$$$\frac{1}{- k^{2} + r^{2}}$$$ 关于$$$k$$$的积分

求$$$\int \frac{1}{- k^{2} + r^{2}}\, dk$$$。

进行部分分式分解:

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

逐项积分：

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

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

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

设$$$u=k + r$$$。

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

所以，

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

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

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

回忆一下 $$$u=k + r$$$:

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

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

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

设$$$u=- k + r$$$。

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

因此，

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

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

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

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

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

回忆一下 $$$u=- k + r$$$:

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

因此，

$$\int{\frac{1}{- k^{2} + r^{2}} d k} = - \frac{\ln{\left(\left|{k - r}\right| \right)}}{2 r} + \frac{\ln{\left(\left|{k + r}\right| \right)}}{2 r}$$

化简：

$$\int{\frac{1}{- k^{2} + r^{2}} d k} = \frac{- \ln{\left(\left|{k - r}\right| \right)} + \ln{\left(\left|{k + r}\right| \right)}}{2 r}$$

加上积分常数：

$$\int{\frac{1}{- k^{2} + r^{2}} d k} = \frac{- \ln{\left(\left|{k - r}\right| \right)} + \ln{\left(\left|{k + r}\right| \right)}}{2 r}+C$$

$$$\int \frac{1}{- k^{2} + r^{2}}\, dk = \frac{- \ln\left(\left|{k - r}\right|\right) + \ln\left(\left|{k + r}\right|\right)}{2 r} + C$$$A