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The short story is, that I have to calculate some transport coefficients, but using the the MB distribution as my distribution function. What I currently need to solve is:

${{\mathcal{L}}^{\,\left( \alpha \right)}}=\frac{1}{2}\frac{{\bar{N}}}{V}{{\left( \frac{{{h}^{2}}}{2\pi m{{k}_{B}}T} \right)}^{3/2}}\frac{1}{{{k}_{B}}T}\int{{{e}^{-\varepsilon /{{k}_{B}}T}}}{{\left( \varepsilon -\mu \right)}^{\alpha }}{{\varepsilon }^{3/2}}d\varepsilon$

As it shows, it's not that the integral is hard, it's more what I use for $\mu$ that is my concern. According to my teacher I need to use:

${{e}^{-\mu /{{k}_{B}}T}}=\frac{1}{2}\frac{V}{{\bar{N}}}{{\left( \frac{2\pi m{{k}_{B}}T}{{{h}^{2}}} \right)}^{3/2}}$

But I can't see how I can use this instead of $\mu$ without solving for $\mu$. And then I would just end up with a ln() term, which I really can't imagine doing anything good for me. But maybe I'm mistaken ?

Any hint would be appreciated :)

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    $\begingroup$ Notice that the factor in front of the integral contains the form $e^{\mu/k_B T}$. You can use that to write the integral in a slightly simpler form. Other than that, I don't understand your question. Just solve the integral with $\mu$ as a general constant. $\endgroup$
    – Pulsar
    May 24, 2013 at 11:58
  • $\begingroup$ Yup... I know that :) That I get from splitting up the MB distribution $e^{-\frac{\varepsilon - \mu}{k_{B}T}}$. But again, then I have been told, that the $\mu$ in the $(\varepsilon - \mu)^{\alpha}$ part is from that equation as well. And that is where my problem is :/ $\endgroup$ May 24, 2013 at 12:01
  • $\begingroup$ Which quantities are known? Do you know $T$, $m$, $h^2$, $V$, and $\bar{N}$? What are the integration limits? Are they $[\mu,\infty]$? $\endgroup$
    – Pulsar
    May 24, 2013 at 12:12
  • $\begingroup$ I think I figured it out actually. Apparently it was just solving for $\mu$ and that's it :) And yes, I do know everything else. But thanks for help ! $\endgroup$ May 24, 2013 at 12:18

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