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Integration by parts wrong (?)

  1. Aug 3, 2011 #1
    Here I used integration by parts to try to solve an integral (I got it wrong, it seems), I know this has no "simple" solution, but, can anyone explain me exactly what did I do wrong? Here is what I did:


    [itex]\int\frac{e^x}{x}dx=\frac{e^x}{x}-(-1)\int\frac{e^x}{x^2}dx=\frac{e^x}{x}+1(\frac{e^x}{x^2}-(-2)\int\frac{e^x}{x^3}dx)=\frac{e^x}{x}+1!(\frac{e^x}{x^2}+2!\frac{e^x}{x^3}+3!\int\frac{e^x}{x^4}dx)=...=0!\frac{e^x}{x}+1!\frac{e^x}{x^2}+2!\frac{e^x}{x^3}+3!\frac{e^x}{x^4}+...[/itex]

    I just used integration by parts on the "remaining" integrals so I could produce a series, but the series is just...wrong. I would appreciate if you told me what is the error in my reasoning.
    thanks.
     
  2. jcsd
  3. Aug 4, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Aug 4, 2011 #3
    It doesn't, that's the point. Why can I use integration by parts and "say" that [itex]\int\frac{e^x}{x}dx[/itex] equals a sum that does not converge?
     
  5. Aug 4, 2011 #4
    And, thanks :)
     
  6. Aug 5, 2011 #5

    tiny-tim

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    Hi Username007! :smile:
    integration by parts works fine so long as the ∑ is finite …

    ∫ ex/x dx

    = ex(∑n=0k-1 n!/xn+1 - n! ∫ ex/xn+1 dx​

    unfortunately, although we can usually rely on the final integral converging to zero as k -> ∞, in this case it doesn't! :rolleyes:
     
  7. Aug 6, 2011 #6
    So, if the result of integration does not converge we can't establish the equality? (Thanks for you patience btw ^_^)
     
  8. Aug 6, 2011 #7

    tiny-tim

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