Distribution of a product of RVs I have this question which I cannot figure out where I was doing it wrong.
Let $(X,Y)$ be a jointly continuous RV with density function $$f_{(X,Y)}(x,y)=\frac{12.5}{LW}\;\;\text{for}\; \;0.9L\leq x\leq1.1L\;\;\text{and}\;\;0.8W\leq y\leq1.2W$$ with the condition that $L\geq W>0$. Let $Z=XY$. Find $f_Z(z)$.
So here is what I had so far. From distribution theory, we have (in this case, $I(\dot\;)$ is an indicator function):
$$f_Z(z)=\int_{-\infty}^{\infty}\frac{1}{x}f_{X,Y}(x,\frac{z}{x})\mathrm{d}x=\frac{12.5}{LW}I_{[0.72LW,1.32LW]}^{(z)}\left(\int_{0.9L}^{0.83Z/W}\frac{dx}{x}+\int_{1.25Z/W}^{1.1L}\frac{dx}{x}\right)$$ from which I get:
$$f_Z(z)=\frac{12.5}{LW}\left(\ln{\frac{22}{27}}\right)I_{[0.72LW,1.32LW]}^{(z)}$$
However, upon checking, this is not a density function since it does not integrate up to $1$ over its support. Another way I can check it is through the calculation of $E[Z]=E[XY]=LW$, which obviously is not true from my derived distribution of $Z$. I am suspecting that maybe my limits of integration for $X$ is wrog but I cannot figure out whereor why. I hope someone can help. Thanks.
 A: Note that 
$$
f_{X,Y}(x,y)=\frac{12.5}{LW}\mathbf 1_{0.9L\lt x\lt1.1L}\mathbf 1_{0.8W\lt y\lt1.2W},
$$ 
hence
$$
f_Z(z)=\frac{12.5}{LW}\int_\mathbb R\mathbf 1_{0.9L\lt x\lt1.1L}\mathbf 1_{z/(1.2W)\lt x\lt z/(0.8W)}x^{-1}\mathrm dx.
$$
For this integral to be nonzero, one needs that $0.9L\lt z/(0.8W)$ and $z/(1.2W)\lt1.1L$, that is, that $z$ is in $(A,B)$, where
$$
A=0.72LW,\qquad B=1.32LW.
$$
When this holds, the integral defining $f_Z(z)$ is over the interval $(a(z),b(z))$, where
$$
a(z)=\max(0.9L,z/(1.2W)),\qquad b(z)=\min(1.1L,z/(0.8W)),
$$
hence
$$
f_Z(z)=\frac{12.5}{LW}\log\left(\frac{b(z)}{a(z)}\right)\mathbf 1_{A\lt z\lt B}.
$$
The logarithm can be made more explicit, considering separately the cases when $z$ is in $(A,C)$, $(C,D)$ and $(D,B)$, where
$$
C=0.88LW,\qquad D=1.08LW.
$$
For example, on $(A,C)$, $a(z)=0.9L$ and $b(z)=\frac{z}{0.8W}$ hence $$\log\left(\frac{b(z)}{a(z)}\right)=\log\left(\frac{z}{0.72LW}\right),$$ while on $(C,D)$, $a(z)=0.9L$ and $b(z)=1.1L$ hence $$\log\left(\frac{b(z)}{a(z)}\right)=\log\left(\frac{1.1}{0.9}\right),$$ and on $(D,B)$, $a(z)=z/(1.2W)$ and $b(z)=1.1L$ hence $$\log\left(\frac{b(z)}{a(z)}\right)=\log\left(\frac{1.32LW}{z}\right).$$
