# Thread: An improper integral

1. ## An improper integral

So i'm doing some exercises in probabilty theory but I got stuck on some elementary calculus.

I have to integrate $\int_{A_t}2(x+y)dxdy$ with $A_t$ being the set $A_t=\{x; xy\leq t\}$. The limit of the integral will be a function of $t$.

Okay. So I rewrite $A_t=\{(x,y); 0 \leq x \leq \frac{t}{y}; 0 \leq y \leq \infty\}$ (Is this correct?)

Then we have $\int_0^\infty(\int_0^{\frac{t}{y}}(2x+2y)dx)dy$
$=\int_0^\infty [x^2+2xy]_{x=0}^{x=\frac{t}{y}})dy$
$=\int_0^\infty(\frac{t^2}{y^2}+2t)dy$

Which is a integral I don't know how to solve. I tried splitting it up:
$=\int_0^1(\frac{t^2}{y^2}+2t)dy+\int_1^\infty(\fra c{t^2}{y^2}+2t)dy$
$=t^2+2t-lim_{s\rightarrow 0}(-\frac{t^2}{a}+2ta)+....$

??

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

Fernando Revilla

3. 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:

With X1, X2 as my X, Y and u(X1,X2)=X1X2

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