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:

$\displaystyle =\int_0^1(\frac{t^2}{y^2}+2t)dy+\int_1^\infty(\fra c{t^2}{y^2}+2t)dy$

$\displaystyle =t^2+2t-lim_{s\rightarrow 0}(-\frac{t^2}{a}+2ta)+....$

??