Fermi-Dirac-Integral

Definition C.1   Fermi-Dirac-Integral

$$F_j ( x ) := \frac{1}{\Gamma (j+1)} \int_0^{\infty} \frac{u^j}{1+exp(u-x)} \ du$$ (C.1)

Lemma C.1   Für die Ableitung des Fermi-Dirac-Integrals gilt

$$\boxed{ \frac{d}{dx} F_j ( x ) = F_{j-1} (x) }\qquad.$$ (C.2)

Beweis C.1   berechne die Ableitung des Fermi-Dirac-Integrals

$$\frac{d}{dx} F_j ( x ) = \frac{1}{\Gamma (j+1)} \frac{d}{dx} \int_0^{\infty} \frac{u^j}{1+exp(u-x)} \ du$$ (C.3)

$$= \frac{1}{\Gamma (j+1)} \int_0^{\infty} u^j \frac{d}{dx} \frac{1}{1+exp(u-x)} \ du \ \ \cite[Satz \ 19.18]{WUE95b}$$ (C.4)

$$= \frac{1}{\Gamma (j+1)} \int_0^{\infty} u^j \underbrace{(-1)(1+exp(u-x))^{-2}(-exp(u-x)}_{-\frac{d}{du} \frac{1}{1+exp(u-x)}} \ du$$ (C.5)

$$= - \frac{1}{\Gamma (j+1)} \int_0^{\infty} u^j \frac{d}{du} \frac{1}{1+exp(u-x)} \ du$$ (C.6)

$$\overset{part.Integ.}{=} - \frac{1}{\Gamma (j+1)} (\underbrace{\left[ u^j \frac{1}{1+exp(u-x)} \right]^{\infty}_0}_{0}$$

$$- \int_0^{\infty} j u^{j-1} \frac{1}{1+exp(u-x)} \ du \, )$$ (C.7)

$$= \frac{j}{\Gamma (j+1)} \int_0^{\infty} u^{j-1} \frac{1}{1+exp(u-x)} \ du$$ (C.8)

Nebenrechnung ( $\Gamma ( j + 1) = j \Gamma ( j )$ [Bro81])

$$\frac{j}{\Gamma (j + 1)} = \frac{j}{j \Gamma ( j )} = \frac{1}{\Gamma ( j )}$$ (C.9)

also

$$\frac{d}{dx} F_j ( x ) = \frac{1}{\Ga... ...^{j-1} \frac{1}{1+exp(u-x)} \ du \overset{(\ref{DefFermiInt})}{=} F_{j-1} ( x )$$ (C.10)

$\square$