莫比乌斯变换（分式线性变换）

在复变函数中，我们学习过复平面中的莫比乌斯变换，形式为

$f\left(z\right)=\frac{az+b}{cz+d} \quad \left(a,b,c,d \in \mathbb{C},ad-bc \neq0\right)$

它具有许多优美的性质，我们也称其为分式线性变换

莫比乌斯变换（Möbius Transformation）虽然起源于复分析，但其思想和方法可以与微积分的工具结合，用于解决特定类型的积分

例如这类带根号项的分式有理函数积分

$\int \frac{gx+h}{\left(ax^2+bx+c\right)\sqrt{dx^2+ex+f}}\, \mathrm{d}x$

下面我们从最基础的变换手段讲起，最后给出解决这类积分问题的通法

倒代换

通过倒代换 $\displaystyle x=\frac{1}{t}$ 将待求问题的区间从 $[1,+\infty)$ 转换成 $(0,1]$，简化计算

若问题区间为 $[0,+\infty)$，则分段求解，将原积分拆解成

$\begin{align} \int_{0}^{+\infty}f\,\mathrm{d}x &=\int_{0}^1f\,\mathrm{d}x+\int_{1}^{+\infty}f\,\mathrm{d}x\\ &=\int_{0}^1f\,\mathrm{d}x+\int_{0}^1f(\frac{1}{x})\cdot \frac{1}{x^2}\,\mathrm{d}x\\ &=\int_{0}^1 \left(f+\frac{f\left(\frac{1}{x}\right)}{x^2}\right)\,\mathrm{d}x \end{align}$

例1

$\int_{0}^{+\infty} \frac{1}{(x^{2025}+1)(x^2+1)}\,\mathrm{d}x$

对 $[1,+\infty)$ 作倒代换，令 $\displaystyle x=\frac{1}{t}$，则有

$\begin{align} \int_{0}^{+\infty} \frac{1}{(x^{2025}+1)(x^2+1)}\,\mathrm{d}x &=\int_{0}^1\frac{1}{(x^{2025}+1)(x^2+1)}\,\mathrm{d}x+\int_{1}^{+\infty}\frac{1}{(x^{2025}+1)(x^2+1)}\,\mathrm{d}x\\ &=\int_{0}^1\frac{1}{(x^{2025}+1)(x^2+1)}\,\mathrm{d}x+\int_{0}^1\frac{1}{(\displaystyle \frac{1}{t^{2025}}+1)(\displaystyle \frac{1}{t^2}+1)}\cdot \frac{1}{t^2}\,\mathrm{d}t\\ &=\int_{0}^1\left(\frac{1}{(x^{2025}+1)(x^2+1)}+\frac{x^{2025}}{(x^{2025}+1)(x^2+1)}\right)\,\mathrm{d}x\\ &=\int_{0}^1 \frac{\mathrm{d}x}{x^2+1}=\frac{\pi}{4} \end{align}$

例2

$\int_{0}^{+\infty} \frac{4 + x + 2x^2 - x^3}{x^6 + 2x^4 + 2x^2 + 1} \,\mathrm{d}x$

同上题解法

$\begin{align} \int_{0}^{+\infty} \frac{4 + x + 2x^2 - x^3}{x^6 + 2x^4 + 2x^2 + 1} \,\mathrm{d}x&=\int_{0}^1\frac{4 + x + 2x^2 - x^3}{x^6 + 2x^4 + 2x^2 + 1} \,\mathrm{d}x +\int_{0}^1 \frac{4 + \frac{1}{t} + 2\cdot \frac{1}{t^2} - \frac{1}{t^3}}{\frac{1}{t^6} + 2\cdot \frac{1}{t^4} + 2\cdot \frac{1}{t^2} + 1} \cdot \frac{1}{t^2} \,\mathrm{d}t\\ &=\int_{0}^1 \left(\frac{4 + x + 2x^2 - x^3}{x^6 + 2x^4 + 2x^2 + 1}+\frac{4x^4 + x^3 + 2x^2 - x}{x^6 + 2x^4 + 2x^2 + 1}\right)\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{4x^4 + 4x^2 +4}{x^6 + 2x^4 + 2x^2 + 1}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{4\left(x^4+x^2+1\right)}{x^2\cdot \left(x^4+x^2+1\right)+\left(x^4+x^2+1\right)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{x^2+1}=\frac{\pi}{4} \end{align}$

例3

$\int_{0}^{+\infty} \frac{x^2\ln x}{\left(1+x^2\right)^3}\,\mathrm{d}x$ $\begin{align} \int_{0}^{+\infty} \frac{x^2\ln x}{\left(1+x^2\right)^3}\,dx &=\int_{0}^{1} \frac{x^2\ln x}{\left(1+x^2\right)^3}\,\mathrm{d}x+\int_{0}^{1} \frac{-\frac{1}{t^2}\cdot \ln t}{\left(1+\frac{1}{t^2}\right)^3} \cdot \frac{1}{t^2}\,\mathrm{d}t\\ &=\int_{0}^{+\infty} \left(\frac{x^2\ln x}{\left(1+x^2\right)^3}-\frac{x^2\ln x}{\left(1+x^2\right)^3}\right)\,\mathrm{d}x\\ &=0 \end{align}$

退化莫比乌斯变换

通过代换 $\displaystyle x=\frac{1-t}{1+t}$ 将待求积分的上下限从 $\displaystyle \int_{0}^{1}$ 转换成 $\displaystyle -\int_{0}^{1}$

例1

$\int_{0}^{1}\frac{\ln\left(x+1\right)}{x^2+1}\,\mathrm{d}x$

令 $\displaystyle x=\frac{1-t}{1+t}$，则有

$\begin{align} \int_{0}^{1}\frac{\ln\left(x+1\right)}{x^2+1}\,\mathrm{d}x &= -\int_{0}^{1}\frac{\ln\left(\frac{1-t}{1+t}+1\right)}{\left(\frac{1-t}{1+t}\right)^2+1}\cdot \frac{-2}{\left(t+1\right)^2}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\ln 2-\ln \left(x+1\right)}{x^2+1}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\ln 2}{x^2+1}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln \left(x+1\right)}{x^2+1}\,\mathrm{d}x\\ &=\ln 2\cdot \frac{\pi}{4}-\int_{0}^{1}\frac{\ln\left(x+1\right)}{x^2+1}\,\mathrm{d}x \end{align}$

即

$\begin{align} 2\cdot \int_{0}^{1}\frac{\ln\left(x+1\right)}{x^2+1}\,\mathrm{d}x&=\frac{\ln 2\cdot \pi}{4}\\ \int_{0}^{1}\frac{\ln\left(x+1\right)}{x^2+1}\,\mathrm{d}x &=\frac{\ln 2\cdot \pi}{8}\ \end{align}$

例2

$\int_{0}^{1}\frac{1}{1-x^2}\cdot \ln \left(\frac{x+1}{2x}\right)\,\mathrm{d}x$

令 $\displaystyle x=\frac{1-t}{1+t}$，则有

$\begin{align} \int_{0}^{1}\frac{1}{1-x^2}\cdot \ln \left(\frac{x+1}{2x}\right)\,\mathrm{d}x &=\int_{0}^{1}\frac{1}{1-\left(\frac{1-t}{1+t}\right)^2}\cdot \frac{1}{\left(1+t\right)^2} \cdot \ln \left(\frac{\frac{1-t}{1+t}+1}{2\cdot \frac{1-t}{1+t}}\right)\,\mathrm{d}t\\ &=-\frac{1}{2}\cdot \int_{0}^{1}\frac{\ln \left(1-x\right)}{x}\,\mathrm{d}x\\ &=-\frac{1}{2}\cdot \int_{0}^{1}\frac{1}{x}\cdot \sum_{k=1}^\infty \frac{\left(-1\right)}{k}\cdot x^k\,\mathrm{d}x\\ &=\frac{1}{2}\cdot \sum_{k=1}^\infty\int_{0}^{1}x^{k-1}\,\mathrm{d}x\\ &=\frac{1}{2}\cdot \sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi ^2}{12} \end{align}$

例3

$\int \frac{\arctan x}{\left(x+1\right)^2}\,\mathrm{d}x$

令 $\displaystyle x=\frac{1-t}{1+t}$，则有

$\begin{align} \int \frac{\arctan x}{\left(x+1\right)^2}\,\mathrm{d}x &= \int \frac{\arctan \left(\frac{1-t}{1+t}\right)}{\left(\left(\frac{1-t}{1+t}\right)+1\right)^2}\cdot \frac{(-2)}{\left(t+1\right)^2}\,\mathrm{d}t\\ &=-\frac{1}{2}\cdot \int \arctan \left(\frac{1-t}{1+t}\right)\,\mathrm{d}t\\ &=-\frac{1}{2}\cdot \int \left(\frac{\pi}{4}-\arctan t\right)\,\mathrm{d}t\\ &=-\frac{\pi}{8}x+\frac{1}{2}\cdot x\cdot \arctan x-\frac{1}{4}\ln \left(x^2+1\right)+C \end{align}$

莫比乌斯变换

作如下分式线性变换

$x=\frac{at+b}{ct+d} \quad or \quad t=\frac{ax+b}{cx+d}$

例1

$\int_{0}^{1}\frac{x^{m-1}\cdot \left(1-x\right)^{n-1}}{\left(t+x\right)^{m+n}}\,\mathrm{d}x$

令 $\displaystyle x=\frac{t\left(1-z\right)}{t+z}$，则有

$\begin{align} \int_{0}^{1}\frac{x^{m-1}\cdot \left(1-x\right)^{n-1}}{\left(t+x\right)^{m+n}}\,\mathrm{d}x&=\int_{0}^{1}\left(\frac{x}{t+x}\right)^m\cdot \left(\frac{1-x}{t+x}\right)^n\cdot \frac{\mathrm{d}x}{x\cdot \left(1-x\right)}\\ &=\int_{0}^{1} \left( \left[\frac{\frac{t\left(1-z\right)}{t+z}}{t+\frac{t\left(1-z\right)}{t+z}}\right]^m \cdot \left[\frac{1-\frac{t\left(1-z\right)}{t+z}}{t+\frac{t\left(1-z\right)}{t+z}}\right]^n \cdot \frac{1}{\frac{t\left(1-z\right)}{t+z}\cdot \left(1-\frac{t\left(1-z\right)}{t+z}\right)}\cdot \frac{-t\left(t+1\right)}{\left(t+z\right)^2} \right) \,\mathrm{d}z\\ &=\int_{0}^{1} \left(\left[\frac{\left(1-z\right)}{t+1}\right]^m\cdot \left[\frac{z}{t}\right]^n\cdot \frac{1}{z\left(z-1\right)}\right)\,\mathrm{d}z\\ &=\frac{1}{\left(t+1\right)^m\cdot t^n}\cdot \int_{0}^{1} \left(1-z\right)^{m-1} \cdot z^{n-1}\,\mathrm{d}z\\ &=\frac{\mathrm{B}\left(m,n\right)}{\left(t+1\right)^m\cdot t^n} \end{align}$

例2

$\int \frac{1}{\left(x+1\right)\sqrt{x^2+x+1}}\,\mathrm{d}x$

令 $\displaystyle t=\frac{1-x}{1+x}$，则 $x=\displaystyle \frac{1+t}{1-t}$，原式化为

$\begin{align} \int \frac{1}{\left(x+1\right)\sqrt{x^2+x+1}}\,\mathrm{d}x&=\int \frac{1}{\left(\frac{1+t}{1-t}+1\right)\sqrt{\left(\frac{1+t}{1-t}\right)^2+\left(\frac{1+t}{1-t}\right)+1}}\cdot \frac{2}{\left(t-1\right)^2}\,\mathrm{d}x\\ &=\int \frac{1}{\sqrt{t^2+3}}\,\mathrm{d}t\\ \end{align}$

由双元法，立刻得出原函数为

$\int \frac{1}{\sqrt{t^2+3}}\,\mathrm{d}t=\ln \left(x+\sqrt{x^2+3}\right)+C$

例3

$\int \frac{1}{\left(x-2\right)^2\left(x-3\right)^3}\,\mathrm{d}x$

对于形如此种形式的积分

$\int \frac{1}{\left(x-a\right)^m\left(x-b\right)^n}\,\mathrm{d}x \quad \left(m>n\right)$

一般令 $t=\displaystyle \frac{x-b}{x-a}$，将幂次高的项放分母

对此题而言，只需令 $t=\displaystyle \frac{x-2}{x-3}$，则有

$\begin{align} \int \frac{1}{\left(x-2\right)^2\left(x-3\right)^3}\,\mathrm{d}x &=\int \frac{1}{\left(\frac{x-2}{x-3}\right)^2\cdot \left(x-3\right)^5}\,\mathrm{d}x\\ &=\int \frac{1}{t^2\cdot \left(\frac{1}{t-1}\right)^5}\cdot \frac{(-1)}{(t-1)^2}\,dt\\ &=\int \frac{-\left(t-1\right)^3}{t^2}\,\mathrm{d}t\\ &=\int \left(\frac{1}{t^2}-\frac{3}{t}+3-t\right)\,\mathrm{d}t\\ &=-\frac{1}{x}-3\ln x +3x-\frac{x^2}{2}+C \end{align}$

莫比乌斯变换一般化

系数推导

对于一般的带根号项的分式有理函数积分，如

$\int \frac{gx+h}{\left(ax^2+bx+c\right)\sqrt{dx^2+ex+f}}\,\mathrm{d}x$

不妨令

$\begin{align} x &= \frac{\alpha t+\beta}{t+1}\\ \rightarrow \mathrm{d}x &= \frac{\alpha-\beta}{\left(t+1\right)^2}\,\mathrm{d}t \end{align}$

带入原积分，对于 $(ax^2+bx+c)$，我们观察其一次项系数

$\begin{align} &\left[a\left(\frac{\alpha t + \beta}{t+1}\right)^2+b\left(\frac{\alpha t+\beta}{t+1}\right)+c\right]\cdot \left(t+1\right)^2\\ &=a\left(\alpha t+\beta\right)^2+b\left(\alpha t+\beta\right)\cdot \left(t+1\right)+c\cdot \left(t+1\right)^2\\ &=a\left(\alpha^2t^2+2\alpha\beta t+\beta^2\right)+b\left(\alpha^2t^2+(\alpha+\beta) t+\beta\right)+c\left(t^2+2t+1\right)\\ &=\left(a\cdot \alpha^2+b\cdot \alpha^2+c\right)\cdot t^2+\left(2a\cdot \alpha\cdot \beta+b\left(\alpha+\beta\right)+2c\right)\cdot t+\left(a\cdot \beta^2+b\cdot \beta+c\right) \end{align}$

同理，对于 $\left(dx^2+ex+f\right)$，其一次项系数应该与 $(ax^2+bx+c)$ 类似，即

$\begin{cases} 2a\cdot \alpha\cdot \beta+b\left(\alpha+\beta\right)+2c\\ \\ 2d\cdot \alpha\cdot \beta+e\left(\alpha+\beta\right)+2f \end{cases}$

令

$u=\alpha\cdot \beta \quad \quad v=\alpha+\beta$

当上述两项的一次项系数都为零时

$\begin{cases} 2a\cdot \alpha\cdot \beta+b\left(\alpha+\beta\right)=-2c\\ \\ 2d\cdot \alpha\cdot \beta+e\left(\alpha+\beta\right)=-2f \end{cases}$

记

$\begin{align} \begin{vmatrix} 2a & b \\ 2d & e \end{vmatrix} &=2\cdot \begin{vmatrix} a & b \\ d & e \end{vmatrix} =D\\ \begin{vmatrix} -2c & b \\ -2f & e \end{vmatrix} &=-2\cdot \begin{vmatrix} c& b \\ f & e \end{vmatrix} =D_1\\ \begin{vmatrix} 2a & -2c \\ 2d & -2f \end{vmatrix} &= -4\cdot \begin{vmatrix} a & c \\ d & f \end{vmatrix} =D_2 \end{align}$

则

$\begin{cases} u=\displaystyle \frac{D_1}{D}\\ \\ v=\displaystyle \frac{D_2}{D} \end{cases}$

特征方程为

$\begin{align} \left(x-\alpha\right)\left(x-\beta\right) &=x^2-\left(\alpha+\beta\right)\cdot x+\alpha\cdot \beta=0\\ &=x^2-\frac{D_2}{D}\cdot x+\frac{D_1}{D}=0\\ &=D\cdot x^2-D_2\cdot x+D_1=0 \end{align}$

即

$\begin{vmatrix} a & b \\ d & e \end{vmatrix} \cdot x^2+2\cdot\begin{vmatrix} a & c \\ d & f \end{vmatrix}\cdot x+\begin{vmatrix} b & c \\ e & f \end{vmatrix}=0$

解出该方程的两根即为莫比乌斯变换的两个系数 $\alpha,\beta$

例1

$\int \frac{1}{\left(x^2-x+1\right)\sqrt{x^2+x+1}}\,\mathrm{d}x$

求出其对应的特征方程系数

$\begin{align} \begin{vmatrix} a & b \\ d & e \end{vmatrix} &= 2\\ \begin{vmatrix} a & c \\ d & f \end{vmatrix} &= 0\\ \begin{vmatrix} b & c \\ e & f \end{vmatrix} &= -2 \end{align}$

则特征方程为

$\begin{align} x^2&-1=0\\ \rightarrow &x= \pm1 \end{align}$

则令 $x=\displaystyle \frac{t-1}{t+1}$，代入原式得

$\begin{align} &\int \frac{1}{\left(x^2-x+1\right)\sqrt{x^2+x+1}}\,\mathrm{d}x \\ &=\int \frac{1}{\left[\left(\frac{t-1}{t+1}\right)^2-\left(\frac{t-1}{t+1}\right)+1\right]\cdot \sqrt{\left(\frac{t-1}{t+1}\right)^2+\left(\frac{t-1}{t+1}\right)+1}}\cdot \frac{\left(-2\right)}{\left(t+1\right)^2}\,\mathrm{d}t\\ &=\int \frac{2\left(t+1\right)}{\left[(t-1)^2-\left(t-1\right)\left(t+1\right)+\left(t+1\right)^2\right]\cdot \sqrt{(t-1)^2+\left(t-1\right)\left(t+1\right)+\left(t+1\right)^2}}\,\mathrm{d}t\\ &=2\cdot \left[\int \frac{t\cdot \mathrm{d}t}{\left(t^2+3\right)\cdot \sqrt{3t^2+1}}+\int \frac{\mathrm{d}t}{\left(t^2+3\right)\cdot \sqrt{3t^2+1}}\right] \end{align}$

对于+号左边，直接凑微分

$\begin{align} \int \frac{t\cdot \mathrm{d}t}{\left(t^2+3\right)\cdot \sqrt{3t^2+1}} &=\frac{1}{6}\cdot \int \frac{\mathrm{d}\left(3t^2+1\right)}{\left(t^2+3\right)\cdot \sqrt{3t^2+1}}\\ &=\frac{1}{3}\cdot \int \frac{\mathrm{d}\left(\sqrt{3t^2\\+1}\right)}{t^2+3}\\ &=\int \frac{\mathrm{d}\left(\sqrt{3t^2\\+1}\right)}{3t^2+1+8}\\ &=\frac{1}{2\sqrt{2}} \cdot \arctan\frac{\sqrt{3t^2+1}}{2\sqrt{2}}+C \end{align}$

对于+号右边，可用双元法

令

$x=\sqrt{3}t \quad \quad y=\sqrt{3t^2+1}$

则

$\begin{align} \int \frac{\mathrm{d}t}{\left(t^2+3\right)\cdot \sqrt{3t^2+1}} &= \sqrt{3}\cdot \int \frac{\mathrm{d}x}{\left(x^2+9\right)\cdot y}\\ &=\sqrt{3}\cdot \int \frac{1}{\left(x^2+9\right)}\cdot \left[\left(-x^2\right)\cdot \mathrm{d}\left(\frac{y}{x}\right)\right]\\ &=\sqrt{3}\cdot \int \left(\frac{9}{x^2+9}-1 \right)\cdot \mathrm{d}\left(\frac{y}{x}\right)\\ &=9\sqrt{3}\cdot \int \frac{y^2-x^2}{x^2+9\cdot \left(y^2-x^2\right)}\,\mathrm{d}\left(\frac{y}{x}\right)-\sqrt{3}\cdot \ln \frac{y}{x}\\ &=9\sqrt{3}\cdot \int \frac{\left(\frac{y}{x}\right)^2-1}{9\cdot \left(\frac{y}{x}\right)^2-8}\,\mathrm{d}\left(\frac{y}{x}\right)-\sqrt{3}\cdot \ln \frac{y}{x}\\ &=\sqrt{3}\cdot \ln \frac{y}{x}+\sqrt{3}\cdot \int \frac{\mathrm{d}t}{8-9t^2}-\sqrt{3}\cdot \ln \frac{y}{x}\\ &=\frac{\sqrt{6}}{24}\cdot \ln \frac{3\cdot \sqrt{3t^2\\+1}+2\sqrt{6}\cdot t}{3\cdot \sqrt{3t^2\\+1}-2\sqrt{6}\cdot t}+C \end{align}$

故最终结果为

$\int \frac{1}{\left(x^2-x+1\right)\sqrt{x^2+x+1}}\,\mathrm{d}x=\sqrt{2} \cdot \arctan\frac{\sqrt{3x^2+1}}{2\sqrt{2}}+\frac{\sqrt{6}}{12}\cdot \ln \frac{3\cdot \sqrt{3x^2\\+1}+2\sqrt{6}\cdot x}{3\cdot \sqrt{3x^2\\+1}-2\sqrt{6}\cdot x}+C$

例2

$\int \frac{1}{\left(x^2+2\right)\sqrt{2x^2-2x+5}}\,\mathrm{d}x$

求出其对应的特征方程系数

$\begin{align} \begin{vmatrix} a & b \\ d & e \end{vmatrix} &= -2\\ \begin{vmatrix} a & c \\ d & f \end{vmatrix} &= 1\\ \begin{vmatrix} b & c \\ e & f \end{vmatrix} &= 4 \end{align}$

则特征方程为

$\begin{align} x^2&-x-2=0\\ \rightarrow &x= 2 \quad or \,\,\,-1 \end{align}$

则令 $x=\displaystyle \frac{2t-1}{t+1} $，代入原式得

$\begin{align} \int \frac{1}{\left(x^2+2\right)\sqrt{2x^2-2x+5}}\,\mathrm{d}x &= \frac{1}{3}\cdot \int \frac{t+1}{\left(2t^2+1\right)\sqrt{t^2+1}}\,\mathrm{d}t\\ &= \frac{1}{3}\cdot \int \frac{t}{\left(2t^2+1\right)\sqrt{t^2+1}}\,\mathrm{d}t+ \frac{1}{3}\cdot \int \frac{1}{\left(2t^2+1\right)\sqrt{t^2+1}}\,\mathrm{d}t\\ \end{align}$

同例1的分析步骤，对左边的式子凑微分，对右边的式子使用双元法

最终得到答案为

$\int \frac{1}{\left(x^2+2\right)\sqrt{2x^2-2x+5}}\,\mathrm{d}x=\frac{\sqrt{2}}{12}\cdot \ln \frac{\sqrt{2x^2+2}-1}{\sqrt{2x^2+2}+1}-\frac{1}{3}\cdot \arctan \frac{\sqrt{x^2+1}}{x}+C$