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Improper integral using comparison theorem

  1. Oct 15, 2012 #1
    State if the following integral converges or diverges, and justify your claim.

    [itex] \int_{-1}^{1} \frac{e^x}{x+1}\,dx [/itex]

    I tried using the comparison theorem by comparing it to [itex] \frac{1}{x+1} [/itex]. But for the interval (-1,0) the function is smaller for all x. So I could not conclude whether it diverged or converged.

    If the comparison had concluded that the function was larger for all x I could have said it would diverge.

    So my problem is finding the right equation to compare it to.
     
  2. jcsd
  3. Oct 15, 2012 #2

    tiny-tim

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    welcome to pf!

    hi markr2! welcome to pf! :smile:
    why 1 on top? :wink:
     
  4. Oct 15, 2012 #3
    Thanks for your reply, but I found the solution by comparing it to [itex] \frac{e^{-1}}{x+1} [/itex].

    Should I delete this thread now?
     
  5. Oct 15, 2012 #4

    tiny-tim

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  6. Oct 15, 2012 #5

    SammyS

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    Never delete a thread.

    Besides your comparison is the wrong way around.
     
  7. Oct 15, 2012 #6

    Zondrina

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    Notice your integral has a discontinuity at -1.

    [itex]\int_{-1}^{1} \frac{e^x}{x+1} dx[/itex]

    You either want something BIGGER that CONVERGES or you want something SMALLER that DIVERGES.

    This integral looks like it's going to diverge to me, could you think of a smaller function which we can integrate that will DIVERGE here? That is :

    [itex]\int_{-1}^{1} \frac{e^x}{x+1}dx ≥ \int_{-1}^{1} something that diverges \space dx[/itex]
     
  8. Oct 15, 2012 #7

    SammyS

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    Zondrina,

    Use \text{} or \mbox{} to do text with spaces, etc. in Latex.

    [itex]\dots \int_{-1}^{1} \text{something that diverges} \space dx[/itex]
     
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