ゾンマーフェルト展開
{\displaystyle I=\underbrace {\tau \int _{-\infty }^{0}{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x} _{I_{1}}+\underbrace {\tau \int _{0}^{\infty }{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x} _{I_{2}}\,.}
I 1 = τ ∫ − ∞ 0 H ( μ + τ x ) e x + 1 d x = τ ∫ 0 ∞ H ( μ − τ x ) e − x + 1 d x {\displaystyle I_{1}=\tau \int _{-\infty }^{0}{\frac {H(\mu +\tau x)}{e^{x}+1}}\,\mathrm {d} x=\tau \int _{0}^{\infty }{\frac {H(\mu -\tau x)}{e^{-x}+1}}\,\mathrm {d} x\,}
次に、 I 1 {\displaystyle I_{1}}
1 e − x + 1 = 1 − 1 e x + 1 , {\displaystyle {\frac {1}{e^{-x}+1}}=1-{\frac {1}{e^{x}+1}}\,,}
すると、次を得る。