Fermi integral; chemical potential vs temperature

Question

$\displaystyle \int_{0}^{\infty} \dfrac{z^{\frac{1}{2}}}{\exp{(z - \tilde{\mu}) + 1}}dz = \dfrac{2}{3} \tilde{T}^{-\frac{3}{2}}$

I try to find chemical potential as the function of temperature

Answer 1

Rewrite $$I=\int_{0}^{\infty} \dfrac{\sqrt z}{\exp{(z - \tilde{\mu}) + 1}}\,dz =\int_{0}^{\infty} \dfrac{\sqrt z}{1+k\exp{(z ) }}dz $$ where $k=e^{- \tilde{\mu}}$.

This leads to $$I=-\frac{\sqrt{\pi }}{2}\, \text{Li}_{\frac{3}{2}}\left(-\frac{1}{k}\right)=-\frac{\sqrt{\pi }}{2} \, \text{Li}_{\frac{3}{2}}\left(-e^{\tilde{\mu}) }\right)$$ where appears the polylogarithm function.

Have a look here.

For solving the equation, you will need some numerical method such as Newton. I should work fine since the rhs is a quite smooth function. From a practical point of view, I should try to find $\tilde{\mu}$ looking for the zero of $$f(\tilde{\mu})=\log \left(-\frac{ \sqrt{\pi }}{2}\, \text{Li}_{\frac{3}{2}}\left(-e^{\tilde{\mu} }\right)\right)-\log\left( \dfrac{2}{3} \tilde{T}^{-\frac{3}{2}}\right)$$ using $$f'(\tilde{\mu})=\frac{\text{Li}_{\frac{1}{2}}\left(-e^{\tilde{\mu} }\right)}{\text{Li}_{\frac{3}{2}}\left(-e^{\tilde{\mu} }\right)}$$