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Improper Integral

  1. Jan 20, 2006 #1
    Int[(x^3)/((e^x)-1)] [0, infinity]

    What is the trick?

    I thought of by-parts but how would you integrate 1/((e^x)-1)?

    Substitution won't work with
    u = e^x -1
    with x = ln|u+1|

    or it would be rather tough to evaluate u^3

    Someone please give me a hint.
  2. jcsd
  3. Jan 20, 2006 #2


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    You can take:

    [tex]\frac{1}{e^x-1} = \frac{e^{-x}}{1-e^{-x}} [/tex]

    and then use the substitution u=1-e-x.
  4. Jan 20, 2006 #3
    and dU= e^-x dx
    dx = e^x du

    e^-x = 1-u
    e^x = 1/(1-u)

    [(1-u)/u][1/(1-u)] = 1/u

    Now putting limits from 0 to infinity.
    ln|1-e^-x| = ln(1) - ln(1-1) = 0

    That doesn't work.
    Last edited: Jan 20, 2006
  5. Jan 20, 2006 #4


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    Well, that's how you would get the indefinite integral of 1/(ex-1), which is what I thought you wanted, and although this diverges at 0 as 1/x, your original integral will not because of the x3 in the numerator.

    So getting back to the original integral, I don't know how you would find the indefinite integral of this, but the trick to find the integral over the range you specified is this:

    [tex]\int_0^{\infty} \frac{x^3 dx}{e^x-1} = \int_0^{\infty} \frac{x^3 e^{-x} dx}{1-e^{-x}} [/tex]

    [tex]= \int_0^{\infty} x^3 e^{-x} (1+e^{-x}+e^{-2x}+...)dx [/tex]

    This can be turned into a sum over the inverse fourth powers of the natural numbers, whose value is, I think, pi^4/90.
  6. Jan 21, 2006 #5


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    Exactly correct!

    Exactly correct! In general: for all real y>1, (or complex y with real part greater than 1,) let

    [tex]I_y=\int_0^{\infty} \frac{x^{y-1} dx}{e^x-1} = \int_0^{\infty} x^{y-1} e^{-x}\frac{1}{1-e^{-x}}dx, [/tex]

    expanding the fraction as a geometric series gives

    [tex]\frac{1}{1-e^{-x}}=\sum_{k=0}^{\infty} e^{-kx}[/tex]

    and hence

    [tex]I_y= \int_0^{\infty} x^{y-1}e^{-x}\sum_{k=0}^{\infty} e^{-kx}dx = \int_0^{\infty} \sum_{k=1}^{\infty} e^{-kx}x^{y-1}dx = \sum_{k=1}^{\infty} \int_0^{\infty} e^{-kx}x^{y-1} dx [/tex]

    substitute [itex]u=kx[/itex] so that [itex]x=\frac{u}{k},[/itex] and hence [itex]dx=\frac{du}{k}[/itex] to get

    [tex]I_y=\sum_{k=1}^{\infty} \int_0^{\infty} e^{-u}\left( \frac{u}{k} \right) ^{y-1} \frac{du}{k} = \sum_{k=1}^{\infty} \left( \frac{1}{k} \right) ^{y} \int_0^{\infty}e^{-u}u^{y-1}du=\zeta (y)\Gamma (y)[/tex]

    By the way, I coppied this proof from mathworld. :biggrin:

    In particular, we have [tex]\int_0^{\infty} \frac{x^3 dx}{e^x-1} =\int_0^{\infty} \frac{x^{4-1} dx}{e^x-1} = \zeta (4)\Gamma (4) = \frac{\pi ^4}{90}\cdot 3! = \frac{\pi ^4}{15}[/tex]
    Last edited: Jan 21, 2006
  7. Jan 22, 2006 #6
    Is there another and easier way?

    I was never introduced to special functions.
  8. Jan 22, 2006 #7


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    Not really. All you have to do is get it into a form involving the sum:


    You can get to this step as long as you know how to do integrals like:

    [tex]\int_0^{\infty} x^3 e^{-a x} dx[/tex]

    Once you have it into that form, you can just take it as a given that the above sum is [itex]\pi^4/90[/itex]. The fact that pi is involved gives you a clue there probably isn't an easier way, unless there were some lucky trig substitution (which I'm pretty sure there isn't).
  9. Nov 15, 2010 #8
    It was very useful to calculate the Stefan-Boltzmann constant.
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