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Electron Heat Capacity Integral

  1. Nov 12, 2015 #1
    1. The answer to this problem is easy when plugged into mathematica it's (pi^2)/3. I am trying to integrate it by hand however and can't figure out how to start it. I also can't find any other attempts of it online (our professor says we can just look it up if we can find it).

    [(x^2*E^x)/(E^x + 1)^2, {x, -Infinity, Infinity}]


    2. No equations

    3. I've tried U-sub with setting U= (e^x+1) and then tried some integration by parts but I'm not getting there.
     

    Attached Files:

  2. jcsd
  3. Nov 13, 2015 #2

    fzero

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    This actually turns out to be very complicated to do and I am having trouble giving hints that you can follow without giving too much of the answer away, so please bear with me. At least using Mathematica seems like a legitimate solution to the problem and I don't believe that many people would expect an undergrad to come up with the solution below on their own.

    First, integrals of functions of ##x^n## times exponentials can often be done by replacing ##e^x## by ##e^{a x}## and then noting that ##d/da(e^{ax}) = x e^{ax}##, so we try to replace the powers of ##x## with derivatives of another expression. Then we can exploit this by bringing the derivative outside of the integral. For example
    $$\int dx ~ x e^x = \left[ \frac{d}{da} \int dx~e^{ax} \right]_{a=1},$$
    which you should be able to verify by doing both integrals explicitly.

    In your case, we can use
    $$ \frac{x^2 e^x}{(e^x+1)^2} = \left[ \frac{d^2}{da^2} \ln ( 1+ e^{ax})\right]_{a=1}.$$
    Furthermore, we can determine the indefinite integral
    $$ \int dx \ln ( 1+ e^{ax}) $$
    in terms of the dilogarithm function (see for instance https://en.wikipedia.org/wiki/Spence's_function)
    $$\text{Li}_2(z) = - \int^z_0 \frac{du}{u} \ln ( 1-u).$$

    The big difficulty here is that the dilogarithm is infinite as ##z\rightarrow -\infty##, so the naive substitution for your integral over the whole real axis will result in a divergent integral. (The dilogarithm is also usually not defined for ##1 \leq z < \infty##, but I believe that the proper substitutions keep us on the negative real axis.) However, I believe that it is possible to show that the definite integral
    $$ F(a) =\int_{-\infty}^0 dx \ln ( 1+ e^{ax}) $$
    exists. So we should break your original integral into two parts, then the answer can be expressed as the appropriate derivative of ##F(a)+F(-a)##.

    It will probably be important to use the results (https://en.wikipedia.org/wiki/Spence's_function#Special_values) ##\text{Li}_2(-1)=-\pi^2/12## and ##\text{Li}_2(0)=0.##
     
  4. Nov 14, 2015 #3
    Thanks but this is math I am completely unfamiliar with. It ended up being doable also with a contour integral.

    SOLVED.
     

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  5. Nov 14, 2015 #4
    I drew it terribly above but you catch my drift.
     
  6. Nov 14, 2015 #5

    fzero

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    Sure, I didn't seriously consider suggesting the contour integral because it is a bit rare to find someone comfortable with the method. I probably should have asked first. It's good that you were able to work it out yourself that way.
     
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