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A bose-einstein integral

  1. May 3, 2014 #1


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    1. The problem statement, all variables and given/known data

    I am trying to evuluate the value of the integral:
    [itex] J= \int_{0}^{∞} \frac{x^{3}}{e^{x}-1}dx[/itex]

    Could you please supply me with the method used for that? I thought of breaking the integral from 0 to 1 and from 1 to infinity. That way I could expand the exponential to taylor series for the small values of x, while the second would drop the 1 from the denominator, so it would be like exp[-x]x^3....(am I correct in the idea?).. However with that idea I'm left with a singularity at point 0. Would that mean I need to expand to Laurent series?
    (it's not "really" homework, I am just playing around with black body radiation and radiation energy density to show that p[rad]~T^4 which we were given in class).
    Last edited: May 3, 2014
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  3. May 3, 2014 #2


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    There should be no singularity at x=0, as the numerator goes to 0 faster than the denominator. I don't see how you get exp[-x]x^3.
  4. May 3, 2014 #3


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    Hmm Let me write it in formula:
    [itex] \int_{0}^{∞} \frac{x^3}{e^{x}-1} dx= \int_{0}^{1} \frac{x^3}{e^{x}-1} dx+\int_{1}^{∞} \frac{x^3}{e^{x}-1} dx[/itex]
    so here I'm saying that the first integrand has a singular point at x=0 because the denominator vanishes... Of course you are right about it going slower to 0 for the nominator (I got confused by thinking that exp is a faster function that any power but I lost it hehe). In fact I would expand the exponential to taylor series:
    [itex] e^{x} ≈1+x[/itex] and solve to get
    [itex]\int_{0}^{1} \frac{x^3}{x}dx= \frac{x^{3}}{3}|^{1}_{0}=1/3[/itex]
    then I would try to say that the 2nd integrand, for x>1 the denominator is dominated by the exponential rather than the "-1" term... Or in other words I would write:
    [itex]\int_{1}^{∞} \frac{x^3}{e^{x}-1} dx =\int_{1}^{∞} \frac{x^3}{e^{x}} dx=\int_{1}^{∞} e^{-x}x^3 dx[/itex]

    Unfortunately I don't think that this approach can give me the correct result, since the correct result needs π factors... probably numerically I might be close to the correct answer, since I guess it is a game of approximations I use, against a strict result (?)
    Last edited: May 3, 2014
  5. May 3, 2014 #4

    Ray Vickson

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    [tex] f(x) = \frac{x^3}{e^x-1} = x^3 e^{-x} \frac{1}{1 - e^{-x}}
    = \sum_{n=1}^{\infty} x^3 e^{-nx} [/tex]
    Integrate term-by-term. You will get an infinite series with a known sum.
  6. May 3, 2014 #5
    You do have the right idea about using the taylor series but splitting the integral won't be a very good idea.

    $$J=\int_0^{\infty} \frac{x^3}{e^x-1}=\int_0^{\infty} \frac{x^3e^{-x}}{1-e^{-x}}\,dx$$

    Now, from the following series expansion,
    $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k\,\,\,\,\,\, (|x|<1)$$
    we can write:
    $$\frac{1}{1-e^{-x}}=\sum_{k=0}^{\infty} e^{-kx}$$
    $$J=\sum_{k=0}^{\infty} \int_0^{\infty} x^3e^{-(k+1)x}\,dx$$

    Using integration by parts, you can eliminate ##x^3##. I hope this helps.
  7. May 3, 2014 #6
    I think I found another method which doesn't involve the use of series expansions.

    $$J=\int_0^{\infty} \frac{x^3}{e^x-1}\,dx=\frac{1}{2}\left(\int_0^{\infty} \frac{x^3}{e^{x/2}-1}\,dx-\frac{x^3}{e^{x/2}+1}\,dx\right)=\frac{1}{2}\left(I_1-I_2\right)$$

    Proceeding with ##I_1## first, use the substitution ##x/2=t## to get:
    $$I_1=\int_0^{\infty} \frac{x^3}{e^{x/2}-1}\,dx=16\int_0^{\infty} \frac{t^3}{e^t-1}\,dt=16J$$
    With ##I_2##, use the same substitution to get:
    $$I_2=\int_0^{\infty} \frac{x^3}{e^{x/2}+1}\,dx=16\int_0^{\infty} \frac{t^3}{e^t+1}\,dt$$
    From the formulation of dirichilet eta function,
    $$\eta(s)=\frac{1}{\Gamma(s)}\int_0^{\infty} \frac{t^{s-1}}{e^t+1}\,dt$$
    $$\Rightarrow \eta(4)=\frac{1}{\Gamma(4)}\int_0^{\infty} \frac{t^3}{e^t+1}\,dt$$
    $$\Rightarrow \int_0^{\infty} \frac{t^3}{e^t+1}\,dt =\eta(4)\Gamma(4)$$
    $$J=\frac{1}{2}\left(16J-16\eta(4)\Gamma(4)\right) \Rightarrow J=\frac{8}{7}\eta(4)\Gamma(4)$$
    Since ##\eta(4)=7\pi^4/720## and ##\Gamma(4)=3!##, the final result is:
  8. May 3, 2014 #7


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    both, thanks for the answers.
    Pranav, Could you also check Post #3? and give me your "complains"?
    One complain I would state for that is the arbitrariness of the function as going from 0 to 1 (I mean exponential would need more powers to be closer to the needed one, like 2nd or 3rd order), the errors at that points ([itex]1^{-}[/itex]) would be over 100%....
  9. May 3, 2014 #8
    I am not sure if I can help with that. I can solve the integrals but I am not very good at doing approximations. I hope someone else can help with that.
  10. May 3, 2014 #9


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    nevermind, thanks anyway Shino kun, and the 2nd way was really nice :)
  11. May 3, 2014 #10
    Glad to help! :)
  12. May 5, 2014 #11


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    Well my approximation yields:
    [itex]\frac{1}{3}+\frac{16}{e} \approx 6.219401 [/itex]
    While the correct answer is:
    [itex]\frac{\pi^{4}}{15} \approx 6.4939394[/itex]
    So the error coming from the approximation is ~4.2%, so I guess it's not that bad...
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