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Finite Integral, not sure why

  1. Apr 10, 2012 #1
    [itex]\int^{\infty}_{1}[/itex][itex]\frac{1}{e^{t}-1}dt[/itex]

    [itex]= -ln(e - 1) + 1 [/itex]

    Not sure how to get the +1 part from infinity, seems like it should be infinity, i.e.
    [itex] ln(e^{\infty} -1) = ??? [/itex]


    Any help appreciated, thanks.
     
  2. jcsd
  3. Apr 10, 2012 #2

    CompuChip

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    What anti-derivative did you find?
    What is the limit as [itex]t \to \infty[/itex]?
     
  4. Apr 10, 2012 #3


    I don't know how you solve this integral, but I must make a substitution:

    [itex]e^t=u\Longrightarrow t=\log u \Longrightarrow dt=\frac{du}{u}[/itex] , so the integral becomes:

    [itex]\int^{\infty}_e \frac{du}{u(u-1)}=\int^\infty_e\frac{du}{u-1}-\int^\infty_e\frac{du}{u}= 1-\log (e-1)[/itex] , after evaluating the limit in infinity...

    DonAntonio
     
  5. Apr 10, 2012 #4
    Eventually your answer will boil down to [itex]\lim\limits_{a \rightarrow \infty} (-ln|a| + ln|a-1| + ...)[/itex] (as you can see from DonAntonio's work above).

    You need to take this limit, can you think of a way to combine the two natural logarithms in order to do this?
     
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