \left[\,v + \frac{v^2}{2x^2}\,\right]\; dx\\
&= \tfrac12 \left[\,x^2+x\,\right]_0^v + \left[\, vx - \frac{v^2}{2x}\,\right]_v^1\\
&= (\tfrac12 v^2 + \tfrac12 v) + ((v - \tfrac12 v^2) - (v^2 - \tfrac12 v))\\
&= -v^2 +2v
= 1-(v-1)^2 
\end{align}
The density function $f_{XY}(v)$ is then given by:
$$\frac{d}{dv} P(XY<v) = \frac{d}{dv} \left(\, 1 - (v-1)^2 \,\right) = 1 - 2(v-1) = 3 - 2v$$
