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Math Help - An improper integral

  1. #1
    CSM
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    Question 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)+....

    ??
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The question is rather unclear. Could you quote it completely?. Thanks.


    Fernando Revilla
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  3. #3
    CSM
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    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
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  4. #4
    CSM
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    "AND ZERO OTHERWISE".... Damn! I did not read that properly.
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