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

  • Thread starter Gregg
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  • #1
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Homework Statement



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


The Attempt at a Solution



No clue, tried a series expansion but it shed no light.
 
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Answers and Replies

  • #2
459
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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
 
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  • #3
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[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.
 
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  • #4
1,838
7
1/[1+exp(y)] + 1/[1+exp(-y)] =

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

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

:approve:
 
  • #5
459
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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.
 
  • #6
1,838
7
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.
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:

http://en.wikipedia.org/wiki/Bernoulli_number

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.
 
  • #7
1,838
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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