Derivation of Fermi level for T>0 I am working through the derivation of the Fermi level $ \mu_0$ for T>0. However, at one point in the notes I have, it states without any explanation that:
$$ \int_0^\infty F'(\epsilon) f_{FD}(\epsilon) d\epsilon=F(\mu)+\frac{\pi^2}{6\beta^2}F''(\mu)$$
where $f_{FD}$ is the Fermi Dirac distribution and $\mu$ is the fermi level. Can anyone explain/show why this should be true? I understand it as an integration by parts and as $$[F(\epsilon)f_{FD}(\epsilon)]_0^\infty = 0$$ then 
$$ \int_0^\infty F'(\epsilon) f_{FD}(\epsilon) d\epsilon = -\int_0^\infty F(\epsilon) f'_{FD}(\epsilon) d\epsilon $$
but I am unsure as to how to reach the result from here.
Thank you.
 A: Let's assume that we are looking for average value of some quantity $G(E)$ at $T << T_{F}$.
For some dispersion relation which states that $d^{3}\mathbf p = f(E)dE$ $F{'}(E) = G(E)f(E)$ and $\epsilon = \beta E$ we'll give
$$
\tag 1 \langle G(E)\rangle = A(V, \beta)\int \limits_{0}^{\infty}  \frac{F{'}(\epsilon )d \epsilon}{e^{\epsilon - \beta \mu } + 1}.
$$
Then let's use relation $T << T_{F}$ and integrate $(1)$ by parts:
$$
\langle G(E)\rangle = A(V, \beta)F(0) + A(V, \beta )\int \limits_{0}^{\infty} \frac{F(\epsilon ) e^{\epsilon - \beta \mu}}{\left( 1 + e^{\epsilon - \beta \mu}\right)^{2}}d\epsilon = \left|y = \epsilon - \beta \mu \right| = 
$$
$$
\tag 2 =A(V, \beta) F(0) + A(V, \beta)\int \limits_{-\beta \mu}^{\infty}F(y + \beta \mu)\frac{e^{y}}{\left( e^{y} + 1\right)^{2}}dy.
$$
Let's analyze the second summand. Function $g(y) = \frac{e^{y}}{\left( 1+e^{y}\right)^{2}}$ is symmetric under replacing $y \to -y$ and has strong peak near $y = 0$. So we can make the lower bound of integration to be equal to $-\infty$ and then expand $F(y + \beta \mu )$ near $y = 0$ (the lower the temperature, the bigger is $\beta \mu = t$):
$$
F(y + t) = F(t) + F'(t)y + \frac{F{''}(t)}{2}y^{2} + ...
$$
After substituting of these euqalities into $(2)$ it takes the form
$$
\langle G(E)\rangle = A(V, \beta) F(0) + A(V, \beta)\left[ F(t) \int \limits_{-\infty}^{\infty} \frac{e^{y}dy}{\left(e^{y} + 1 \right)^{2}} + \frac{F{''}(t)}{2}\int \limits_{-\infty}^{\infty} \frac{y^{2}e^{y}dy}{\left(e^{y} + 1 \right)^{2}} + ...\right]
$$
Here I've left only summands with even degrees of $y$ from the expansion of $F(y + t)$, because integration of summands with odd degrees of $y$ is equal to zero ($g(y)$ is symmetric). The first integral is equal to $1$. The second integral (as it can be shown by using complex analisys) is equal to $\frac{\pi^{2}}{3}$, and so on. By using these equalities you can get the statement from your question.
