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Math Help - improper intergrals

  1. #1
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    improper intergrals

    i am trying to see if i am right on this answer i was asked to see if
    the intergral of e^x/(e^x-1) from -1 to 1 is convergent or divergent i
    think it is divergent because 0 lies between the the limits given and it would
    make the denominator equal to 0 which is undefined



    p.s sry i had to write the problem like this because i do not know how to write
    the integral in the message box like others in this forum
    thank you for the help
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  2. #2
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    Quote Originally Posted by momo55 View Post
    i am trying to see if i am right on this answer i was asked to see if
    the intergral of e^x/(e^x-1) from -1 to 1 is convergent or divergent i
    think it is divergent because 0 lies between the the limits given and it would
    make the denominator equal to 0 which is undefined



    p.s sry i had to write the problem like this because i do not know how to write
    the integral in the message box like others in this forum
    thank you for the help

    I am not keen on such improper integral but can we do it in this way ?

     \int_{-1}^1 \frac{e^{x}}{e^x - 1}~dx

     = \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx + \int_{-1}^{-a} \frac{e^{x}}{e^x - 1}~dx  \right)

    Sub .  x = -t on the second integral we have

     = \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx  + \int_a^1 \frac{e^{-x}}{e^{-x} - 1}~dx \right)

     = \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx  + \int_a^1 \frac{1}{1 - e^x}~dx \right)

     = \lim_{a\to 0^+}  \int_a^1 \left(  \frac{e^{x}}{e^x - 1} - \frac{1}{e^x - 1 } \right) ~dx

     =  \lim_{a\to 0^+} \int_a^1 ~dx =  \lim_{a\to 0^+} [x]_a^1 = \lim_{a\to 0^+}  (1-a) = 1
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by simplependulum View Post
    I am not keen on such improper integral but can we do it in this way ?

     \int_{-1}^1 \frac{e^{x}}{e^x - 1}~dx

     = \lim_{a\to 0^+} \left( \int_a^1 \frac{e^{x}}{e^x - 1}~dx + \int_{-1}^{-a} \frac{e^{x}}{e^x - 1}~dx \right)

    ...
    The question is a little controversial... what simplependulum has computed is the so called 'principal value' of the integral...

     \int_{-1}^1 \frac{e^{x}}{e^x - 1}\cdot dx (1)

    For more information see...

    Cauchy Principal Value -- from Wolfram MathWorld

    In general if we don't specify that, is...

     \int_{-1}^1 \frac{e^{x}}{e^x - 1}\cdot dx = \lim_{a \rightarrow 0-} \int_{-1}^{a} \frac{e^{x}}{e^x - 1}\cdot dx + \lim_{b \rightarrow 0 +} \int_{b}^{1} \frac{e^{x}}{e^x - 1}\cdot dx (2)

    ... and in this case both the limits in (2) doesn't exist...

    Kind regards

    \chi \sigma
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  4. #4
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    is it because when you put 0 in the limit the bottom will turn to 0 which will make it undefined? so it diverges ??
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by momo55 View Post
    is it because when you put 0 in the limit the bottom will turn to 0 which will make it undefined? so it diverges ??
    Yes, it is!...

    If You intend to compute the 'principal value' is...

    S= PV \int_{-1}^{1} f(x)\cdot dx = \int_{0}^{1} [f(x) + f(-x)]\cdot dx (1)

    ... and here is...

    f(x) + f(-x) = \frac{e^{x}}{e^{x}-1} + \frac{e^{-x}}{e^{-x} -1} = \frac{e^{x}}{e^{x}-1} - \frac{1}{e^{x}-1} = 1 (2)

    ... so that is...

    S= PV \int_{-1}^{1} \frac{e^{x}}{e^{x}-1}\cdot dx = 1 (3)

    Kind regards

    \chi \sigma
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  6. #6
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    I think that there is much easier way out:

    <br />
\int_{-1}^1 \frac{e^{x}}{e^x - 1}~dx<br />

    First compute the antiderivative.

    Make substitution t=e^{x}-1, dt=e^{x}dx

    <br />
\int \frac{dt}{t}dx = ln|t|=ln|e^{x}-1|<br />

    Because you should find the integral on interval [-1,1] and ln|e^{x}-1| is only defined on e^{x} \neq 1 or x \neq 0, clearly the integral do not converge on [-1,1].


    Regards.
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