
An improper integral
So i'm doing some exercises in probabilty theory but I got stuck on some elementary calculus.
I have to integrate $\displaystyle \int_{A_t}2(x+y)dxdy$ with $\displaystyle A_t$ being the set $\displaystyle A_t=\{x; xy\leq t\}$. The limit of the integral will be a function of $\displaystyle t$.
Okay. So I rewrite $\displaystyle A_t=\{(x,y); 0 \leq x \leq \frac{t}{y}; 0 \leq y \leq \infty\}$ (Is this correct?)
Then we have $\displaystyle \int_0^\infty(\int_0^{\frac{t}{y}}(2x+2y)dx)dy$
$\displaystyle =\int_0^\infty [x^2+2xy]_{x=0}^{x=\frac{t}{y}})dy$
$\displaystyle =\int_0^\infty(\frac{t^2}{y^2}+2t)dy$
Which is a integral I don't know how to solve. I tried splitting it up:
$\displaystyle =\int_0^1(\frac{t^2}{y^2}+2t)dy+\int_1^\infty(\fra c{t^2}{y^2}+2t)dy$
$\displaystyle =t^2+2tlim_{s\rightarrow 0}(\frac{t^2}{a}+2ta)+....$
??

The question is rather unclear. Could you quote it completely?. Thanks.
Fernando Revilla

Sure. Thanks for replying;
Suppose X and Y are continuous random variables with joint pdf f (x,y)=2(x+y) if 0<x<y<1 and zero otherwise.
i)Find the joint pdf of the variables S=X and T=XY.
ii)Find the marginal pdf of T.
So I used this theorem:
http://i55.tinypic.com/30ldw12.jpg
With X1, X2 as my X, Y and u(X1,X2)=X1X2

"AND ZERO OTHERWISE".... Damn! I did not read that properly.