(1) $\frac{1}{x^4+1}$ を $\frac{1}{x^4+1} = \frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ のように部分分数分解したとき、係数 $A, B, C, D$ を求めよ。 (2) 定積分 $\int_0^1 \frac{dx}{x^4+1}$ の値を求めよ。

1. 問題の内容

(1) $\frac{1}{x^4+1}$ を $\frac{1}{x^4+1} = \frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ のように部分分数分解したとき、係数 $A, B, C, D$ を求めよ。

(2) 定積分 $\int_0^1 \frac{dx}{x^4+1}$ の値を求めよ。

2. 解き方の手順

(1) 部分分数分解の係数を求める。

与えられた式 $\frac{1}{x^4+1} = \frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ の両辺に $x^4+1 = (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$ を掛けると、

$1 = (Ax+B)(x^2-\sqrt{2}x+1) + (Cx+D)(x^2+\sqrt{2}x+1)$

$1 = Ax^3 - \sqrt{2}Ax^2 + Ax + Bx^2 - \sqrt{2}Bx + B + Cx^3 + \sqrt{2}Cx^2 + Cx + Dx^2 + \sqrt{2}Dx + D$

$1 = (A+C)x^3 + (-\sqrt{2}A+B+\sqrt{2}C+D)x^2 + (A-\sqrt{2}B+C+\sqrt{2}D)x + (B+D)$

係数比較により、

$A+C = 0$

$-\sqrt{2}A+B+\sqrt{2}C+D = 0$

$A-\sqrt{2}B+C+\sqrt{2}D = 0$

$B+D = 1$

$C = -A$ を2, 3番目の式に代入

$-\sqrt{2}A+B-\sqrt{2}A+D = 0 \implies -2\sqrt{2}A + B + D = 0$

$A-\sqrt{2}B-A+\sqrt{2}D = 0 \implies -\sqrt{2}B + \sqrt{2}D = 0 \implies B = D$

$B+D=1$ より $2B = 1$, つまり $B = \frac{1}{2}$

$D = \frac{1}{2}$

$-2\sqrt{2}A + \frac{1}{2} + \frac{1}{2} = 0 \implies -2\sqrt{2}A + 1 = 0 \implies A = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

$C = -A = -\frac{\sqrt{2}}{4}$

したがって、$A = \frac{\sqrt{2}}{4}, B = \frac{1}{2}, C = -\frac{\sqrt{2}}{4}, D = \frac{1}{2}$

(2) 定積分の計算

$\int_0^1 \frac{dx}{x^4+1} = \int_0^1 \frac{\frac{\sqrt{2}}{4}x + \frac{1}{2}}{x^2+\sqrt{2}x+1} + \frac{-\frac{\sqrt{2}}{4}x + \frac{1}{2}}{x^2-\sqrt{2}x+1} dx$

$= \frac{\sqrt{2}}{4} \int_0^1 \frac{x + \sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{-x + \sqrt{2}}{x^2-\sqrt{2}x+1} dx$

$= \frac{\sqrt{2}}{8} \int_0^1 \frac{2x + \sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{\sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{-2x + \sqrt{2}}{x^2-\sqrt{2}x+1} + \frac{\sqrt{2}}{x^2-\sqrt{2}x+1} dx$

$= \frac{\sqrt{2}}{8} [\ln(x^2+\sqrt{2}x+1) - \ln(x^2-\sqrt{2}x+1)]_0^1 + \frac{1}{4} \int_0^1 \frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}} + \frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}} dx$

$= \frac{\sqrt{2}}{8} [\ln(2+\sqrt{2}) - \ln(2-\sqrt{2})] + \frac{1}{4} [\sqrt{2}\arctan(\sqrt{2}x+1) + \sqrt{2}\arctan(\sqrt{2}x-1)]_0^1$

$= \frac{\sqrt{2}}{8} \ln(\frac{2+\sqrt{2}}{2-\sqrt{2}}) + \frac{\sqrt{2}}{4} [\arctan(1+\sqrt{2}) - \arctan(-1+\sqrt{2})]$

$= \frac{\sqrt{2}}{8} \ln(\frac{(2+\sqrt{2})^2}{4-2}) + \frac{\sqrt{2}}{4} [\arctan(1+\sqrt{2}) - \arctan(\sqrt{2}-1)]$

$= \frac{\sqrt{2}}{8} \ln(\frac{4+4\sqrt{2}+2}{2}) + \frac{\sqrt{2}}{4} [\arctan(1+\sqrt{2}) - \arctan(\sqrt{2}-1)]$

$= \frac{\sqrt{2}}{8} \ln(3+2\sqrt{2}) + \frac{\sqrt{2}}{4} \arctan(\frac{1+\sqrt{2}-\sqrt{2}+1}{1+(1+\sqrt{2})(\sqrt{2}-1)}) = \frac{\sqrt{2}}{8} \ln(3+2\sqrt{2}) + \frac{\sqrt{2}}{4} \arctan(2) = \frac{\pi\sqrt{2}}{8} + \frac{\sqrt{2}}{4}\arctan(-1+2^{.5})-\frac{\sqrt{2}}{8}\ln(\sqrt{2}-1)$

$\int_0^1 \frac{dx}{x^4+1} = \frac{\pi + \ln(1+\sqrt{2})}{2\sqrt{2}}$

$\frac{\sqrt{2}}{4}(\arctan(\sqrt{2}+1) + \arctan(\sqrt{2}-1)) = \frac{\pi\sqrt{2}}{4} = \frac{ \pi+ \ln(1+2^{.5})}{8} * 2^{.5}$

3. 最終的な答え

(1) $A = \frac{\sqrt{2}}{4}, B = \frac{1}{2}, C = -\frac{\sqrt{2}}{4}, D = \frac{1}{2}$

(2) $\frac{\pi\sqrt{2}}{8} + \frac{\sqrt{2}}{4}\arctan(-1+\sqrt{2})-\frac{\sqrt{2}}{8}\ln(\sqrt{2}-1)$

$\frac{\pi\sqrt{2}}{4} = \frac{\pi\sqrt{2}}{4}$