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Can't figure out this improper integral

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    integral of x*e^x from 0 to infinity

    2. Relevant equations

    N/a

    3. The attempt at a solution

    i first integrated the integrand and got (xe^x - e^x) then i use limit as R approaches infinity and used 0 and R as my limits. when calculating i keep getting 1 instead of infinity. is there something i did wrong? thanks forhelping
     
  2. jcsd
  3. Dec 11, 2008 #2
    What do you mean "when calculating I keep getting 1"?
    You should get:

    Lim (R-->inf) {(e^R)*(R-1)+1} = inf.
    How can you ever get 1?
     
  4. Dec 11, 2008 #3
    i plugged in infinity wihtout factoring out e^R so it was (infinity * e^(infinity) - e^(infinity) plus one. so i thought the infinity would subtract to zero plus 1 for the other limit. why do you have to factor out e^R though. that's the part i don't get
     
  5. Dec 11, 2008 #4
    :)
    First of all, you don't substruct infinity from infinity :).
    Secondly, the final expression should be:
    lim (R-->inf) {R*e^R -e^R - 0 + 1} = lim (R-->inf) {(e^R)(R-1) + 1}
    Notice that we both got the same answer, only I took out e^R. This way it can be easily seen that this limit gives you kind of "inf * inf + 1", which is of course inf.
    Even if you don't do it this way, and you look at the first limit - the left-most element, R*e^R is definitely much larger than e^R as R-->inf, so that you're assumption that they cancel each other is far from the truth.
     
  6. Dec 11, 2008 #5
    thansk there is also another question where it asks for integral from negative infinity to 1 for (x^2)(e^3x) the final answer that i got was (5/27)e^3 -(infinity -infinity-infinity). when plugging in infinity to the integral, all i get are infinities. what would i do with those
     
  7. Dec 11, 2008 #6

    Defennder

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    Well you don't actually get all infinities. If you look at all those expressions you'll see that all of those whose terms have infinity in them converge to 0. That's because the limit of [tex]\frac{x^2}{e^{3x}}[/tex] as x approaches infinity is 0 because the e^3x diverges faster than the x^n term.
     
  8. Dec 11, 2008 #7
    so are you saying that the denominator will have a slightly bigger infinity then the numerator so it equals to 0?
     
  9. Dec 11, 2008 #8

    Defennder

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    Yes, something like that.
     
  10. Dec 12, 2008 #9

    HallsofIvy

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    Certainly, before you try to do an improper integral, you should learn the definition of an improper integral.
    [tex]\int_0^\infty f(t)dt= \lim_{a\rightarrow \infty} \int_0^a f(t)dt[/tex]
    That says nothing about "evaluating at t= [itex]\infty[/itex]" because such a thing is not defined.
     
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