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自己紹介・記録議論

K²+K'² を分母にもつ関数の定積分

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$$\newcommand{BA}[0]{\begin{align*}} \newcommand{BE}[0]{\begin{equation}} \newcommand{bl}[0]{\boldsymbol} \newcommand{BM}[0]{\begin{matrix}} \newcommand{D}[0]{\displaystyle} \newcommand{EA}[0]{\end{align*}} \newcommand{EE}[0]{\end{equation}} \newcommand{EM}[0]{\end{matrix}} \newcommand{h}[0]{\boldsymbol{h}} \newcommand{k}[0]{\boldsymbol{k}} \newcommand{L}[0]{\left} \newcommand{l}[0]{\boldsymbol{l}} \newcommand{m}[0]{\boldsymbol{m}} \newcommand{n}[0]{\boldsymbol{n}} \newcommand{R}[0]{\right} \newcommand{vep}[0]{\varepsilon} $$

定義・記法

$\hspace{5pt}$以下，当記事での定義および記法を示します．

$\BA\D \beta_r^{}&=2^{-2r}\binom{2r}{r}\\ K(x)&=\frac{\pi}{2}\sum_{n=0}^\infty \beta_n^2x^{2n}\\ K'(x)&=K\L(\sqrt{1-x^2}\R)\\ \frac{1}{\sqrt{1-2tx+t^2}}&=\sum_{n=0}^\infty t^nP_n^{}(x)\\ \frac{1}{1-2tx+t^2}&=\sum_{n=0}^\infty t^nU_n^{}(x)\\ G&=\sum_{n=0}^\infty \cfrac{(-1)^n}{(2n+1)^2} \EA$

また，分子の$K(x),K'(x)$の数に応じて，積分の"重み"を決定します．すなわち$\D {K(x)}^a{K'(x)}^b$の場合を重み$a+b-1$とします．

使用アプリケーション

$\hspace{5pt}$$\Large\color{Red}{\bf Wolfram}\color{RedOrange}{\bf Alpha}$
$\hspace{8pt}$PARI/GP

(数値実験 １)　重み$0$

$x^n$

$\BA\D \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{0}\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{2}\,dx&=\frac{5}{64}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{4}\,dx&=\frac{11}{256}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{6}\,dx&=\frac{469}{16384}\\ \EA$

$P_n^{}(x)$

$\BA\D \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,P_{0}^{}(x)\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,P_{2}^{}(x)\,dx&=-\frac{1}{128}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,P_{4}^{}(x)\,dx&=-\frac{23}{2048}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,P_{6}^{}(x)\,dx&=\frac{499}{262144}\\ \EA$

$U_n^{}(x)$

$\BA\D \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{0}^{}(x)\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{4}^{}(x)\,dx&=0\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{8}^{}(x)\,dx&=0\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{12}^{}(x)\,dx&=0\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{2}^{}(x)\,dx&=\frac{1}{16}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{6}^{}(x)\,dx&=-\frac{5}{256}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{10}^{}(x)\,dx&=-\frac{11}{1024}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{14}^{}(x)\,dx&=\frac{469}{65536}\\ \EA$

★ 

$\BA\D \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{4n}^{}(x)\,dx&=\frac{1}{4}\,\delta_{0,n}^{}\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{4n}\,dx&=4(-1)^n\int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{8n+2}^{}(x)\,dx\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{4n+2}\,dx&=4(-1)^{n+1}\int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,U_{8n+6}^{}(x)\,dx\\ \int_0^1 \cfrac{K'(x)}{K(x)^2+K'(x)^2}\,x^{2n}\,dx&=[x^{2n}]\frac{1}{x^2}\L(1-\frac{\pi}{2}\frac{1}{K(x)}\R) \EA$

(数値実験 ２)　重み$1$

$x^n$

$\BA\D \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{1}\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{3}\,dx&=\frac{19}{128}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{5}\,dx&=\frac{41}{384}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{7}\,dx&=\frac{5491}{65536}\\ \EA$

$\BA\D \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{1}\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{3}\,dx&=\frac{13}{128}\\ \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{5}\,dx&=\frac{23}{384}\\ \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{7}\,dx&=\frac{2701}{65536}\\ \EA$

$P_n^{}(x)$

$\BA\D \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{1}^{}(x)\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{3}^{}(x)\,dx&=-\frac{31}{256}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{5}^{}(x)\,dx&=\frac{53}{1024}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{7}^{}(x)\,dx&=-\frac{38327}{1048576}\\ \EA$

$\BA\D \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{1}^{}(x)\,dx&=\frac{1}{4}\\ \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{3}^{}(x)\,dx&=-\frac{1}{256}\\ \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{5}^{}(x)\,dx&=\frac{11}{1024}\\ \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{7}^{}(x)\,dx&=-\frac{2633}{1048576}\\ \EA$

$U_n^{}(x)$

$\BA\D \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{1}^{}(x)\,dx&=\frac{1}{2}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{5}^{}(x)\,dx&=\frac{1}{6}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{9}^{}(x)\,dx&=\frac{1}{10}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{13}^{}(x)\,dx&=\frac{1}{14}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{3}^{}(x)\,dx&=\frac{3}{16}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{7}^{}(x)\,dx&=\frac{51}{512}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{11}^{}(x)\,dx&=\frac{35}{512}\\ \int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{15}^{}(x)\,dx&=\frac{13683}{262144}\\ \EA$

★ 

$\BA\D \int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\L(x^{2n+1}-4U_{4n+3}^{}(x)\R)\,dx&=\frac{1}{n+1} \EA$

(数値実験 ３)　重み$2$

$x^n$

$\BA\D \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,x^{0}\,dx&=\beta_0^2\zeta(2)\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,x^{2}\,dx&=\beta_1^2\zeta(2)-\frac{1}{8}\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,x^{4}\,dx&=\beta_2^2\zeta(2)-\frac{49}{512}\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,x^{6}\,dx&=\beta_3^2\zeta(2)-\frac{1405}{18432}\\ \EA$

$P_n^{}(x)$

$\BA\D \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,P_{0}^{}(x)\,dx&=\frac{1}{2}\beta_0^3\zeta(2)\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,P_{2}^{}(x)\,dx&=\frac{3}{16}-\frac{1}{2}\beta_1^3\zeta(2)\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,P_{4}^{}(x)\,dx&=\frac{1}{2}\beta_2^3\zeta(2)-\frac{205}{4096}\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,P_{6}^{}(x)\,dx&=\frac{3605}{98304}-\frac{1}{2}\beta_3^3\zeta(2)\\ \EA$

$\BA\D \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,P_{0}^{}(x)\,dx&=\beta_0^3\zeta(2)\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,P_{2}^{}(x)\,dx&=-\L(\beta_1^3\zeta(2)+\frac{3}{16}\R)\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,P_{4}^{}(x)\,dx&=\beta_2^3\zeta(2)+\frac{205}{4096}\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,P_{6}^{}(x)\,dx&=-\L(\beta_3^3\zeta(2)+\frac{3605}{98304}\R)\\ \EA$

$U_n^{}(x)$

$\BA\D \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{0}^{}(x)\,dx&=\frac{1}{2}\beta_0^2\zeta(2)\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{4}^{}(x)\,dx&=\frac{1}{2}\beta_1^2\zeta(2)+\frac{1}{32}\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{8}^{}(x)\,dx&=\frac{1}{2}\beta_2^2\zeta(2)+\frac{49}{2048}\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{12}^{}(x)\,dx&=\frac{1}{2}\beta_3^2\zeta(2)+\frac{1405}{73728}\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,U_{0}^{}(x)\,dx&=\beta_0^2\zeta(2)\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,U_{4}^{}(x)\,dx&=\beta_1^2\zeta(2)-\frac{1}{32}\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,U_{8}^{}(x)\,dx&=\beta_2^2\zeta(2)-\frac{49}{2048}\\ \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\,U_{12}^{}(x)\,dx&=\beta_3^2\zeta(2)-\frac{1405}{73728}\\ \EA$

$\BA\D \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{2}^{}(x)\,dx&=\frac{1}{2}\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{6}^{}(x)\,dx&=\frac{2}{9}\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{10}^{}(x)\,dx&=\frac{32}{225}\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{14}^{}(x)\,dx&=\frac{128}{1225}\\ \EA$

★ 

$\BA\D \int_0^1 \cfrac{K'(x)^3}{K(x)^2+K'(x)^2}\L(4U_{4n}^{}(x)-x^{2n}\R)\,dx&=\cfrac{\pi^2}{2}\beta_n^2\\ \int_0^1 \cfrac{K(x)^2K'(x)}{K(x)^2+K'(x)^2}\,U_{4n+2}^{}(x)\,dx&=\cfrac{2}{(2n+1)^2\beta_n^2}\\ \EA$

(数値実験 ４)　重み$3$

$U_n^{}(x)$

$\BA\D\\ \int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{1}^{}(x)\,dx&=\frac{3}{2}\zeta(3)\\ \int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{5}^{}(x)\,dx&=\frac{3}{2}\frac{1}{2}\zeta(3)-\frac{3}{16}\\ \int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{9}^{}(x)\,dx&=\frac{3}{2}\frac{11}{32}\zeta(3)-\frac{363}{2048}\\ \int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{13}^{}(x)\,dx&=\frac{3}{2}\frac{17}{64}\zeta(3)-\frac{5833}{36864}\\ \EA$

その他

重み$0$

$\BA\D \int_0^1 \cfrac{K'(x)^4K(x)}{\L(K(x)^2+K'(x)^2\R)^2}\,dx&=\frac{9}{16}\zeta(3)\\ \int_0^1 \cfrac{K'(x)^6K(x)}{\L(K(x)^2+K'(x)^2\R)^3}\,dx&=\frac{45}{64}\zeta(3)-\frac{93}{256}\zeta(5)\\ \int_0^1 \cfrac{K'(x)^{8}K(x)}{\L(K(x)^2+K'(x)^2\R)^4}\,dx&=\frac{105}{128}\zeta(3)-\frac{159}{256}\zeta(5)\\ \EA$

$\BA\D \int_0^1 \cfrac{K'(x)^4}{K(x)\L(K(x)^2+K'(x)^2\R)}\,dx&=\frac{21}{8}\zeta(3)\\ \int_0^1 \cfrac{K'(x)^6}{K(x)\L(K(x)^2+K'(x)^2\R)^2}\,dx&=\frac{33}{16}\zeta(3)\\ \int_0^1 \cfrac{K'(x)^8}{K(x)\L(K(x)^2+K'(x)^2\R)^3}\,dx&=\frac{87}{64}\zeta(3)+\frac{93}{256}\zeta(5)\\ \int_0^1 \cfrac{K'(x)^{10}}{K(x)\L(K(x)^2+K'(x)^2\R)^4}\,dx&=\frac{69}{128}\zeta(3)+\frac{63}{64}\zeta(5)\\ \int_0^1 \cfrac{K'(x)^{12}}{K(x)\L(K(x)^2+K'(x)^2\R)^5}\,dx&=-\frac{393}{1024}\zeta(3)+\frac{8547}{4096}\zeta(5)-\frac{9525}{32768}\zeta(7) \EA$

$\BA\D \int_0^1 \cfrac{K'(x)^4}{K(x)^3}\,dx&=\frac{93}{4}\zeta(5)\\ \int_0^1 \cfrac{K'(x)^6}{K(x)^3\L(K(x)^2+K'(x)^2\R)}\,dx&=-\frac{21}{8}\zeta(3)+\frac{93}{4}\zeta(5)\\ \EA$

重み$3$

$\BA\D \int_0^1 \cfrac{3K(x)^4-4K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}x\,dx&=0 \EA$

重み$4$

$\BA\D \int_0^1 \cfrac{K(x)^5+K(x)^2K'(x)^3-K(x)K'(x)^4}{K(x)^2+K'(x)^2}\,dx&=0\\ \int_0^1 \cfrac{K(x)^4K'(x)}{K(x)^2+K'(x)^2}\,dx-\int_0^1 \cfrac{K(x)^5-K(x)^4K'(x)}{K(x)^2+K'(x)^2}x\,dx-\int_0^1 \cfrac{K(x)^3K'(x)^2-K(x)K'(x)^4}{K(x)^2+K'(x)^2}x^2\,dx&=0\\ \int_0^1 \cfrac{K(x)^5+2K(x)^3K'(x)^2+K(x)K'(x)^4-K'(x)^5}{K(x)^2+K'(x)^2}\,dx&=0\\ \int_0^1 \cfrac{12K(x)^5-2K'(x)^5}{K(x)^2+K'(x)^2}\,dx-\int_0^1 \cfrac{3K(x)^4K'(x)}{K(x)^2+K'(x)^2}x\,dx&=0\\ \int_0^1 \cfrac{32(x)^5-5K'(x)^5}{K(x)^2+K'(x)^2}\,dx-\int_0^1 \cfrac{8K(x)^4K'(x)}{K(x)^2+K'(x)^2}x\,dx&=0\\ \int_0^1 \cfrac{11K'(x)^5}{K(x)^2+K'(x)^2}\,dx-\int_0^1 \cfrac{96K(x)^4K'(x)+39K'(x)^5}{K(x)^2+K'(x)^2}x^2\,dx&=0 \EA$

$\BA\D \int_0^1 \cfrac{K'(x)^6}{K(x)\L(K(x)^2+K'(x)^2\R)}\,dx&=\frac{\Gamma\L(\frac{1}{4}\R)^{8}}{160\pi^2} \EA$

$\D\lambda K(x)^2+ \mu K'(x)^2$

重み$2$

$\BA\D \int_0^1 \cfrac{K(x)^3}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=6G-\frac{7}{2}\zeta(3)\\ \int_0^1 \cfrac{K(x)K'(x)^2}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=-16G+14\zeta(3)\\ \EA$

$\BA\D \int_0^1 \cfrac{K(x)K'(x)^2}{K(x)^2+\frac{1}{3^2}K'(x)^2}\,dx&=\frac{19}{8}\zeta(3)\\ \int_0^1 \cfrac{K(x)K'(x)^2}{\frac{1}{3^2}K(x)^2+K'(x)^2}\,dx&=\frac{11}{8}\zeta(3) \EA$

重み$4$

$\BA\D \int_0^1 \cfrac{K(x)^5}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+\frac{3}{32}\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \int_0^1 \cfrac{K(x)^4K'(x)}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=\frac{5}{2}\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx-\frac{5}{24}\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \int_0^1 \cfrac{K(x)^3K'(x)^2}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=-4\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+\frac{9}{8}\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \int_0^1 \cfrac{K(x)^2K'(x)^3}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=-10\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+\frac{5}{2}\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \int_0^1 \cfrac{K(x)K'(x)^4}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=16\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx-2\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \int_0^1 \cfrac{K'(x)^5}{K(x)^2+\frac{1}{2^2}K'(x)^2}\,dx&=40\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx-5\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \EA$

$\BA\D 6\int_0^1 \cfrac{K(x)^5}{\frac{1}{2^2}K(x)^2+K'(x)^2}\,dx+12\int_0^1 \cfrac{K(x)^4K'(x)}{\frac{1}{2^2}K(x)^2+K'(x)^2}\,dx&=-1536\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+305\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ 2\int_0^1 \cfrac{K(x)^3K'(x)^2}{\frac{1}{2^2}K(x)^2+K'(x)^2}\,dx+4\int_0^1 \cfrac{K(x)^2K'(x)^3}{\frac{1}{2^2}K(x)^2+K'(x)^2}\,dx&=128\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx-23\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \int_0^1 \cfrac{K(x)K'(x)^4}{\frac{1}{2^2}K(x)^2+K'(x)^2}\,dx+2\int_0^1 \cfrac{K'(x)^5}{\frac{1}{2^2}K(x)^2+K'(x)^2}\,dx&=-16\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+6\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \EA$

$\BA\D 729\int_0^1 \cfrac{K(x)^5}{K(x)^2+\frac{1}{3^2}K'(x)^2}\,dx&=-731\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+380\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ 648\int_0^1 \cfrac{K(x)^3K'(x)^2}{K(x)^2+\frac{1}{3^2}K'(x)^2}\,dx&=5848\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx-853\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ 36\int_0^1 \cfrac{K(x)K'(x)^4}{K(x)^2+\frac{1}{3^2}K'(x)^2}\,dx&=-2924\int_0^1 \cfrac{K(x)^5}{K(x)^2+K'(x)^2}\,dx+629\int_0^1 \cfrac{K'(x)^5}{K(x)^2+K'(x)^2}\,dx\\ \EA$

$\BA\D \EA$

投稿日：1月21日
更新日：1月31日

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K²+K'² を分母にもつ関数の定積分