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Integrating exponential

  1. Aug 24, 2008 #1
    if f(x) = pi*xe^(-x^2)
    integrating this function if the lower boundary is 0 and the upper boundary is infinity is the answer pi*(2e-1). is this right?
    Last edited: Aug 24, 2008
  2. jcsd
  3. Aug 24, 2008 #2
    Can you think of an appropriate substitution that may help with this integral?
  4. Aug 24, 2008 #3
    i think i worked out.
    Is the answer (2e-1)*pi
  5. Aug 24, 2008 #4
    Could you show your work? I'm pretty sure that isn't the correct answer, but it is difficult to tell if you don't show how you arrived at it.
  6. Aug 24, 2008 #5
    sure i used substitution method.
    let u = -x^2
    then du/dx=-2x, dx=du/-2x
    this gives the integral of xe^u/-2xdu
    =pi/2 integrate e^u lower boundary 0 upper boundary infinity
    then separating the two integral, one integral with lower boundary 0 and upper boundary 1 and the second integral with lower boundary 1 and the upper boundary infinty.
    the first integral gives e^1-e^0 and the second integral we get e^1.
    then adding them together we get
    (pi/2) * (e-1+e) is that right?
  7. Aug 24, 2008 #6


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    It's hard to follow your work. Add some empty lines to make reading easier. Using Latex helps a whole lot.

    It shouldn't be e^u, it's slightly off. And what do you mean by "separating the integral"? You mean integration by parts? It's not needed here.
  8. Aug 24, 2008 #7


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    More simply, if u= -x2, then du= -2x dx or -(1/2)du= dx

    [itex]xe^{-x^2}dx[/itex] becomes [itex]e^{-x^2}(xdx)= -(1/2)e^u du[/itex]

    When x= 0, u= 0 and when x="infinity", u is -"infinity" The integral becomes
    [tex]\int_0^\infty xe^{-x^2} dx= -\frac{1}{2}\int_0^{-\infty} e^u du[/tex]
    [tex]= \frac{1}{2}\int_{-\infty}^0 e^u du[/itex]

    I can see no reason to "separate" at x= 1. In any case, since x=1 is an upper limit for one integral and a lower limit for the other, those terms will subtract, not add, and will cancel.

    And there certainly is no reason to have a "pi" in there!
    Last edited by a moderator: Aug 24, 2008
  9. Aug 24, 2008 #8
    So the answer is -1/2?
  10. Aug 24, 2008 #9


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    No, I just looked back and realized you had a "pi" in the original integral that you then dropped.

    Including that pi,
    [itex]\frac{1}{2}\pi \int_{-\infty}^0 e^u du= \frac{\pi}{2}e^u\right|_{-infty}^0= \frac{\pi}{2}[/tex]
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