11
自己紹介・記録議論
K²+K'² を分母にもつ関数の定積分
495
0
この著者は初心者として投稿しています。間違いや考慮が足りていない点が含まれている可能性が高いです。見つけたらコメント欄で優しく指摘してあげましょう。
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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$1$
$x^n$
$\BA\D
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{1}\,dx&=\frac{1}{4}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{3}\,dx&=\frac{19}{128}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{5}\,dx&=\frac{41}{384}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,x^{7}\,dx&=\frac{5491}{65536}\\
\EA$
$\BA\D
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{1}\,dx&=\frac{1}{4}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{3}\,dx&=\frac{13}{128}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{5}\,dx&=\frac{23}{384}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,x^{7}\,dx&=\frac{2701}{65536}\\
\EA$
$P_n^{}(x)$
$\BA\D
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{1}^{}(x)\,dx&=\frac{1}{4}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{3}^{}(x)\,dx&=-\frac{31}{256}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{5}^{}(x)\,dx&=\frac{53}{1024}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,P_{7}^{}(x)\,dx&=-\frac{38327}{1048576}\\
\EA$
$\BA\D
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{1}^{}(x)\,dx&=\frac{1}{4}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{3}^{}(x)\,dx&=-\frac{1}{256}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{5}^{}(x)\,dx&=\frac{11}{1024}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\,P_{7}^{}(x)\,dx&=-\frac{2633}{1048576}\\
\EA$
$U_n^{}(x)$
$\BA\D
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{1}^{}(x)\,dx&=\frac{1}{2}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{5}^{}(x)\,dx&=\frac{1}{6}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{9}^{}(x)\,dx&=\frac{1}{10}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{13}^{}(x)\,dx&=\frac{1}{14}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{3}^{}(x)\,dx&=\frac{3}{16}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{7}^{}(x)\,dx&=\frac{51}{512}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{11}^{}(x)\,dx&=\frac{35}{512}\\
\int_0^1 \cfrac{K(x)^2}{K(x)^2+K'(x)^2}\,U_{15}^{}(x)\,dx&=\frac{13683}{262144}\\
\EA$
★
$\BA\D
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\L(x^{2n+1}-4U_{4n+3}^{}(x)\R)\,dx&=\frac{1}{n+1}\\
\int_0^1 \cfrac{K'(x)^2}{K(x)^2+K'(x)^2}\frac{x}{1-z^2x^2}\,dx&=\frac{1}{z^2}\L(\ln\frac{4}{z}-\frac{\pi}{2}\frac{K'(z)}{K(z)}\R)
\EA$
(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$3$
$U_n^{}(x)$
$\BA\D\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{1}^{}(x)\,dx&=\frac{3}{2}\zeta(3)\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{5}^{}(x)\,dx&=\frac{3}{2}\frac{1}{2}\zeta(3)-\frac{3}{16}\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{9}^{}(x)\,dx&=\frac{3}{2}\frac{11}{32}\zeta(3)-\frac{363}{2048}\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,U_{13}^{}(x)\,dx&=\frac{3}{2}\frac{17}{64}\zeta(3)-\frac{5833}{36864}\\
\EA$
$\BA\D\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,P_{1}^{}(x)\,dx&=\frac{3}{4}\zeta(3)\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,P_{3}^{}(x)\,dx&=-\frac{3}{4}\frac{1}{4}\zeta(3)\\
\int_0^1 \cfrac{K(x)^2K'(x)^2}{K(x)^2+K'(x)^2}\,P_{5}^{}(x)\,dx&=\frac{3}{4}\frac{53}{256}\zeta(3)-\frac{189}{4096}\\
\EA$
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：5月24日
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