有理関数の積分

$$$整式P(x),Q(x)に対して次の積分を計算する．$

${\displaystyle I={\displaystyle \int \frac{Q(x)}{P(x)}dx}}$

$$\begin{gathered} Q(x)=Q_1(x)P(x)+R(x) \\ {Q}_1(x)={\displaystyle \sum _{q=0}^t{d}_q{x}^q}とすると， \end{gathered}$$${\displaystyle I={\displaystyle \int Q_1(x)dx+{\displaystyle \int \frac{R(x)}{P(x)}dx}}}(degP>degR)$

$右辺第一項は自明．$

$n次多項式P(x)は複素数の範囲で次のように因数分解できる．（代数学の基本定理）$$P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$$=a_n(x-{\alpha }_1)(x-{\alpha }_2)...(x-{\alpha }_n)({\alpha }_i\in \mathbb{C})$$$$$ $(x-\alpha )(x-\bar{\alpha })=x^2-(\alpha +\bar{\alpha })x+\alpha \bar{\alpha }$$$$$ $=x^2+\beta x+\gamma (\beta ,\gamma \in \mathbb{R})$$P(x)は{\alpha }_i,{\beta }_j,{\gamma }_j\in \mathbb{R}を用いて次のように表される．$$\therefore P(x)=a_n(x-{\alpha }_1{)}^{n_1}...(x-{\alpha }_k{)}^{n_k}(x^2+{\beta }_{1}x+{\gamma }_1)...(x^2+{\beta }_{l}x+{\gamma }_l)$$$$$$({\beta }_j^2-4{\gamma }_j<0)$

$$\begin{gathered} P(x)=a_n(x-{\alpha }_1{)}^{n_1}...(x-{\alpha }_k{)}^{n_k}(x^2+{\beta }_{1}x+{\gamma }_1)...(x^2+{\beta }_{l}x+{\gamma }_l) \\ より\frac{R(x)}{P(x)}を部分分数分解して， \end{gathered}$$

$\frac{R(x)}{P(x)}={\displaystyle \frac{1}{a_n}(\sum _{i=1}^k\frac{a_{i_1}}{x-{\alpha }_i}+\frac{a_{i_2}}{(x-{\alpha }_i{)}^2}+...+\frac{a_{i_{n_1}}}{(x-{\alpha }_i{)}^{n_1}}+\sum _{j=1}^l\frac{b_{j}x+c_j}{x^2+{\beta }_{j}x+{\gamma }_j})}$$={\displaystyle \frac{1}{a_n}(\sum _{i=1}^k\sum _{m=1}^{n_i}\frac{a_{i_m}}{(x-{\alpha }_i{)}^m}+\sum _{j=1}^l\frac{b_{j}x+c_j}{x^2+{\beta }_{j}x+{\gamma }_j})}$${\displaystyle I_2:={\displaystyle \int \frac{R(x)}{P(x)}dx}}$$={\displaystyle \frac{1}{a_n}\left(\sum _{i=1}^k\sum _{m=1}^{n_i}{a}_{i_m}{\displaystyle \int \frac{1}{(x-{\alpha }_i{)}^m}dx+\sum _{j=1}^l{\displaystyle \int \frac{b_{j}x+c_j}{x^2+{\beta }_{j}x+{\gamma }_j}dx}}\right)}$$={\displaystyle \frac{1}{a_n}(\sum _{i=1}^k\sum _{m=1}^{n_i}\frac{a_{i_m}}{1-m}(x-{\alpha }_i{)}^{1-m}+\sum _{j=1}^l{\displaystyle \int \frac{b_{j}x+c_j}{x^2+{\beta }_{j}x+{\gamma }_j}dx})}$$積分定数は一時的に省略$$$${\displaystyle I_3:={\displaystyle \int \frac{b_{j}x+c_j}{x^2+{\beta }_{j}x+{\gamma }_j}dx}}$$={\displaystyle \frac{b_j}{2}{\displaystyle \int \frac{2x+{\beta }_j+(\frac{2}{b_j}{c}_j-{\beta }_j)}{x^2+{\beta }_{j}x+{\gamma }_j}dx}}$$={\displaystyle \frac{b_j}{2}{\displaystyle \int \frac{2x+{\beta }_j}{x^2+{\beta }_{j}x+{\gamma }_j}dx+(c_j-\frac{b_j{\beta }_j}{2}){\displaystyle \int \frac{dx}{{(x+\frac{{\beta }_j}{2})}^2+{\left(\frac{\sqrt{4{\gamma }_j-{\beta }_j^2}}{2}\right)}^2}}}}$$={\displaystyle \frac{b_j}{2}\log |x^2+{\beta }_{j}x+{\gamma }_j|+\frac{2c_j-b_j{\beta }_j}{\sqrt{4{\gamma }_j-{\beta }_j^2}}\arctan \frac{2x+{\beta }_j}{\sqrt{4{\gamma }_j-{\beta }_j^2}}}$$積分定数は一時的に省略$

$\therefore I_2={\displaystyle \frac{1}{a_n}(\sum _{i=1}^k\sum _{m=1}^{n_i}\frac{a_{i_m}}{1-m}(x-{\alpha }_i{)}^{1-m}+\sum _{j=1}^l(\frac{b_j}{2}\log |x^2+{\beta }_{j}x+{\gamma }_j|+\frac{2c_j-b_j{\beta }_j}{\sqrt{4{\gamma }_j-{\beta }_j^2}}\arctan \frac{2x+{\beta }_j}{\sqrt{4{\gamma }_j-{\beta }_j^2}}))}$$積分定数は一時的に省略$

${\displaystyle {\displaystyle \int \frac{Q(x)}{P(x)}dx}}$

$={\displaystyle \sum _{q=0}^t\frac{d_q}{q+1}{x}^{q+1}}$${\displaystyle +\frac{1}{a_n}\sum _{i=1}^k\sum _{m=1}^{n_i}\frac{a_{i_m}}{1-m}(x-{\alpha }_i{)}^{1-m}}$${\displaystyle +\frac{1}{a_n}\sum _{j=1}^l(\frac{b_j}{2}\log |x^2+{\beta }_{j}x+{\gamma }_j|+\frac{2c_j-b_j{\beta }_j}{\sqrt{4{\gamma }_j-{\beta }_j^2}}\arctan \frac{2x+{\beta }_j}{\sqrt{4{\gamma }_j-{\beta }_j^2}})+C}$