关于1/(x^n+1) 的不定积分
在知乎上看到许多此类问题，
来写个小总结(=ﾟДﾟ=)
先来计算
\int \frac{1}{1+x^{2n}}{\rm d}x\quad(n\in\mathbb{N^*})\\
记多项式 1+x^{2n} 的 2n 个根为 ：
a_{k}=\cos\frac{2k-1}{2n}\pi+i\sin\frac{2k-1}{2n}\pi,(k=1,2,\cdots,2n)\\
则有
|a_k|=1,a_k^{2n}=-1\\
\bar{a}_k=a_{2n-k+1},a_k\bar{a}_k=1,a_k+\bar{a}_k=2\cos\frac{2k-1}{2n}\pi\\
可设 \frac{1}{1+x^{2n}}=\sum_{k=1}^{2n}\frac{A_k}{x-a_k}
1=\sum_{k=1}^{2n}\frac{A_k(1+x^{2n})}{x-a_k}{\rm d}x\\
令 x\to a_i, 并利用洛必达法则可得
\begin{align*}1&=\lim _{x \rightarrow a_{i}} \sum_{k=1}^{2 n} \frac{A_{k}\left(1+x^{2 n}\right)}{x-a_{k}}=\lim _{x \rightarrow a_{i}} \frac{A_{i}\left(1+x^{2 n}\right)}{x-a_{i}}\\ &=\lim _{x \rightarrow a_{i}}\left(2 n A_{i} x^{2 n-1}\right)=2 n A_{i} \frac{a_{i}^{2 n}}{a_{i}}=-\frac{2 n A_{i}}{a_{i}} \quad(i=1,2, \cdots, 2 n)\end{align*}\\
即 A_{k}=-\frac{a_{k}}{2 n} \quad(k=1,2, \cdots, 2 n) ，于是
\begin{align*} &\frac{1}{1+x^{2 n}}=-\frac{1}{2 n} \sum_{k=1}^{2 n} \frac{a_{k}}{x-a_{k}}=-\frac{1}{2 n} \sum_{k=1}^{n}\left(\frac{a_{k}}{x-a_{k}}+\frac{\bar{a}_{k}}{x-\bar{a}_{k}}\right)\\ =&-\frac{1}{2 n} \sum_{k=1}^{n} \frac{\left(a_{k}+\bar{a}_{k}\right) x-2 a_{k} \bar{a}_{k}}{x^{2}-\left(a_{k}+\bar{a}_{k}\right) x+a_{k} \bar{a}_{k}}=\frac{1}{n} \sum_{k=1}^{n} \frac{1-x \cos \frac{2 k-1}{2 n} \pi}{x^{2}-2 x \cos \frac{2 k-1}{2 n} \pi+1} \end{align*}\\
于是
\begin{aligned} \int \frac{\mathrm{d} x}{1+x^{2 n}}=& \frac{1}{n} \sum_{k=1}^{n} \int \frac{1-x \cos \frac{2 k-1}{2 n} \pi}{x^{2}-2 x \cos \frac{2 k-1}{2 n} \pi+1} \mathrm{~d} x=-\frac{1}{2 n} \sum_{k=1}^{n}\left(\cos \frac{2 k-1}{2 n} \pi \int \frac{2 x-2 \cos \frac{2 k-1}{2 n} \pi}{x^{2}-2 x \cos \frac{2 k-1}{2 n} \pi+1} \mathrm{~d} x\right) \\ &+\frac{1}{n} \sum_{k=1}^{n}\left[\sin ^{2} \frac{2 k-1}{2 n} \pi \int \frac{\mathrm{d} x}{\left(x-\cos \frac{2 k-1}{2 n} \pi\right)^{2}+\sin ^{2} \frac{2 k-1}{2 n} \pi}\right] \\ =&-\frac{1}{2 n} \sum_{k-1}^{n}\left[\left(\cos \frac{2 k-1}{2 n} \pi \right)\cdot\ln \left|x^{2}-2 x \cos \frac{2 k-1}{2 n} \pi+1\right|\right] \\ &+\frac{1}{n} \sum_{k=1}^{n}\left(\left(\sin \frac{2 k-1}{2 n} \pi\right)\cdot \arctan \frac{x-\cos \frac{2 k-1}{2 n} \pi}{\sin \frac{2 k-1}{2 n} \pi}\right)+C \end{aligned}\\
对于
\int \frac{1}{1+x^{2n+1}}{\rm d}x\quad(n\in\mathbb{N^*})\\
注意到奇数时分解会比偶数时多出一项
同偶数情形可设
\frac{1}{1+x^{2n+1}}=\frac{a}{1+x}+\frac{2}{2n+1}\sum_{k=1}^{n}\frac{1-x \cos \frac{2 k-1}{2 n+1} \pi}{x^{2}-2 x \cos \frac{2 k-1}{2 n+1} \pi+1}\\
\frac{1+x}{1+x^{2n+1}}=a+(1+x)\frac{2}{2n+1}\sum_{k=1}^{n}\frac{1-x \cos \frac{2 k-1}{2 n+1} \pi}{x^{2}-2 x \cos \frac{2 k-1}{2 n+1} \pi+1}\\
a=\lim_{x\to-1}\frac{1+x}{1+x^{2n+1}}=\lim_{x\to-1}\frac{1}{(2n+1)x^{2n}}=\frac{1}{2n+1}\\
则
\frac{1}{1+x^{2n+1}}=\frac{1}{(2n+1)(1+x)}+\frac{2}{2n+1}\sum_{k=1}^{n}\frac{1-x \cos \frac{2 k-1}{2 n+1} \pi}{x^{2}-2 x \cos \frac{2 k-1}{2 n+1} \pi+1}\\
同理有
\begin{align*} &\int \frac{1}{1+x^{2n+1}}{\rm d}x\\ =&\frac{\ln|1+x|}{2n+1}-\frac{1}{2n+1}\sum_{k-1}^{n}\left[\left(\cos \frac{2 k-1}{2 n+1} \pi\right)\cdot \ln \left|x^{2}-2 x \cos \frac{2 k-1}{2 n+1} \pi+1\right|\right]\\ +&\frac{2}{2n+1} \sum_{k=1}^{n}\left(\left(\sin \frac{2 k-1}{2 n+1} \pi\right) \cdot\arctan \frac{x-\cos \frac{2 k-1}{2 n+1} \pi}{\sin \frac{2 k-1}{2 n+1} \pi}\right)+C \end{align*}\\
对于
\int\frac{1}{1+x^{2^n}}{\rm d}x\quad(n\geqslant2)\\
类型的不定积分有如下公式[1]
\int\frac{1}{1+x^{2^{n}}} {\rm d} x=\sum_{k=1}^{2^{n-2}} \frac{a_{k}}{2^{n+1}}\left[\log \left(\frac{x^{2}+a_{k} x+1}{x^{2}-a_{k} x+1}\right)+2 \tan ^{-1}\left(\frac{a_{k} x}{1-x^{2}}\right)\right]+C\\
其中有 a_k 如下形式
\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \cdots \pm \sqrt{2}}}}\\
a_k 中 2 的个数为 n-1
例如
n=2 时 , a_k 中 2 的个数为 1
a_1=\sqrt{2}\\
\int\frac{1}{1+x^{4}} {\rm d} x=\frac{\sqrt{2}}{8}\left[\log \left(\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right)+2 \tan ^{-1}\left(\frac{\sqrt{2} x}{1-x^{2}}\right)\right]+C\\
n=3 时 , a_k 中 2 的个数为 2
a_{1}=\sqrt{2+\sqrt{2}}, a_{2}=\sqrt{2-\sqrt{2}}\\
\begin{align*}\int\frac{1}{1+x^{8}} {\rm d} x&=\frac{\sqrt{2+\sqrt{2}}}{16}\left[\log \left(\frac{x^{2}+\sqrt{2+\sqrt{2}}x+1}{x^{2}-\sqrt{2+\sqrt{2}}x+1}\right)+2 \tan ^{-1}\left(\frac{\sqrt{2+\sqrt{2}} x}{1-x^{2}}\right)\right]\\ &+\frac{\sqrt{2-\sqrt{2}}}{16}\left[\log \left(\frac{x^{2}+\sqrt{2-\sqrt{2}}x+1}{x^{2}-\sqrt{2-\sqrt{2}}x+1}\right)+2 \tan ^{-1}\left(\frac{\sqrt{2-\sqrt{2}} x}{1-x^{2}}\right)\right]+C \end{align*}\\
n=4 时 , a_k 中 2 的个数为 3
\begin{array}{c} a_{1}=\sqrt{2+\sqrt{2+\sqrt{2}}}, \quad a_{2}=\sqrt{2+\sqrt{2-\sqrt{2}}} \\ a_{3}=\sqrt{2-\sqrt{2+\sqrt{2}}}, \quad a_{4}=\sqrt{2-\sqrt{2-\sqrt{2}}} \end{array}\\
\int\frac{1}{1+x^{16}} {\rm d} x=\cdots\cdots\\
这个积分也可以用Gauss Hypergeometric Function表示
\int \frac{1}{x^n+1} \, {\rm d}x=x \, _2F_1\left(1,\frac{1}{n};1+\frac{1}{n};-x^n\right)\\
具体推导见@Miaplacidus的回答
然后就是相关的定积分
\color{red}{\int_{0}^{+\infty}\frac{{\rm d}x}{a^n+x^n}=\frac{a\pi}{na^n\sin\frac{\pi}{n}}}\\
令 y=\left(\frac{x}{a}\right)^{n}, 不难得到
\int_{0}^{+\infty}\frac{{\rm d}x}{a^n+x^n}=\frac{a}{na^n}\int_{0}^{+\infty}\frac{y^{\frac{1}{n}-1}}{1+y}\,{\rm d}y\\
利用余元公式
\int_{0}^{+\infty}\frac{x^{p-1}}{1+x}\,dx=\frac{\pi}{\sin(p\pi )}\\
即可得到结果
证明可参考以下文章
\color{red}{ \begin{align*} \int_0^1 \frac{1}{1+x^p}{\rm d}x&=\frac{1}{2p}\left[\psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)\right] \end{align*} }\\
其中 \psi(z) 是 Digamma Function
\color{blue}{\rm Proof:}
置 x^p=t , 则
\begin{align*} &2p\int_0^1 \frac{dx}{1+x^p} \\ =&2 \int_0^1 \frac{t^{\frac{1}{p}-1}}{1+t} dt= 2\int_0^1 \sum_{n=0}^\infty (-1)^n t^{\frac{1}{p}+n-1} dt\\ =& 2\sum_{n=0}^\infty\int_0^1 (-1)^n t^{\frac{1}{p}+n-1} dt\\ =&2\sum_{n=0}^\infty\int_0^1 (-1)^n t^{\frac{1}{p}+n-1} dt\\ =&2\sum_{n=0}^\infty\frac{(-1)^n}{\frac{1}{p}+n}\\ =& \frac{1}{\frac{1}{2p}} - \frac{1}{\frac{1}{2p} + \frac12} + \sum_{n=1}^\infty\left(\frac{1}{\frac{1}{2p}+n} - \frac{1}{\frac{1}{2p} + n + \frac12}\right)\\ =&\psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right) \end{align*}\\
最后一步利用了
\psi(z)=-\gamma-\frac{1}{z}+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+z}\right)\\
电子版
链接: https://pan.baidu.com/s/10RNPSBF6BNtg2I2RhmSdHA
提取码: 578c
参考
- ^Notebooks of Srinivsu Rmanujan, Vol. 2, Tata Institute of Fundamental Research, Bombay, 1957.
