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Integral of a function with exponent

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    As part of a problem I need to evaluate,

    [itex]\int[/itex][itex]^{\infty}_{0}[/itex]x2e-[itex]\lambda[/itex]x2

    2. Relevant equations

    ∫e-λx2 = 1/2 [itex]\sqrt{\pi/\lambda}[/itex]
    ( with integration limits from 0 to ∞ )

    3. The attempt at a solution
    I've tried integrating by parts, also tried changing variables and combinations of these. been stuck on this integral for a couple of days, would appreciate some help.
     
  2. jcsd
  3. Oct 20, 2011 #2

    phyzguy

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    What happened when you integrated by parts?
     
  4. Oct 20, 2011 #3
    when integrated by parts we have to integrate ∫e-λx2 dx as a function of x , but since its strictly increasing function of x i think it will diverge for the limit ∞
     
  5. Oct 20, 2011 #4
    its decreasing if lambda is positive.

    Your integral is the third integral down from this link. As to solving for it, you might have to do some research, but this will be a nice start for you.

    http://en.wikipedia.org/wiki/Lists_...integrals_lacking_closed-form_antiderivatives
     
  6. Oct 20, 2011 #5

    phyzguy

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    That's not right. The function exp(-lambda x^2) decreases rapidly as x increases, so its integral from 0 to infinity is a known number. In fact you are given this value as part of the problem, no?
     
  7. Oct 20, 2011 #6
    thanks ! but i am talking about the function f(x) = [itex]\int e^{-λx^{2}}dx[/itex] when integrated from 0 to ∞

    what's the value of the indefinite integral [itex]\int e^{-λx^{2}}dx[/itex] ?
     
  8. Oct 21, 2011 #7

    HallsofIvy

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    I don't see why integrating by parts wouldn't work. You did take [itex]u= x[/itex], [itex]dv= xe^{\lambda x^2}[/itex], didn't you?
     
  9. Oct 21, 2011 #8
    yea i got
     
  10. Oct 21, 2011 #9
    Ok when I do by-parts, I get


    -∫(2x).(∫e-λx2dx)dx + (x2)(∫e-λx2dx)|[itex]^{\infty}_{0}[/itex]

    But I'm unsure how to proceed. I only know the integral ∫e-λx2dx when the limits are 0 and ∞, and I don't think I can simply apply the formula I posted. Even if I could, the function would be a constant and then i have both terms going to infinity.
     
  11. Oct 21, 2011 #10
  12. Oct 21, 2011 #11
    use what HallsofIvy posted. You will get the right answer.
     
  13. Oct 21, 2011 #12
    ok i'll use u=x and dv = xe-λx2

    Then I need to find v = ∫xe-λx2

    for this I again integrate by parts

    then v = -∫e-λx2 + x∫e-λx2 |[itex]^{\infty}_{0}[/itex]

    The first term I know is 1/2([itex]\pi[/itex]/λ)1/2 but now the second term goes to infinity?
     
  14. Oct 21, 2011 #13
    nuh uhh. if u = x^2, du / dx = 2x. dx = du/2x, getting rid of one of the x's
     
  15. Oct 21, 2011 #14
    that second integral definitely converges...as I think was stated earlier in this thread a couple of times.

    There is nothing wrong with using IBP here, but it's not necessary if you look at my above post.
     
  16. Oct 21, 2011 #15
    Ah I see, so now I get

    v = -(1/2λ)e-λx2

    and my integral becomes

    (1/2λ)∫[itex]^{\infty}_{0}[/itex]e-λx2dx - x(1/2λ)e-λx2 |[itex]^{∞}_{0}[/itex]

    and the second term goes to zero as x tends to ∞ because of the decaying exponent term.

    giving my intergral as (1/4λ) ([itex]\frac{\pi}{\lambda}[/itex])1/2

    Is that right?
     
  17. Oct 21, 2011 #16
    If I remember correctly from that wikipedia link I sent you, yes.
     
  18. Oct 21, 2011 #17

    SammyS

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    For the integral in the equation for v, [itex]v=\int xe^{-\lambda x^2}dx\,,[/itex] you don't need to use integration by parts. Instead, use substitution. Let u = -λx2 → du = -2λ x dx .
     
  19. Oct 21, 2011 #18
    thanks a lot, you've been a big help
     
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