KBHM

EllipticKを含む14個の積分

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命題.1
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1+x}}dx=\frac13\int_0^1\frac{K(x)}{\sqrt{1-x}}dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{48\sqrt{2}\pi} \end{align}

解法

先ず,
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1-x}}dx=\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_n^2\int_0^1\frac{x^{2n}}{\sqrt{1-x}}dx=\pi\sum_{n=0}^{\infty}\frac{\beta_n^2}{(4n+1)\beta_{2n}}=\pi\ _3F_2\left[\begin{matrix} \frac12,\frac12,\frac12\\ \frac34,\frac54 \end{matrix};1 \right] \end{align}
ですが Whippleの和公式 から,
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1-x}}dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{16\sqrt2 \pi} \end{align}
となります.また,Landen Transformから,
\begin{eqnarray} \int_0^1\frac{K(x)}{\sqrt{1+x}}dx&=&\int_0^1K\left(\frac{2x}{1+x^2}\right)\frac{\sqrt{1+x^2}}{1+x}\frac{2(1-x^2)}{(1+x^2)^2}dx\ \left(x\to\frac{2x}{1+x^2}\right)\\ &=&2\int_0^1\frac{K(x^2)}{\sqrt{1+x^2}}(1-x)dx=\int_0^1\frac{K(x)}{\sqrt{1+x}}\left(\frac1{\sqrt{x}}-1\right)dx\\ \int_0^1\frac{K(x)}{\sqrt{1-x}}dx&=&\int_0^1\frac{K(x)}{\sqrt{1+x}}\left(\frac{1}{\sqrt{x}}+1\right)dx \end{eqnarray}
なので,
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1-x}}dx=3\int_0^1\frac{K(x)}{\sqrt{1+x}}dx \end{align}
がわかります.

命題.2
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1-x^2}}dx=\frac{\pi^3}{4\Gamma(\frac34)^4}=\frac{\Gamma(\frac14)^4}{16\pi} \end{align}

解法

\begin{eqnarray} \int_0^1\frac{K(x)}{\sqrt{1-x^2}}dx&=&\frac{\pi^2}{4}\sum_{n=0}^{\infty}\beta_n^3\\ &=&\frac{\pi^2}{4}\int_0^1\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^2P_{2n}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}\beta_nP_{2n}(2x-1)\right)dx\\ &=&\frac{\pi^2}{4}\int_0^1\frac{1}{\pi}\frac1{(x(1-x))^{\frac12}}\cdot\frac{\Gamma(\frac14)^2}{4\pi^{\frac32}}\frac1{(x(1-x))^{\frac14}}dx\\ &=&\frac{\Gamma(\frac14)^2}{16\sqrt{\pi}}\int_0^1x^{-\frac34}(1-x)^{-\frac34}dx\\ &=&\frac{\Gamma(\frac14)^4}{16\pi} \end{eqnarray}

命題.3
\begin{eqnarray} \int_0^1\frac{K(x)^2}{\sqrt{1-x^2}}dx&=&2\int_0^1K(x)K'(x)dx=\frac{\pi^3}{4}\sum_{n=0}^{\infty}\beta_n^4 \end{eqnarray}

解法

\begin{eqnarray} \int_0^1\frac{K(x)^2}{\sqrt{1-x^2}}dx&=&\frac4\pi\int_0^1\frac{dx}{\sqrt{1-x^2}}\int_0^1tK(t)K'(t)\frac{dt}{1-x^2t^2}\\ &=&\frac{4}{\pi}\int_0^1tK(t)K'(t)dt\int_0^1\frac{dx}{(1-x^2t^2)\sqrt{1-x^2}}\\ &=&2\int_0^1tK(t)K'(t)\frac{dt}{\sqrt{1-t^2}}=2\int_0^1K(x)K'(x)dx\\ &=&\frac{\pi^2}{4}\int_0^1\kappa(x)\kappa(1-x)\cdot\frac{1}{\sqrt{1-x}}dx\\ &=&\frac{\pi^2}{4}\int_0^1\left(\frac{\pi}{2}\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(2x-1)\right)\cdot\left(2\sum_{n=0}^{\infty}P_{n}(2x-1)\right)dx\\ &=&\frac{\pi^3}{4}\sum_{n=0}^{\infty}\beta_n^4 \end{eqnarray}

命題.4
\begin{eqnarray} \int_0^1\frac{xK(x)^2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}dx&=&\frac{\pi^4}{4}\sum_{n=0}^{\infty}\beta_n^4 \end{eqnarray}

解法

\begin{eqnarray} \int_0^1\frac{xK(x)^2}{\sqrt{1-x^2}}\ln\frac1{1-x^2}dx&=&2\int_0^1\frac{4(1-x^2)}{(1+x^2)^3}\frac{1+x^2}{1-x^2}K\left(\frac{2x}{1+x^2}\right)^2\ln\frac{1+x^2}{1-x^2}dx\\ &=&8\int_0^1K(x)^2\tanh^{-1}xdx\\ &=&8\int_0^1K(x)\cdot\left(\frac{\pi}2\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n\beta_n)^2}\sum_{k=0}^{n-1}(4k+1)\beta_k^4\right)dx\\ &=&4\pi\sum_{n=1}^{\infty}\frac{1}{(2n\beta_n)^4}\left(\sum_{k=0}^{n-1}\beta_k^2\right)\sum_{k=0}^{n-1}(4k+1)\beta_k^4\\ &=&2\pi^2\sum_{n=1}^{\infty}\frac1{(2n\beta_n)^2}\sum_{k=0}^{n-1}(4k+1)\beta_k^4\sum_{m=0}^{\infty}\beta_m^2\left(\frac{1}{n+m}+\frac{1}{n-m-\frac12}\right)\\ &=&2\pi^2\sum_{m=0}^{\infty}\beta_m^2\sum_{n=0}^{\infty}\frac1{(2n\beta_n)^2}\left(\frac{1}{n+m}-\frac{1}{m-n+\frac12}\right)\sum_{k=0}^{n-1}(4k+1)\beta_k^4\\ &=&\frac{\pi^4}{4}\sum_{m=0}^{\infty}\beta_m^4 \end{eqnarray}

命題.5
\begin{eqnarray} \int_0^1K(x)^2dx&=&\frac12\int_0^1K'(x)^2dx=\frac{\pi^4}{32}\sum_{n=0}^{\infty}(4n+1)\beta_n^6 \end{eqnarray}

解法

\begin{eqnarray} \int_0^1K(x)^2dx&=&\int_0^12xK(x^2)^2dx\\ &=&\int_0^1\frac{2x}{(1+x^2)^2}K\left(\frac{2x}{1+x^2}\right)^2dx\\ &=&\frac12\int_0^1\frac{x}{\sqrt{1-x^2}}K(x)^2dx=\frac12\int_0^1K'(x)^2dx\\ &=&\frac12\int_0^1\left(\frac{\pi^2}{4}\sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3P_{2n}(x)\right)^2dx\\ &=&\frac{\pi^4}{32}\sum_{n=0}^{\infty}(4n+1)\beta_n^6 \end{eqnarray}

命題.6
\begin{eqnarray} \int_0^1xK(x)^3dx&=&\frac{6}{5}\int_0^1xK(x)K'(x)^2dx=\frac{\Gamma(\frac14)^8}{640\pi^2} \end{eqnarray}

楕円積分の3乗の積分について を参照

余談

この論文 に書かれている計算を追ってみましょう.
p10からの内容を見てみましょう.
\begin{eqnarray} P_{v}(x):&=&\ _2F_1\left[\begin{matrix} -v,v+1\\ 1 \end{matrix};\frac{1-x}{2} \right]\\ T_{u,v}:&=&\int_{-1}^1P_u(x)P_{v}(x)P_{v}(-x)dx \end{eqnarray}
に対して,
\begin{align} I(z):=\frac{1}{\sin^2\pi z}\left(T_{z,z}-\frac{1+2\cos(\pi z)}{3}\frac{\pi\Gamma(\frac{1+z}{2})\Gamma(1+\frac{3z}{2})}{\Gamma(\frac{1-z}{2})^2\Gamma(1+\frac{z}{2})^3\Gamma(\frac{3+3z}{2})}\right) \end{align}
を定義し,複素積分を考えようとしています.
\begin{align} I(v)=\oint_C\frac{I(z)}{z-v}\frac{dz}{2\pi i}=0\ ...(1) \end{align}
が示せれば,
\begin{align} \int_{-1}^1P_{v}(x)^2P_{v}(-x)dx=\frac{1+2\cos(\pi v)}{3}\frac{\pi\Gamma(\frac{1+v}{2})\Gamma(1+\frac{3v}{2})}{\Gamma(\frac{1-v}{2})^2\Gamma(1+\frac{v}{2})^3\Gamma(\frac{3+3v}{2})} \end{align}
となり,$v\to-\frac12$で,
\begin{align} \int_{0}^1xK(x)K'(x)^2dx=\frac{\Gamma(\frac14)^8}{768\pi^2} \end{align}
がわかります.
この論文では$(1)$を10p~12pにかけて示しているようで,その計算方法はかなり大変そうです.

命題.7
\begin{eqnarray} \int_0^1xK(x)^2dx&=&\frac74\zeta(3) \end{eqnarray}

解法

\begin{eqnarray} \int_0^1xK(x)^2dx&=&\frac{\pi^2}{8}\int_0^1\kappa(x)^2dx\\ &=&\frac{\pi^2}{8}\int_0^1\left(\frac4\pi\sum_{n=0}^{\infty}\frac{1}{2n+1}P_n(2x-1)\right)^2dx\\ &=&2\sum_{n=0}^{\infty}\frac1{(2n+1)^3}=\frac74\zeta(3) \end{eqnarray}

命題.8
\begin{eqnarray} \int_0^1xK(x)K'(x)dx&=&\frac{\pi^3}{16} \end{eqnarray}

解法

\begin{eqnarray} \int_0^1xK(x)K'(x)dx&=&\frac{\pi^2}{8}\int_0^1\kappa(x)\kappa(1-x)dx\\ &=&\frac{\pi^2}{8}\int_0^1\left(\frac4\pi\sum_{n=0}^{\infty}\frac{1}{2n+1}P_n(2x-1)\right)\cdot\left(\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}P_n(2x-1)\right)dx\\ &=&2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{16} \end{eqnarray}

命題.9
\begin{eqnarray} \int_0^1x(2x^2-1)K(x)^4dx&=&\frac{93}{16}\zeta(5) \end{eqnarray}

解法

\begin{align} \frac{d}{dx}\frac{\pi K'(x)}{K(x)}=-\frac{\pi^2}{2}\frac{1}{x(1-x^2)K(x)^2} \end{align}
より,
\begin{align} \int_0^1F\left(\frac{\pi K'(x)}{K(x)}\right)\left(\frac{\pi^2}{2}\frac{1}{x(1-x^2)K(x)^2}\right)dx=\int_0^{\infty}F(x)dx \end{align}
となります. sn(u,k) などの Lambert Series まとめ の,
\begin{eqnarray} \mathrm{sd}^2\left(K(x)z,x\right)&=&\frac1{x^2(1-x^2)K(x)^2}\left(K(x)E(x)-(1-x^2)K(x)+\pi^2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n\cos(\pi nz)}{\sinh(n\pi\frac{K'(x)}{K(x)})}\right)\\ &=&(K(x)z)^2+\frac{2x^2-1}{3}\left(K(x)z\right)^4+... \end{eqnarray}
より,$z^4$の係数比較から,
\begin{eqnarray} \frac{\pi^2}{2}\frac1{x(1-x^2)K(x)^2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{\sinh(n\pi\frac{K'(x)}{K(x)})}[z^4]\cos(\pi nz)&=&\frac{1}{6}x(2x^2-1)K(x)^4\\ \frac{\pi^2}{2}\frac1{x(1-x^2)K(x)^2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{\sinh(n\pi\frac{K'(x)}{K(x)})}\frac{\pi^4n^4}{24}&=& \end{eqnarray}
なので,
\begin{eqnarray} \int_0^1x(2x^2-1)K(x)^sK'(x)^{4-s}dx&=&\frac1{\pi^{4-s}}\int_0^1x(2x^2-1)K(x)^4\left(\pi\frac{K'(x)}{K(x)}\right)^{4-s}dx\\ &=&\frac1{\pi^{4-s}}\int_0^1\left(\pi\frac{K'(x)}{K(x)}\right)^{4-s}\frac{\pi^2}{2}\frac1{x(1-x^2)K(x)^2}\frac{\pi^4}{4}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n^5}{\sinh(n\pi\frac{K'(x)}{K(x)})}dx\\ &=&\frac{\pi^s}{4}\int_0^{\infty}x^{4-s}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n^5}{\sinh(nx)}dx...(1)\\ &=&\sum_{n=1}^{\infty}(-1)^{n-1}n^5\int_0^{\infty}\frac{x^{4-s}}{\sinh(nx)}dx...(2)\\ \end{eqnarray}
とできます.ここで注意すべきなのは積分区間が$(0,\infty)$であることです.
通常,
\begin{align} \int_0^{\infty}\frac{x^{s-1}}{\sinh(nx)}dx=2\sum_{m=0}^{\infty}\int_0^{\infty}x^{s-1}e^{-(2m+1)nx}dx=2\frac{\Gamma(s)}{n^s}\sum_{m=0}^{\infty}\frac1{(2m+1)^s} \end{align}
と計算できますが今回はこの変形を用いることはできません.
$(1)\to(2)$は,
\begin{align} \lim_{\varepsilon\to+0}\int_{\varepsilon}^{\infty}\lim_N\sum^N dx=\lim_{\varepsilon\to+0}\lim_N\sum^N\int_{\varepsilon}^{\infty}dx \end{align}
であり,
\begin{align} \lim_{\varepsilon\to+0}\int_{\varepsilon}^{\infty}\lim_N\sum^N dx=\lim_N\sum^N\lim_{\varepsilon\to+0}\int_{\varepsilon}^{\infty}dx \end{align}
ではないからです.次の変形が正しいです.
\begin{eqnarray} \lim_{\varepsilon\to+0}\int_{\varepsilon}^{\infty}x^{4-s}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n^5}{\sinh(nx)}dx&=&\lim_{\varepsilon\to+0}\sum_{n=1}^{\infty}(-1)^{n-1}n^5\int_{\varepsilon}^{\infty}\frac{x^{4-s}}{\sinh(nx)}dx\\ &=&\lim_{\varepsilon\to+0}\sum_{n=1}^{\infty}(-1)^{n-1}n^5\sum_{m=0}^{\infty}2\int_{\varepsilon}^{\infty}x^{4-s}e^{-(2m+1)nx}dx\\ &=&\lim_{\varepsilon\to+0}\sum_{n=1}^{\infty}(-1)^{n-1}n^5\sum_{m=0}^{\infty}2e^{-(2m+1)n\varepsilon}\int_{0}^{\infty}(x+\varepsilon)^{4-s}e^{-(2m+1)nx}dx\\ \end{eqnarray}
$s=0,1,2,3$のとき,
\begin{eqnarray} \lim_{\varepsilon\to+0}\int_{\varepsilon}^{\infty}x^{4-s}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n^5}{\sinh(nx)}dx&=&\lim_{\varepsilon\to+0}\sum_{n=1}^{\infty}(-1)^{n-1}n^5\sum_{m=0}^{\infty}2e^{-(2m+1)n\varepsilon}\int_{0}^{\infty}\left(x^{4-s}+O(\varepsilon)\right)e^{-(2m+1)nx}dx\\ &=&\lim_{\varepsilon\to+0}\sum_{n=1}^{\infty}(-1)^{n-1}n^5\sum_{m=0}^{\infty}2e^{-(2m+1)n\varepsilon}\int_{0}^{\infty}x^{4-s}e^{-(2m+1)nx}dx\\ &=&\lim_{\varepsilon\to+0}\sum_{n=1}^{\infty}(-1)^{n-1}n^5\sum_{m=0}^{\infty}2e^{-(2m+1)n\varepsilon}\frac{\Gamma(5-s)}{(2m+1)^{5-s}n^{5-s}}\\ &=&\lim_{\varepsilon\to+0}2\sum_{n=1}^{\infty}(-1)^{n-1}n^se^{-n\varepsilon}\sum_{m=0}^{\infty}\frac{\Gamma(5-s)}{(2m+1)^{5-s}}\\ &=&2\left(\lim_{x\to1-0}\sum_{n=1}^{\infty}(-1)^{n-1}n^s x^{n}\right)\cdot\left(1-\frac1{2^{5-s}}\right)\Gamma(5-s)\zeta(5-s) \end{eqnarray}
となります.
また,$4-s\to+0$では,$x^{4-s}\leq(x+\varepsilon)^{4-s}\leq x^{4-s}+\varepsilon^{4-s}$から同様の結果を得ます.

次に,$\mathrm{Re}\ s\gt0$で
\begin{align} \lim_{x\to1-0}\sum_{n=1}^{\infty}(-1)^{n-1}n^{-s}x^{n}=\eta(s)=\left(1-2^{1-s}\right)\zeta(s) \end{align}
であり,
\begin{align} h(-s,x):=\sum_{n=1}^{\infty}(-1)^{n-1}n^sx^n \end{align}
が$ 0 \leq x\lt1$で収束することと,$n^s=e^{s\ln n}$であることから,
$\lim_{x\to1-0}h(s,x)$は解析関数であり,解析接続の一意性により
\begin{align} \lim_{x\to1-0}\sum_{n=1}^{\infty}(-1)^{n-1}n^{-s}x^{n}=\eta(s) \end{align}
となります.
以上から,
\begin{align} \int_0^1x(2x^2-1)K(x)^sK'(x)^{4-s}dx=\frac{\pi^s}{2}(2^{s+1}-1)(1-2^{s-5})\Gamma(5-s)\zeta(5-s)\zeta(-s) \end{align}
であり,
\begin{eqnarray} \int_0^1x(2x^2-1)K'(x)^4&=&-\frac{93}{16}\zeta(5)\\ \int_0^1x(2x^2-1)K(x)K'(x)^3dx&=&-\frac{45}{64}\pi\zeta(4)\\ \int_0^1x(2x^2-1)K(x)^2K'(x)^2dx&=&0\\ \int_0^1x(2x^2-1)K(x)^3K'(x)dx&=&\frac{3}{64}\pi^3\zeta(2)\\ \int_0^1x(2x^2-1)K(x)^4dx&=&\lim_{s\to0}\frac{31}{4}\pi^4\zeta(1-s)\zeta(-4-s)=\lim_{s\to0}\frac{31}{4}\pi^4\frac{\zeta(s)}{2^s\pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)}2^{-s-4}\pi^{-s-5}\sin(\frac{-\pi s}{2})\Gamma(5+s)\zeta(5+s)=\frac{93}{16}\zeta(5) \end{eqnarray}
となります.

命題.10
\begin{eqnarray} \int_0^1\frac{K'(x)}{\sqrt{1-x}}dx&=&\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{32\pi}+\frac12\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2} \end{eqnarray}

解法

\begin{eqnarray} \int_0^1\frac{K'(x)}{\sqrt{1-x}}dx&=&\sum_{n=0}^{\infty}\beta_n\int_0^1x^nK'(x)dx\\ &=&\frac{\pi^2}{4}\sum_{n=0}^{\infty}\beta_{2n}\beta_n^2+\frac12\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2}\\ &=&\frac{\pi^2}{4}\ _3F_2\left[\begin{matrix} \frac12,\frac14,\frac34\\1,1 \end{matrix};1\right]+\frac12\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2}\\ &=&\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{32\pi}+\frac12\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2} \end{eqnarray}
最後にWhippleの和公式を用いました.

命題.11
\begin{eqnarray} \int_0^1\frac{\kappa(x)^2}{(x(1-x))^{\frac13}}dx&=&\frac{\sqrt{3}\Gamma(\frac13)^9}{2^{3}2^{\frac13}\pi^5} \end{eqnarray}

未解決

\begin{eqnarray} \int_0^1\frac{\kappa(x)^2}{(x(1-x))^{\frac13}}dx&=&\frac1{2^{\frac43}}\int_0^1\frac{x^{-\frac13}}{\sqrt{1-x}}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2+\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx\\ &=&\frac1{2^{\frac43}}\int_0^1\frac{x^{-\frac13}}{\sqrt{1-x}}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx+\frac1{2^{\frac{1}{3}}}\int_0^1\frac{x^{-\frac13}}{\sqrt{1-x}}\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2dx\\ &=&\frac1{2^{\frac43}}\sum_{n\geq0}\beta_n\int_0^1x^{n-\frac13}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx+\frac1{2^{\frac13}}\sum_{n\geq0}\beta_n^3\int_0^1x^{n-\frac13}(1-x)^{-\frac12}dx\\ &=&\frac{\Gamma(\frac23)^3}{2^{\frac13}\pi^{\frac12}\Gamma(\frac76)^3}\ _4F_3\left[\begin{matrix} \frac12,\frac23,\frac23,\frac23\\ \frac76,\frac76,\frac76 \end{matrix};1 \right]+\frac{\pi^{\frac12}\Gamma(\frac23)}{2^{\frac13}\Gamma(\frac76)}\ _4F_3\left[\begin{matrix} \frac12,\frac12,\frac12,\frac23\\ 1,1,\frac76 \end{matrix};1 \right] \end{eqnarray}
となりますがここから先はよくわかりません.$\kappa(x)^2$を含む積分では他にも特殊な計算方法があるのでしょうか.

命題.12
\begin{eqnarray} \int_0^1K(x)K'(x)^2dx&=&\frac12\int_0^1K'(x)^3dx=\frac{\Gamma(\frac14)^8}{256\pi^2} \end{eqnarray}

解法

\begin{eqnarray} \int_0^1xK(x)^aK'(x)^{3-a}dx&=&\int_0^1\frac{2x}{1+x^2}K\left(\frac{2x}{1+x^2}\right)^aK'\left(\frac{2x}{1+x^2}\right)^{3-a}\frac{2(1-x^2)}{(1+x^2)^2}dx\\ &=&2\int_0^12x(1-x^2)\left(\frac1{1+x^2}K\left(\frac{2x}{1+x^2}\right)\right)^a\left(\frac{2}{1+x^2}K'\left(\frac{2x}{1+x^2}\right)\right)^{3-a}\frac{dx}{2^{3-a}}\\ &=&\frac1{2^{2-a}}\int_0^12x(1-x^2)K(x^2)^aK'(x^2)^{3-a}dx\\ &=&\frac1{2^{2-a}}\int_0^1(1-x)K(x)^aK'(x)^{3-a}dx\\ \therefore\int_0^1xK(x)^aK'(x)^{3-a}dx&=&\frac1{1+2^{2-a}}\int_0^1K(x)^aK'(x)^{3-a}dx \end{eqnarray}
よって,
\begin{eqnarray} \int_0^1K'(x)^3dx&=&5\int_0^1xK'(x)^3dx=\frac{\Gamma(\frac14)^8}{128\pi^2}\\ \int_0^1K(x)K'(x)^2dx&=&3\int_0^1xK(x)K'(x)^2dx=\frac{\Gamma(\frac14)^8}{256\pi^2} \end{eqnarray}
となります.

命題.13
\begin{eqnarray} \int_0^1\frac{\kappa(x)\kappa(1-x)}{x}\ln\frac1{1-x}dx&=&7\zeta(3) \end{eqnarray}

解法

\begin{eqnarray} \int_0^1\frac{\kappa(x)\kappa(1-x)}{x}\ln\frac1{1-x}dx&=&\int_0^1\kappa(x)\kappa(1-x)\int_0^1\frac{dt}{1-xt}dx\\ &=&\int_0^1\int_0^1\frac{\kappa(x)\kappa(1-x)}{1-xt}dxdt\\ &=&\frac{\pi}{2}\int_0^1\kappa(t)^2dt\\ &=&\frac{4}{\pi}\int_0^1xK(x)^2dx=\frac{7\zeta(3)}{\pi} \end{eqnarray}

命題.14
\begin{eqnarray} \int_0^1\frac{\sin^{-1}x}{x\sqrt{1-x^2}}K(x)dx&=&\frac{7}{2}\zeta(3) \end{eqnarray}

解法

\begin{eqnarray} \int_0^1\frac{\sin^{-1}x}{x\sqrt{1-x^2}}K(x)dx&=&\sum_{n=1}^{\infty}\frac1{2n\beta_n}\int_0^1x^{2n-2}K(x)dx\\ &=&\sum_{n=1}^{\infty}\frac{1}{2n-1}\int_0^1x^{2n-2}\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx\\ &=&\int_0^1\frac{\left(\tanh^{-1}x\right)^2}{x\sqrt{1-x^2}}dx\\ &=&\int_0^{\infty}\frac{x^2}{\sinh x}dx\ (x\to\tanh x)\\ &=&\sum_{n=0}^{\infty}\frac{2\Gamma(3)}{(2n+1)^3}=\frac72\zeta(3) \end{eqnarray}

命題.15
\begin{eqnarray} \int_0^1\frac{\tanh^{-1}x\tanh^{-1}\sqrt{1-x^2}}{x(1-x^2)}dx&=&\frac{\pi^2\ln2}{2} \end{eqnarray}

解法

\begin{eqnarray} \int_0^1\frac{\tanh^{-1}x\tanh^{-1}\sqrt{1-x^2}}{x(1-x^2)}dx&=&\int_{0\lt s\lt x\lt t\lt1}\frac{ds}{1-s^2}\frac{dx}{x(1-x^2)}\frac{dt}{t\sqrt{1-t^2}}\\ &=&\sum_{n,m\geq0}\int_{0\lt s\lt x\lt t\lt1}s^{2m}ds\ x^{2n-1}dx\frac{dt}{t\sqrt{1-t^2}}\\ &=&\sum_{n,m\geq0}\frac{1}{2m+1}\frac1{2m+2n+1}\int_0^1\frac{t^{2n+2m}}{\sqrt{1-t^2}}dt\\ &=&\frac{\pi}{2}\sum_{n\geq m\geq0}\frac{\beta_n}{(2m+1)(2n+1)}\\ &=&\frac{\pi}2\left(\sum_{n\geq0}\frac{\beta_n}{(2n+1)^2}+\sum_{n\gt m\geq0}\frac{\beta_n}{(2m+1)(2n+1)}\right)\\ &=&\frac{\pi}{2}\left(\int_{0\lt s\lt t\lt1}\sum_{n\geq0}\beta_n s^{2n}ds\frac{dt}{t}+\int_0^1\sum_{n\gt m\geq0}\frac{\beta_n}{2m+1}x^{2n}dx\right)\\ &=&\frac{\pi}2\left(\int_{0\lt s\lt t\lt1}\frac{ds}{\sqrt{1-s^2}}\frac{dt}{t}+\int_0^1\frac{\frac12\ln\frac1{1-x^2}}{\sqrt{1-x^2}}dx\right)\\ &=&\frac{\pi}2\left(\frac{\pi}{2}\ln2+\frac{\pi}2\ln2\right)\\ &=&\frac{\pi^2\ln2}{2} \end{eqnarray}

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