Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finite Integral, not sure why

  1. Apr 10, 2012 #1

    [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


    User Avatar
    Science Advisor
    Homework Helper

    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...

  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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook