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Homework Help: Evaluate Integral of x^2/(1 + e^sin(x))

  1. Jan 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate [tex] \int^2_{-2} {x^2dx\over {1+e^\sin x}} [/tex]

    3. The attempt at a solution

    No clue, tried a series expansion but it shed no light.
    Last edited: Jan 5, 2010
  2. jcsd
  3. Jan 4, 2010 #2
    This seems to be true

    [tex] \int_{-a}^a{x^2dx \over {1+e^{\sin x} } } = {a^3\over 3} [/tex]

    Not sure how to get there though
    Last edited: Jan 5, 2010
  4. Jan 5, 2010 #3
    [tex] \int_{-2}^{2}\frac{x^2}{1+e^{\sin x}} dx[/tex]

    is the same as

    [tex] \int_{0}^{2} \frac{x^2}{1+e^{\sin x}} dx+ \int_{0}^{2}\frac{x^2}{1+e^{\sin -x}}dx [/tex]

    Series expansion shows that

    [tex] \frac{x^2}{1+e^{\sin x}} = \frac{x^2}{2}-\frac{x^3}{4}+\frac{x^5}{16}-\frac{7 x^7}{480} \cdots [/tex]

    [tex] \int \frac{x^2}{1+e^{\sin x}} dx= \frac{x^3}{6}-\frac{x^4}{16}+\frac{x^6}{96}-\frac{7 x^8}{3840}+\cdots [/tex]

    [tex] \frac{x^2}{1+e^{\sin -x}} = \frac{x^2}{2}+\frac{x^3}{4}-\frac{x^5}{16}+\frac{7 x^7}{480} \cdots [/tex]

    [tex] \int \frac{x^2}{1+e^{\sin -x}}dx = \frac{x^3}{6}+\frac{x^4}{16}-\frac{x^6}{96}+\frac{7 x^8}{3840}-\cdots [/tex]

    [tex] \int \frac{x^2}{1+e^{\sin x}} dx+ \int \frac{x^2}{1+e^{\sin -x}}dx = \frac{x^3}{3} [/tex]

    That's as close as I can get. Before integration, changing x for -x preserves only the first term with an even index. Not really sure how to show that. I doubt this is the way the question is supposed to be answered.
    Last edited: Jan 5, 2010
  5. Jan 5, 2010 #4
    1/[1+exp(y)] + 1/[1+exp(-y)] =

    1/[1+exp(y)] + exp(y)/[exp(y)+1] =

    [1+exp(y)]/[1+exp(y)] = 1

  6. Jan 6, 2010 #5
    hmmm so

    x^2/(1+e^sinx) + x^2/(1+e^sin-x) = x^2. No need for expansion... I think it's a bit sneaky to expect you to notice the symmetry/assymetry in the integral etc.
  7. Jan 6, 2010 #6
    Most people who know this would have noted it via the well known fact that the Bernouilli numbers B_n for odd n are all zero except for n = 1:


    This fact follows most easily by using the generating function of the Bernoulli numbers:

    [tex]\frac{x}{\exp(x) - 1}=\sum_{n=0}^{\infty}\frac{B_{n}}{n!}x^n[/tex]

    and then using the same symmetry.
  8. Jan 6, 2010 #7
    Or via the similar symmetry of the Fermi distribution function around the Fermi energy. That function has a plus 1 in the numerator, so its more similar to this problem.
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