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How do I integrate this?

  1. Mar 31, 2014 #1
    I'm trying to derive Saha's equation, and I've come up against this definite integral, which I can't seem to find anywhere and may not even be doable; I'm not sure.

    \begin{equation}

    \int\limits_0^\infty \frac{x^{2}}{e^{x}+1}dx

    \end{equation}

    Can anyone help? Thanks!
     
  2. jcsd
  3. Mar 31, 2014 #2

    maajdl

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  4. Mar 31, 2014 #3

    D H

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    It's doable. That's essentially the Fermi-Dirac integral F2(0).
     
  5. Mar 31, 2014 #4

    arildno

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    Presumably, traipsing along a different curve in the complex plane, using Cauchy's theorem, might give you the answer.
     
  6. Mar 31, 2014 #5
    Here is a more elementary solution:

    Rewrite the given integral as:
    $$I=\int_0^{\infty} \frac{x^2e^{-x}}{1+e^{-x}}\,dx$$
    Use the substitution ##e^{-x}=t## to obtain:
    $$I=\int_0^1 \frac{\ln^2 t}{1+t}\,dt$$
    Next, use integration by parts in the following way:
    $$I=\left(\ln^2t \ln(1+t)\right|_0^1-2\int_0^1 \frac{2\ln t\ln(1+t)}{t}\,dt$$
    Notice that the first term is zero, so we are left with:
    $$I=-2\int_0^1 \frac{\ln t \ln(1+t)}{t}\,dt$$
    Since,
    $$\ln(1+t)=-\sum_{k=1}^{\infty} \frac{(-1)^kt^k}{k}$$
    Hence,
    $$I=2\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\int_0^1 \frac{t^k\ln t}{t}\,dt$$
    It is not difficult to show that
    $$\int_0^1 \frac{t^k\ln t}{t}\,dt =-\frac{1}{k^2}$$
    From above, we get:
    $$I=-2\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$$
    It can be easily shown that
    $$\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}=-\frac{3}{4}\zeta(3)$$
    Hence, the final answer is:
    $$I=\frac{3}{2}\zeta(3)$$
    which agrees with post #2.

    I hope that helps.
     
  7. Mar 31, 2014 #6

    arildno

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    Cool, Pranav-Arora!
    :smile:
     
  8. Mar 31, 2014 #7
    Thank you! :)
     
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