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．$

$~~~~~~~ {\displaystyle I= \bigintsss{ \cfrac{ Q(x) }{ P(x) } ~dx }}$

$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，$${\displaystyle I= \bigintsss{ Q_1(x)~dx } +\bigintsss{ \cfrac{ 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）$$P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_1x+a_0$$~~~~~~=a_{n} (x-\alpha_1)(x-\alpha_2)...(x-\alpha_{n})~~~(\alpha_i\in\mathbb{C})$$~~~~~~~~~~$$~~~~~~~~~~$ $\textcolor{magenta}{ (x-\alpha)(x-\overline{\alpha})=x^2-(\alpha+\overline{\alpha})x+\alpha\overline{\alpha} }$$~~~~~~~~~~$$~~~~~~~~~~$ $~~~~~~~~~~~~~~~~~~~~\textcolor{magenta}{ =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)$$~~~~~~~~~~$$~~~~~~~~~~$$\textcolor{magenta}{(\beta^2_j-4\gamma_j<0)}$

$P(x)=a_{n} (x-\alpha_1)^{n_1} ...(x-\alpha_k)^{n_k} (x^2+\beta_1 x+\gamma_1)...(x^2+\beta_l x+\gamma_l) \\より\cfrac{R(x)}{P(x)}を部分分数分解して，$

$\cfrac{R(x)}{P(x)}= {\displaystyle \cfrac{1}{a_n} \left( \sum_{i=1}^{k} \cfrac{a_{i_1}}{x-\alpha_i} +\cfrac{a_{i_2}}{(x-\alpha_i)^2} +...+ \cfrac{a_{i_{n_1}}}{(x-\alpha_i)^{n_1}} +\sum_{j=1}^{l} \cfrac{b_j x+c_j}{x^2+\beta_j x+\gamma_j} \right) }$$= {\displaystyle \cfrac{1}{a_n} \left( \sum_{i=1}^{k} \sum_{m=1}^{n_i} \cfrac{a_{i_m}}{(x-\alpha_i)^{m}} +\sum_{j=1}^{l} \cfrac{b_j x+c_j}{x^2+\beta_j x+\gamma_j} \right) }$${\displaystyle I_2:=\bigintsss{\cfrac{R(x)}{P(x)}}~dx}$$= {\displaystyle \cfrac{1}{a_n} \left( \sum_{i=1}^{k} \sum_{m=1}^{n_i} a_{i_m} \bigintsss{\cfrac{1}{(x-\alpha_i)^m}}~dx +\sum_{j=1}^{l} \bigintsss{\cfrac{b_j x+c_j}{x^2+\beta_j x+\gamma_j}}~dx \right)}$$= {\displaystyle \cfrac{1}{a_n} \left( \sum_{i=1}^{k} \sum_{m=1}^{n_i} \cfrac{a_{i_m}}{1-m} (x-\alpha_i)^{1-m} +\sum_{j=1}^{l} \bigintsss{\cfrac{b_j x+c_j}{x^2+\beta_j x+\gamma_j}}~dx \right)}$$~~~~~~~~~~~~~~~~~~~~~\textcolor{magenta}{積分定数は一時的に省略}$$~$${\displaystyle I_3:=\bigintsss{\cfrac{b_j x+c_j}{x^2+\beta_j x+\gamma_j}}~dx}$$= {\displaystyle \cfrac{b_j}{2} \bigintsss{ \cfrac{2x+\beta_j+\left(\cfrac{2}{b_j}~c_j-\beta_j\right)}{x^2+\beta_j x+\gamma_j} }~dx }$$= {\displaystyle \cfrac{b_j}{2} \bigintsss{ \cfrac{2x+\beta_j}{x^2+\beta_j x+\gamma_j} }~dx +\left(c_j-\cfrac{b_j \beta_j}{2}\right) \bigintsss{ \cfrac{dx}{\left(x+\cfrac{\beta_j}{2}\right)^2+\left(\cfrac{\sqrt{4\gamma_j-\beta_j^2}}{2}\right)^2} } }$$= {\displaystyle \cfrac{b_j}{2} \log \left| x^2+\beta_j x+\gamma_j \right| +\cfrac{2c_j-b_j\beta_j}{\sqrt{4\gamma_j-\beta_j^2}} \arctan \cfrac{2x+\beta_j}{\sqrt{4\gamma_j-\beta_j^2}} }$$~~~~~~~~~~~~~~~~~~~~~\textcolor{magenta}{積分定数は一時的に省略}$

$\therefore I_2= {\displaystyle \cfrac{1}{a_n} \left( \sum_{i=1}^{k} \sum_{m=1}^{n_i} \cfrac{a_{i_m}}{1-m} (x-\alpha_i)^{1-m} +\sum_{j=1}^{l} \left(\cfrac{b_j}{2} \log \left| x^2+\beta_j x+\gamma_j \right| +\cfrac{2c_j-b_j\beta_j}{\sqrt{4\gamma_j-\beta_j^2}} \arctan \cfrac{2x+\beta_j}{\sqrt{4\gamma_j-\beta_j^2}} \right) \right)}$$~~~~~~~~~~~~~~~~~~~~~\textcolor{magenta}{積分定数は一時的に省略}$

${\displaystyle \bigintsss{ \cfrac{Q(x)}{P(x)} }~dx }$

$={\displaystyle \sum_{q=0}^{t} \cfrac{d_q}{q+1} x^{q+1}}$$ {\displaystyle +\cfrac{1}{a_n} \sum_{i=1}^{k} \sum_{m=1}^{n_i} \cfrac{a_{i_m}}{1-m} (x-\alpha_i)^{1-m} }$$ {\displaystyle +\cfrac{1}{a_n} \sum_{j=1}^{l} \left(\cfrac{b_j}{2} \log \left| x^2+\beta_j x+\gamma_j \right| +\cfrac{2c_j-b_j\beta_j}{\sqrt{4\gamma_j-\beta_j^2}} \arctan \cfrac{2x+\beta_j}{\sqrt{4\gamma_j-\beta_j^2}} \right)+C }$