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Double integral problem

  1. Nov 27, 2012 #1
    Evaluate the integral ∫(2,∞) ∫(2/x,∞) 1/(y^2)*e^(-x/y) dydx by changing the order of integration.

    I get ∫(1,∞) ∫(2y,∞) 1/(y^2)*e^(-x/y)dxdy

    etc. etc. etc.

    I get to ∫(1,∞) (e^(-2)/y) dy

    Which is (ln∞-ln1)/e^2 = ∞

    Does this thing not converge?
     
  2. jcsd
  3. Nov 27, 2012 #2

    SammyS

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    Sketch the region of integration.

    You have errors in your integration limits for your integral, [itex]\displaystyle \int\int \frac{1}{y^2}e^{-x/y}\,dx\,dy\ .[/itex]

    It looks to me like you will need to write that as the sum of two integrals.
     
    Last edited: Nov 27, 2012
  4. Nov 27, 2012 #3
    You're right, it should be ∫(2,2y) instead of ∫(2y,∞)

    I'm still getting ∞
     
  5. Nov 27, 2012 #4

    SammyS

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    I also get ∞. However, the limits of integration are not what you have.

    The original integral has y going from y = 2/x to y = ∞.

    If y = 2/x , then x = 2/y not what you have, which is x = 2y.

    Furthermore:

    When y ≥ 1 , x goes from x = 2 to x = +∞ .

    And when 0 ≤ y ≤ 1 , x goes from x = 2/y to x = +∞ .
     
  6. Nov 27, 2012 #5
    My bad, it was supposed to be x/2, not 2/x.

    I think it's infinity but I can't integrate e^(-2/y)/y.
     
  7. Nov 27, 2012 #6

    SammyS

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    O.K.

    Then you do get [itex]\displaystyle \int_{1}^{\infty}\,\int_{2}^{2y} \frac{1}{y^2}e^{-x/y}\,dx\,dy \ .[/itex]

    I also get that this does not converge.

    Yes. That does not integrate to an elementary function.
     
  8. Nov 28, 2012 #7
    Surely
    $$
    \int_{2}^{\infty}\int_{x/2}^{\infty}\frac 1{y^2}e^{-x/y} \, \mathrm dy \, \mathrm dx =
    \int_{1}^{\infty} \frac{e^{1/y}}{y^2} \int_{2y}^{\infty}e^{-x} \, \mathrm dx \, \mathrm dy
    $$
    is integrable. I get ##e^{-2}## as its value.
     
  9. Nov 28, 2012 #8

    SammyS

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    Except that [itex]\ \ e^{-x/y}\ne e^{-x}e^{1/y}\ .[/itex]
     
  10. Nov 28, 2012 #9
    I'd still try moving ##1/y^2## forwards:
    $$
    \int_2^\infty\int_{x/2}^\infty\frac1{y^2}e^{-x/y}\,\mathrm dy\,\mathrm dx = \int_2^\infty\frac1{y^2}\int_{2y}^\infty e^{(-1/y)\cdot x}\,\mathrm dx\,\mathrm dy.
    $$
     
    Last edited: Nov 28, 2012
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