$$$\frac{2 z}{- \epsilon_{k}^{2} + z^{2}}$$$ 关于$$$\epsilon_{k}$$$的积分

求$$$\int \frac{2 z}{- \epsilon_{k}^{2} + z^{2}}\, d\epsilon_{k}$$$。

对 $$$c=2 z$$$ 和 $$$f{\left(\epsilon_{k} \right)} = \frac{1}{- \epsilon_{k}^{2} + z^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(\epsilon_{k} \right)}\, d\epsilon_{k} = c \int f{\left(\epsilon_{k} \right)}\, d\epsilon_{k}$$$：

$${\color{red}{\int{\frac{2 z}{- \epsilon_{k}^{2} + z^{2}} d \epsilon_{k}}}} = {\color{red}{\left(2 z \int{\frac{1}{- \epsilon_{k}^{2} + z^{2}} d \epsilon_{k}}\right)}}$$

进行部分分式分解:

$$2 z {\color{red}{\int{\frac{1}{- \epsilon_{k}^{2} + z^{2}} d \epsilon_{k}}}} = 2 z {\color{red}{\int{\left(\frac{1}{2 z \left(\epsilon_{k} + z\right)} + \frac{1}{2 z \left(- \epsilon_{k} + z\right)}\right)d \epsilon_{k}}}}$$

逐项积分：

$$2 z {\color{red}{\int{\left(\frac{1}{2 z \left(\epsilon_{k} + z\right)} + \frac{1}{2 z \left(- \epsilon_{k} + z\right)}\right)d \epsilon_{k}}}} = 2 z {\color{red}{\left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \int{\frac{1}{2 z \left(\epsilon_{k} + z\right)} d \epsilon_{k}}\right)}}$$

对 $$$c=\frac{1}{2 z}$$$ 和 $$$f{\left(\epsilon_{k} \right)} = \frac{1}{\epsilon_{k} + z}$$$ 应用常数倍法则 $$$\int c f{\left(\epsilon_{k} \right)}\, d\epsilon_{k} = c \int f{\left(\epsilon_{k} \right)}\, d\epsilon_{k}$$$：

$$2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + {\color{red}{\int{\frac{1}{2 z \left(\epsilon_{k} + z\right)} d \epsilon_{k}}}}\right) = 2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + {\color{red}{\left(\frac{\int{\frac{1}{\epsilon_{k} + z} d \epsilon_{k}}}{2 z}\right)}}\right)$$

设$$$u=\epsilon_{k} + z$$$。

则$$$du=\left(\epsilon_{k} + z\right)^{\prime }d\epsilon_{k} = 1 d\epsilon_{k}$$$ (步骤见»)，并有$$$d\epsilon_{k} = du$$$。

所以，

$$2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \frac{{\color{red}{\int{\frac{1}{\epsilon_{k} + z} d \epsilon_{k}}}}}{2 z}\right) = 2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 z}\right)$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 z}\right) = 2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2 z}\right)$$

回忆一下 $$$u=\epsilon_{k} + z$$$:

$$2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2 z}\right) = 2 z \left(\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}} + \frac{\ln{\left(\left|{{\color{red}{\left(\epsilon_{k} + z\right)}}}\right| \right)}}{2 z}\right)$$

对 $$$c=\frac{1}{2 z}$$$ 和 $$$f{\left(\epsilon_{k} \right)} = \frac{1}{- \epsilon_{k} + z}$$$ 应用常数倍法则 $$$\int c f{\left(\epsilon_{k} \right)}\, d\epsilon_{k} = c \int f{\left(\epsilon_{k} \right)}\, d\epsilon_{k}$$$：

$$2 z \left({\color{red}{\int{\frac{1}{2 z \left(- \epsilon_{k} + z\right)} d \epsilon_{k}}}} + \frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z}\right) = 2 z \left({\color{red}{\left(\frac{\int{\frac{1}{- \epsilon_{k} + z} d \epsilon_{k}}}{2 z}\right)}} + \frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z}\right)$$

设$$$u=- \epsilon_{k} + z$$$。

则$$$du=\left(- \epsilon_{k} + z\right)^{\prime }d\epsilon_{k} = - d\epsilon_{k}$$$ (步骤见»)，并有$$$d\epsilon_{k} = - du$$$。

积分变为

$$2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} + \frac{{\color{red}{\int{\frac{1}{- \epsilon_{k} + z} d \epsilon_{k}}}}}{2 z}\right) = 2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} + \frac{{\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}}{2 z}\right)$$

对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} + \frac{{\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}}{2 z}\right) = 2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} + \frac{{\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}}{2 z}\right)$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 z}\right) = 2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2 z}\right)$$

回忆一下 $$$u=- \epsilon_{k} + z$$$:

$$2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2 z}\right) = 2 z \left(\frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z} - \frac{\ln{\left(\left|{{\color{red}{\left(- \epsilon_{k} + z\right)}}}\right| \right)}}{2 z}\right)$$

因此，

$$\int{\frac{2 z}{- \epsilon_{k}^{2} + z^{2}} d \epsilon_{k}} = 2 z \left(- \frac{\ln{\left(\left|{\epsilon_{k} - z}\right| \right)}}{2 z} + \frac{\ln{\left(\left|{\epsilon_{k} + z}\right| \right)}}{2 z}\right)$$

化简：

$$\int{\frac{2 z}{- \epsilon_{k}^{2} + z^{2}} d \epsilon_{k}} = - \ln{\left(\left|{\epsilon_{k} - z}\right| \right)} + \ln{\left(\left|{\epsilon_{k} + z}\right| \right)}$$

加上积分常数：

$$\int{\frac{2 z}{- \epsilon_{k}^{2} + z^{2}} d \epsilon_{k}} = - \ln{\left(\left|{\epsilon_{k} - z}\right| \right)} + \ln{\left(\left|{\epsilon_{k} + z}\right| \right)}+C$$

$$$\int \frac{2 z}{- \epsilon_{k}^{2} + z^{2}}\, d\epsilon_{k} = \left(- \ln\left(\left|{\epsilon_{k} - z}\right|\right) + \ln\left(\left|{\epsilon_{k} + z}\right|\right)\right) + C$$$A