$$$\frac{2 z}{- \epsilon_{k}^{2} + z^{2}}$$$ 對 $$$\epsilon_{k}$$$ 的積分

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

套用常數倍法則 $$$\int c f{\left(\epsilon_{k} \right)}\, d\epsilon_{k} = c \int f{\left(\epsilon_{k} \right)}\, d\epsilon_{k}$$$，使用 $$$c=2 z$$$ 與 $$$f{\left(\epsilon_{k} \right)} = \frac{1}{- \epsilon_{k}^{2} + z^{2}}$$$：

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

進行部分分式分解:

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

逐項積分：

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

套用常數倍法則 $$$\int c f{\left(\epsilon_{k} \right)}\, d\epsilon_{k} = c \int f{\left(\epsilon_{k} \right)}\, d\epsilon_{k}$$$，使用 $$$c=\frac{1}{2 z}$$$ 與 $$$f{\left(\epsilon_{k} \right)} = \frac{1}{\epsilon_{k} + z}$$$：

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

令 $$$u=\epsilon_{k} + z$$$。

則 $$$du=\left(\epsilon_{k} + z\right)^{\prime }d\epsilon_{k} = 1 d\epsilon_{k}$$$ (步驟見»)，並可得 $$$d\epsilon_{k} = du$$$。

因此，

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=\epsilon_{k} + z$$$：

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

套用常數倍法則 $$$\int c f{\left(\epsilon_{k} \right)}\, d\epsilon_{k} = c \int f{\left(\epsilon_{k} \right)}\, d\epsilon_{k}$$$，使用 $$$c=\frac{1}{2 z}$$$ 與 $$$f{\left(\epsilon_{k} \right)} = \frac{1}{- \epsilon_{k} + z}$$$：

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

令 $$$u=- \epsilon_{k} + z$$$。

則 $$$du=\left(- \epsilon_{k} + z\right)^{\prime }d\epsilon_{k} = - d\epsilon_{k}$$$ (步驟見»)，並可得 $$$d\epsilon_{k} = - du$$$。

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=-1$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=- \epsilon_{k} + z$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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