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Need help with E integral

  1. Mar 2, 2008 #1
    Im doing a chapter on Improper integrals, and one of my problems is i need to compute an integral if it converges. I really don't know where to start with this one. Anyone point me in the right direction?

    integral of x/(e^x) where the lower limit is 0 and the upper is infinite

    really any help would be appreciated, I got a bit of a cold and seem to be forgetting every bit of math
  2. jcsd
  3. Mar 2, 2008 #2
    ook wait, i think i figured it out

    i stat off by calculating the integral, which turns out to be

    -( (x+1)/e^x )

    then you end up getting 1 after you plug in the limits, because when its infinity it becomes zero and when 0 it becomes -1, so when you subtract negetive 1 it becomes 1, and 1 is the answer
  4. Mar 4, 2008 #3
    You have x/(e^x) = [tex]xe^{-x}[/tex]

    So [tex]\int_{0}^{\infty}xe^{-x}dx[/tex]

    Integration by parts gives [tex]{\lim }\limits_{x \to \infty } -e^{-x}(x+1) + 1[/tex]

    [tex]{\lim }\limits_{x \to \infty } \frac{-(x+1) }{e^x} + 1[/tex]

    Applying l'Hôpital's rule gives [tex]{\lim }\limits_{x \to \infty } \frac{-1 }{e^x} + 1[/tex] = 0 + 1
    = 1
  5. Mar 4, 2008 #4
    As an aside, in case any one cares:

    [tex]\int_0^\infty x^n e^{-x}\,dx = n![/tex]

    For integer n. However, the integral form is also well-defined for real, and even complex n, except at the negative integers where it diverges. Look up "gamma function" for more details.
  6. Mar 4, 2008 #5
    I somehow recognized it before I read the rest of your post. Math is fascinating
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