秋の楕円積分祭り

被積分関数について$k=0$周りでべき級数展開して計算するだけ。

$f(x)\coloneqq ,R\coloneqq\sqrt{1-k^{2}\sin^{2}{x}}$とおくと
\begin{eqnarray} \frac{\partial R^{s}f(x)}{\partial x}&=&R^{s}f^{'}(x)-k^{2}sR^{s-2}f(x)\sin{x}\cos{x}\\ &=&\{R^{2}f^{'}(x)-k^{2}sf(x)\sin{x}\cos{x}\}R^{s-2}\\ &=&R^{s-2}\{f^{'}(x)(1-k^{2}\sin^{2}{x})-k^{2}sf(x)\sin{x}\cos{x}\\ &=&R^{s-2}\{f^{'}(x)-k^{2}\{f^{'}(x)\sin{x}+sf(x)\cos{x}\}\sin{x} \end{eqnarray}

[1]
\begin{eqnarray} E^{'}(k)&=&\int_{0}^{\frac{\pi}{2}}\frac{-k\sin^{2}{x}}{\sqrt{1-k^{2}\sin^{2}{x}}}dx\\ &=&\frac{1}{k}\int_{0}^{\frac{\pi}{2}}\frac{1-k^{2}\sin^{2}{x}-1}{\sqrt{1-k^{2}\sin^{2}{x}}}dx\\ &=&\frac{E(k)-K(k)}{k} \end{eqnarray}
[2]
\begin{eqnarray} K^{'}(k)&=&\int_{0}^{\frac{\pi}{2}}\frac{k\sin^{2}{x}}{(1-k^{2}\sin^{2}{x})^{\frac{3}{2}}}dx\\ &=&-\frac{K(k)}{k}+\frac{1}{k}\int_{0}^{\frac{\pi}{2}}\frac{1}{(1-k^{2}\sin^{2}{x})^{\frac{3}{2}}}dx\\ &=&-\frac{K(k)}{k}+\frac{k}{k^{2}-1}\int_{0}^{\frac{\pi}{2}}\frac{k^{2}-1+R^{4}-R^{4}}{k^{2}R^{3}}dx\\ &=&-\frac{K(k)}{k}+\frac{k}{k^{2}-1}[\frac{\sin{x}\cos{x}}{R}]_{0}^{\frac{\pi}{2}}+\frac{1}{k(1-k^{2})}E(k)\\ &=&\frac{E(k)-(1-k^{2})K(k)}{k(1-k^{2})}\\ &=&\frac{E(k)-\overline{k}^{2}K(k)}{k\overline{k}^{2}} \end{eqnarray}

[1]
\begin{eqnarray} K^{''}&=&\frac{1}{k\overline{k}^{2}}E^{'}+(\frac{1}{k\overline{k}^{2}})^{'}E-(\frac{1}{k}K)^{'}\\ &=&\frac{E-K}{k^{2}\overline{k}^{2}}-\frac{1-3k^{2}}{k^{2}\overline{k}^{4}}E+\frac{1}{k^{2}}K-\frac{1}{k}K^{'}\\ &=&\frac{2}{\overline{k}^{4}}E-\frac{1}{\overline{k}^{2}}K-\frac{1}{k}K^{'} \end{eqnarray}
[2]
\begin{eqnarray} \frac{\overline{k}^{4}}{2}K^{''}-k\overline{k}^{2}K^{'}&=&-\frac{\overline{k}^{4}}{2k}K^{'}-\frac{\overline{k}^{2}}{2}K+\overline{k}^{2}K\\ &=&-\frac{\overline{k}^{4}}{2k}K^{'}+\frac{\overline{k}^{2}}{2}K \end{eqnarray}
\begin{equation} 2k\overline{k}^{2}K^{''}+(1-3k^{2})K^{'}-kK=0 \end{equation}

\begin{eqnarray} \int_{0}^{\epsilon}\frac{dx}{\sqrt{(1-x^{2})(1-k^{2}x^{2})}}&\leq&\int_{0}^{\epsilon}\frac{dx}{\sqrt{(1-x)(1-kx)}}\\ &=&\int_{0}^{\epsilon}\frac{dx}{\sqrt{kx^{2}-(k+1)x+1}}\\ &=&\frac{1}{\sqrt{k}}\int_{0}^{\epsilon}\frac{dx}{\sqrt{(x-\frac{k+1}{2k})^{2}-\frac{(1-k)^{2}}{4k^{2}}}}\\ &=&\frac{1}{\sqrt{k}}[\log{|x+\sqrt{x^{2}-\frac{(1-k)^{2}}{4k^{2}}}|}]_{-\frac{k+1}{2k}}^{\epsilon-\frac{k+1}{2k}}\\ &\leq&\frac{1}{\sqrt{k}}\{\log(2\epsilon)-\log\frac{(1-\sqrt{k})^{2}}{2k}\} \end{eqnarray}
ゆえに
\begin{equation} K(\sqrt{1-k^{2}})=O(\log{(1-\sqrt{\sqrt{1-k^{2}}})}) \end{equation}

[1]
\begin{eqnarray} \frac{d}{dk}K(k)E(\overline{k})&=&K^{'}(k)E(\overline{k})-\frac{k}{\overline{k}}K(k)E^{'}(\overline{k})\\ &=&\frac{E(k)-\overline{k}^{2}K(k)}{k\overline{k}^{2}}E(\overline{k})-\frac{k}{\overline{k}}K(k)\frac{E(\overline{k})-K(\overline{k})}{\overline{k}}\\ &=&\frac{1}{k\overline{k}^{2}}E(k)E(\overline{k})-\frac{1}{k\overline{k}^{2}}K(k)E(\overline{k})+\frac{k}{\overline{k}^{2}}K(k)K(\overline{k}) \end{eqnarray}
[2]
\begin{eqnarray} \frac{d}{dk}K(\overline{k})E(k)&=&-\frac{k}{\overline{k}}\frac{E(\overline{k})-k^{2}K(\overline{k})}{\overline{k}k^{2}}E(k)+K(\overline{k})\frac{E(k)-K(k)}{k}\\ &=&-\frac{1}{k\overline{k}^{2}}E(k)E(\overline{k})+\frac{1}{\overline{k}^{2}}E(k)K(\overline{k})-\frac{1}{k}K(k)K(\overline{k}) \end{eqnarray}
[3]
\begin{eqnarray} \frac{d}{dk}K(k)K(\overline{k})&=&\frac{E(k)-\overline{k}^{2}K(k)}{k\overline{k}^{2}}K(\overline{k})-K(k)\frac{k}{\overline{k}}\frac{E(\overline{k})-k^{2}K(\overline{k})}{\overline{k}k^{2}}\\ &=&\frac{1}{k\overline{k}^{2}}E(k)K(\overline{k})-\frac{1}{k\overline{k}^{2}}K(k)E(\overline{k})-(\frac{1}{k}-\frac{k}{\overline{k}^{2}})K(k)K(\overline{k}) \end{eqnarray}
より
\begin{equation} \frac{d}{dk}\{K(k)E(\overline{k})+K(\overline{k})E(k)-K(k)K(\overline{k})\}=0 \end{equation}
[4]$k\rightarrow0$とすると
\begin{eqnarray} \left\{ \begin{array}{l} \lim_{k\rightarrow 0}K(k)E(\overline{k})=K(0)E(1)=\frac{\pi}{2}\\ \lim_{k\rightarrow 0}K(\overline{k})\{E(k)-K(k)\}=\lim_{k\rightarrow 0}O(k^{2}\log{(1-\sqrt{\sqrt{1-k^{2}}})})=0 \end{array} \right. \end{eqnarray}

[1]
\begin{eqnarray} \left\{ \begin{array}{l} \displaystyle K^{'}(k)=\frac{E(k)-\overline{k}^{2}K(k)}{k\overline{k}^{2}}\\ \displaystyle E^{'}(k)=\frac{E(k)-K(k)}{k} \end{array} \right. \end{eqnarray}
を用いる。
[2]
\begin{eqnarray} \frac{d}{dk}(k\overline{k}^{2}K)&=&(\overline{k}^{2}-2k^{2})K+k\overline{k}^{2}K^{'}\\ &=&(\overline{k}^{2}-2k^{2})K+E-\overline{k}^{2}K\\ &=&E-2k^{2}K \end{eqnarray}
\begin{eqnarray} \frac{d}{dk}(kE)&=&E+kE^{'}\\ &=&2E-K \end{eqnarray}
より
\begin{equation} \frac{d}{dk}(kE-2k\overline{k}^{2}K)=(4k^{2}-1)K \end{equation}
[3]
\begin{eqnarray} \frac{d}{dk}(\overline{k}^{2}K)&=&-2kK+\overline{k}^{2}K^{'}\\ &=&-2kK+\frac{E-\overline{k}^{2}K}{k}\\ &=&\frac{E}{k}-\frac{1+k^{2}}{k}K \end{eqnarray}
\begin{equation} \frac{d}{dk}(E-\overline{k}^{2}K)=kK \end{equation}

$x=\beta\cos^{2}{\theta}+\gamma\sin^{2}{\theta}$の様に置く。
\begin{eqnarray} \int_{\beta}^{\gamma}\frac{dx}{\sqrt{(x-\alpha)(x-\beta)(\gamma-x)}}&=&2\int_{0}^{\frac{\pi}{2}}\frac{(\gamma-\beta)\sin{\theta}\cos{\theta}}{\sqrt{(\beta\cos^{2}{\theta}+\gamma\sin^{2}{\theta}-\alpha)(\gamma-\beta)^{2}\sin^{2}{\theta}\cos^{2}{\theta}}}d\theta\\ &=&2\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{\beta\cos^{2}{\theta}+\gamma\sin^{2}{\theta}-\alpha}}\\ &=&2\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{\beta-\alpha+(\gamma-\beta)\sin^{2}{\theta}}}\\ &=&\frac{2}{\sqrt{\gamma-\alpha}}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{\frac{\beta-\alpha}{\gamma-\alpha}+\frac{\gamma-\beta}{\gamma-\alpha}\sin^{2}{\theta}}}\\ &=&\frac{2}{\sqrt{\gamma-\alpha}}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{(1-\frac{\gamma-\beta}{\gamma-\alpha})+\frac{\gamma-\beta}{\gamma-\alpha}\sin^{2}{\theta}}}\\ &=&\frac{2}{\sqrt{\gamma-\alpha}}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1+\frac{\gamma-\beta}{\gamma-\alpha}(\sin^{2}{\theta}}-1)}\\ &=&\frac{2}{\sqrt{\gamma-\alpha}}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-\frac{\gamma-\beta}{\gamma-\alpha}\cos^{2}{\theta}}}\\ &=&\frac{2}{\sqrt{\gamma-\alpha}}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-\frac{\gamma-\beta}{\gamma-\alpha}\sin^{2}{\theta}}}\\ &=&\frac{2}{\sqrt{\gamma-\alpha}}K(\sqrt{\frac{\gamma-\beta}{\gamma-\alpha}}) \end{eqnarray}

[1]$k\sin{x}=\sin{(2y-x)}$とおく。
\begin{eqnarray} k\sin{x}&=&\sin{(2y-x)}\\ &=&\sin{2y}\cos{x}-\cos{2y}\sin{x} \end{eqnarray}
より以下の式を得る。
\begin{equation} \tan{x}=\frac{\sin{2y}}{k+\cos{2y}} \end{equation}
[2]
\begin{eqnarray} \cos{x}&=&\sqrt{\frac{1}{1+\tan^{2}{x}}}\\ &=&\sqrt{\frac{(k+\cos{2y})^{2}}{(k+\cos{2y})^{2}+\sin^{2}{2y}}}\\ &=&\frac{k+\cos{2y}}{\sqrt{1+k^{2}+2k\cos{2y}}} \end{eqnarray}
\begin{eqnarray} \sin{x}&=&\sqrt{\frac{1}{1+\frac{1}{\tan^{2}{x}}}}\\ &=&\sqrt{\frac{\sin^{2}{2y}}{\sin^{2}{2y}+(k+\cos{2y})^{2}}}\\ &=&\frac{\sin{2y}}{\sqrt{1+k^{2}+2k\cos{2y}}} \end{eqnarray}
\begin{eqnarray} dx&=&\cos^{2}{x}\frac{2(1+k\cos{2y})}{(k+\cos{2y})^{2}}dy\\ &=&\frac{2(1+k\cos{2y})}{1+k^{2}+2k\cos{2y}}dy \end{eqnarray}
[3]上記をまとめると
\begin{eqnarray} K(k)&=&\frac{1}{2}\int_{0}^{\pi}\frac{1}{\sqrt{1-k^{2}\sin^{2}{x}}}dx\\ &=&\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-\frac{k^{2}\sin^{2}{2y}}\\{1+k^{2}+2k\cos{2y}}}}\frac{2(1+k\cos{2y})}{1+k^{2}+2k\cos{2y}}dy\\ &=&\int_{0}^{\frac{\pi}{2}}(1+k\cos{2y})\frac{dy}{\sqrt{(1+k^{2}+2k\cos{2y})(1+k^{2}+2k\cos{2y}-k^{2}\sin^{2}{2y})}}\\ &=&\int_{0}^{\frac{\pi}{2}}(1+k\cos{2y})\frac{dy}{\sqrt{(1+k^{2}+2k\cos{2y})(1+2k\cos{2y}+k^{2}\cos^{2}{2y})}}\\ &=&\int_{0}^{\frac{\pi}{2}}\frac{dy}{\sqrt{1+k^{2}+2k\cos{2y}}}\\ &=&\int_{0}^{\frac{\pi}{2}}\frac{dy}{\sqrt{1+k^{2}+2k(1-2\sin^{2}{y})}}\\ &=&\frac{1}{1+k}\int_{0}^{\frac{\pi}{2}}\frac{dy}{\sqrt{1-(\frac{2\sqrt{k}}{1+k})^{2}\sin^{2}{y}}}\\ &=&\frac{1}{1+k}K(\frac{2\sqrt{k}}{1+k}) \end{eqnarray}

[1]
\begin{eqnarray} a_{n}-a_{n+1}&=&a_{n}-\frac{a_{n}+b_{n}}{2}\\ &=&\frac{a_{n}-b_{n}}{2} \end{eqnarray}
\begin{eqnarray} b_{n+1}-b_{n}&=&\sqrt{a_{n}b_{n}}-b_{n}\\ &=&\frac{a_{n}-b_{n}}{\sqrt{a_{n}}+\sqrt{b_{n}}}\sqrt{b_{n}} \end{eqnarray}
\begin{eqnarray} a_{n}-b_{n}&=&\frac{a_{n-1}+b_{n-1}}{2}-\sqrt{a_{n-1}b_{n-1}}\\ &\leq&\frac{a_{n-1}+b_{n-1}}{2} \end{eqnarray}
まず$n=0$の場合は(1),(2),(3)全て成り立つ。
[2]$0,1,2,...,n-1$まで成り立つとすると$n$の場合は上記の計算から直ちに成り立つ事が分かる。

[1]$y\coloneqq b\tan{x}$
\begin{eqnarray} I(a,b)&=&\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sqrt{a^{2}\cos^{2}{x}+b^{2}\sin^{2}{x}}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{bdy}{(b^{2}+y^{2})\sqrt{\frac{b^{2}}{b^{2}+y^{2}}(a^{2}+y^{2})}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dy}{\sqrt{(a^{2}+y^{2})(b^{2}+y^{2})}} \end{eqnarray}
[2]$z=\frac{1}{2}(y-\frac{ab}{y})$とおく。
\begin{eqnarray} \left\{ \begin{array}{l} z^{2}=\frac{1}{4}(y^{2}+\frac{a^{2}b^{2}}{y^{2}})-\frac{ab}{2}\\ y=z+\sqrt{ab+z^{2}}\\ dy=\frac{y}{\sqrt{ab+z^{2}}}dz \end{array} \right. \end{eqnarray}
\begin{eqnarray} I(a,b)&=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dy}{\sqrt{(a^{2}+y^{2})(b^{2}+y^{2})}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dy}{\sqrt{y^{4}+(a^{2}+b^{2})y^{2}+a^{2}b^{2}}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dy}{\sqrt{y^{2}(y^{2}+\frac{a^{2}b^{2}}{y^{2}}-2ab)+(a^{2}+2ab+b^{2})y^{2}}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dy}{\sqrt{y^{2}(y^{2}+\frac{a^{2}b^{2}}{y^{2}}-2ab)+(a^{2}+2ab+b^{2})y^{2}}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dy}{y\sqrt{4z^{2}+(a+b)^{2}}}\\ &=&\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{dz}{\sqrt{(ab+z^{2})\{(\frac{a+b}{2})^{2}+z^{2}\}}}\\ &=&\frac{2}{\pi}\int_{0}^{\infty}\frac{dz}{\sqrt{(ab+z^{2})\{(\frac{a+b}{2})^{2}+z^{2}\}}}\\ &=&I(\frac{a+b}{2},\sqrt{ab}) \end{eqnarray}

\begin{eqnarray} \int_{-1}^{1}\frac{dx}{\sqrt{(1-2xt+t^{2})(1-2xu+u^{2})}}&=&\frac{1}{2\sqrt{tu}}\int_{-1}^{1}\frac{dx}{\sqrt{(\frac{1+t^{2}}{2t}-x)(\frac{1+u^{2}}{2u}-x)}}\quad(A\coloneqq\frac{1+t^{2}}{2t},B\coloneqq\frac{1+u^{2}}{2u})\\ &=&\frac{1}{2\sqrt{tu}}\int_{-1}^{1}\frac{dx}{\sqrt{(A-x)(B-x)}}\\ &=&\frac{1}{2\sqrt{tu}}\int_{-1}^{1}\frac{dx}{\sqrt{x^{2}-(A+B)x+AB}}\\ &=&\frac{1}{2\sqrt{tu}}\int_{-1}^{1}\frac{dx}{\sqrt{(\frac{A+B}{2}-x)^{2}-(\frac{A-B}{2})^{2}}}\quad(A+B\gt 2)\\ &=&\frac{1}{2\sqrt{tu}}\int_{\frac{A+B-2}{2}}^{\frac{A+B+2}{2}}\frac{dx}{\sqrt{x^{2}-(\frac{A-B}{2})^{2}}}\\ &=&\frac{1}{2\sqrt{tu}}\int_{\frac{A+B-2}{A-B}}^{\frac{A+B+2}{A-B}}\frac{dx}{\sqrt{x^{2}-1}}\\ &=&\frac{1}{2\sqrt{tu}}\cosh^{-1}{x}|_{\frac{A+B-2}{A-B}}^{\frac{A+B+2}{A-B}}\\ &=&\frac{1}{2\sqrt{tu}}\log{\frac{A+B+2+\sqrt{(A+B+2)^{2}-(A-B)^{2}}}{A+B-2+\sqrt{(A+B-2)^{2}-(A-B)^{2}}}}\\ &=&\frac{1}{2\sqrt{tu}}\log{\frac{A+B+2+2\sqrt{(A+1)(B+1)}}{A+B-2+2\sqrt{(A-1)(B-1)}}}\\ &=&\frac{1}{2\sqrt{tu}}\log{\frac{\frac{(1+t)^{2}}{2t}+\frac{(1+u)^{2}}{2u}+\frac{(1+t)(1+u)}{\sqrt{tu}}}{\frac{(1-t)^{2}}{2t}+\frac{(1-u)^{2}}{2u}+\frac{(1-t)(1-u)}{\sqrt{tu}}}}\\ &=&\frac{1}{2\sqrt{tu}}\log{(\frac{\frac{1+t}{\sqrt{t}}+\frac{1+u}{\sqrt{u}}}{\frac{1-t}{\sqrt{t}}+\frac{1-u}{\sqrt{u}}})^{2}}\\ &=&\frac{1}{\sqrt{tu}}\log{\frac{\frac{1+t}{\sqrt{t}}+\frac{1+u}{\sqrt{u}}}{\frac{1-t}{\sqrt{t}}+\frac{1-u}{\sqrt{u}}}}\\ &=&\frac{1}{\sqrt{tu}}\log{\frac{\sqrt{t}+\sqrt{u}+\sqrt{tu}(\sqrt{t}+\sqrt{u})}{\sqrt{t}+\sqrt{u}-\sqrt{tu}(\sqrt{t}+\sqrt{u})}}\\ &=&\frac{1}{\sqrt{tu}}\log{\frac{1+\sqrt{tu}}{1-\sqrt{tu}}}\\ &=&\frac{2}{\sqrt{tu}}\sum_{n=1}^{\infty}\frac{(tu)^{\frac{2n-1}{2}}}{2n-1}\\ &=&2\sum_{n=0}^{\infty}\frac{(tu)^{n}}{2n+1}\\ &=&\sum_{m,n}t^{m}u^{n}\int_{0}^{1}P_{m}(x)P_{n}(x)dx \end{eqnarray}
より係数比較により
\begin{equation} \int_{-1}^{1}P_{m}(x)P_{n}(x)dx=\frac{2}{2n+1}\delta_{mn} \end{equation}

$\forall x\in[-1,1]:c_{0}P_{0}(x)+c_{1}P_{1}(x)+\cdots+c_{n}P_{n}(x)+\cdots\equiv0$が成り立つとする。すると
\begin{equation} \int_{-1}^{1}P_{m}(x)\sum_{n=0}^{\infty}c_{n}P_{n}(x)dx=\frac{2}{2m+1}c_{m}=0 \end{equation}
なので$c_{m}=0\quad(m=0,1,2,...)$を得る。

[1]$n=0$の場合は$\int_{-1}^{1}P_{0}^{2}dx=2$なので$P_{0}\equiv1$であるから明らか。
[2]そこで、$0,1,2,...,n-1$まで成立すると仮定する。
nの場合を考える。
適当な係数$c_{n}$を定めて$x^{n}-c_{n}P_{n}=Q_{n}\quad\deg(Q)\lt n$が成り立つような多項式を作る。
すると$Q(x)$は仮定より$1,x,...,x^{n-1}$の$\mathbb{R}$係数一次結合で書けるので$1,x,...,x^{n-1}$はそれぞれ$P_{0};P_{0},P_{1};...;P_{0},P_{1},...,P_{n-1}$の一次結合で書けるので証明完了。

[1]曲線$C$内で正則な複素関数$f(z)$に対して留数定理より以下の式が成り立つ。
\begin{equation} f(z)=\frac{1}{i2\pi}\oint_{C}\frac{f(w)}{w-z}dw \end{equation}
両辺$z$で$n$回微分して
\begin{equation} f^{(n)}(z)=\frac{n!}{i2\pi}\oint_{C}\frac{f(w)}{(w-z)^{n+1}}dw \end{equation}
を得る。
[2]$f(z)=\frac{1}{2^{n}n!}(z^{2}-1)^{n}$とおく。すると以下の式が成り立つ。
\begin{equation} \frac{1}{2^{n}n!}\frac{d}{dz}(z^{2}-1)^{n}=\frac{1}{i2\pi}\oint\frac{1}{w-z}\{\frac{w^{2}-1}{2(w-z)}\}^{n}dw \end{equation}
[3]
\begin{eqnarray} \sum_{n=0}^{\infty}f^{(n)}t^{n}&=&\frac{1}{i2\pi}\oint_{C}\frac{dw}{w-z-\frac{w^{2}-1}{2}t}\\ &=&\frac{1}{i2\pi}\oint_{C}\frac{-2}{tw^{2}-2w+2z-t} \end{eqnarray}
[4]被積分関数の$1$位の極は$w=\frac{1\pm\sqrt{1-2zt+t^{2}}}{t}$であり$t\rightarrow0$の時発散しない条件で考えると符号は$-1$の方。
よって$C$を$-1$の符号を持つ$1$位の極のみを囲う様にとると
\begin{eqnarray} \frac{1}{i2\pi}\oint_{C}\frac{-2}{tw^{2}-2w+2z-t}&=&\frac{-2}{t}\lim_{w\rightarrow\frac{1-\sqrt{1-2zt+t^{2}}}{t}}\frac{1}{w-\frac{1+\sqrt{1-2zt+t^{2}}}{t}}\\ &=&\frac{1}{\sqrt{1-2zt+t^{2}}} \end{eqnarray}
[5]上記の計算から直ちに以下の結論を得る。
\begin{equation} f^{(n)}(x)=\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}=P_{n} \end{equation}

[1]$m\lt n$の場合は$x^{m}$は$m$次以下のLegender多項式$P_{0}(x),P_{1}(x)...,P_{m}(x)$の$\mathbb{R}$一次結合で書けるのでLegender多項式の定義より$0$
[2]$m\geq n$の場合は
\begin{eqnarray} 2^{n}n!\int_{-1}^{1}x^{m}P_{n}(x)dx&=&\int_{-1}^{1}x^{m}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}\\ &=&[x^{m}\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}]_{-1}^{1}-m\int_{-1}^{1}x^{m-1}\frac{d^{n-1}}{dx^{n-1}}(x^{2}-1)^{n}\\ &=&(-1)^{m}\frac{m!}{n!}\int_{-1}^{1}x^{m-n}(x^{2}-1)^{n}dx \end{eqnarray}
$m-n\equiv0\mod{2}$の場合
\begin{eqnarray} (-1)^{m}\frac{m!}{n!}\int_{-1}^{1}x^{m-n}(x^{2}-1)^{n}dx&=&2(-1)^{m}\frac{m!}{n!}\int_{0}^{1}x^{m-n}(x^{2}-1)^{n}dx\\ &=&(-1)^{m}\frac{m!}{n!}\int_{0}^{1}x^{\frac{m-n-1}{2}}(x-1)^{n}dx\\ &=&(-1)^{m+n}\frac{m!}{(m-n)!}\int_{0}^{1}x^{\frac{m-n-1}{2}}(1-x)^{n}dx\\ &=&\frac{m!}{(m-n)!}\frac{\Gamma(\frac{m-n+1}{2})\Gamma(n+1)}{\Gamma(\frac{m+n+3}{2})}\\ &=&\frac{m!n!}{(m-n)!}\frac{\Gamma(\frac{m-n+1}{2})}{\Gamma(\frac{m+n+3}{2})} \end{eqnarray}
ゆえに
\begin{equation} \int_{-1}^{1}x^{m}P_{n}(x)dx=\frac{m!}{2^{n}(m-n)!}\frac{\Gamma(\frac{m-n+1}{2})}{\Gamma(\frac{m+n+3}{2})} \end{equation}
[3]$m\equiv1\mod{2}$の場合は奇関数なので$0$

\begin{eqnarray} \int_{0}^{1}\hat{P}_{m}(x)\hat{P}_{n}(x)dx&=&\int_{0}^{1}P_{m}(2x-1)P_{n}(2x-1)dx\\ &=&\frac{1}{2}\int_{-1}^{1}P_{m}(x)P_{n}(x)dx\\ &=&\frac{1}{2n+1}\delta_{m,n} \end{eqnarray}

$K(x)=\sum_{n=0}^{\infty}c_{n}P_{n}(x)$と書けたとする。
\begin{eqnarray} c_{n}&=&\frac{2n+1}{2}\int_{-1}^{1}K(x)P_{n}(x)dx\\ &=&\frac{(2n+1)\pi}{4}\sum_{m=0}^{\infty}\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\int_{-1}^{1}x^{2m}P_{n}(x) \end{eqnarray}
なので$c_{2n-1}=0\quad(n=1,2,...)$。
\begin{eqnarray} c_{2n}&=&\frac{4n+1}{2}\int_{-1}^{1}K(x)P_{2n}(x)dx\\ &=&\frac{(4n+1)\pi}{4}\sum_{m=0}^{\infty}\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\int_{-1}^{1}x^{2m}P_{2n}(x)\\ &=&\frac{(4n+1)\pi}{4}\sum_{m=n}^{\infty}\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\frac{(2m)!}{4^{n}(2m-2n)!}\frac{\Gamma(\frac{2m-2n+1}{2})}{\Gamma(\frac{2m+2n+3}{2})}\\ &=&\frac{(4n+1)\pi}{4}\sum_{m=n}^{\infty}\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\frac{(2m)!}{4^{n}(2m-2n)!}\frac{\frac{(2m-2n)!}{4^{m-n}(m-n)!}\sqrt{\pi}}{\frac{(2m+2n+2)!}{4^{m+n+1}(m+n+1)!}\sqrt{\pi}}\\ &=&\frac{(4n+1)\pi}{4}\sum_{m=n}^{\infty}\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\frac{(2m)!}{4^{n}(2m-2n)!}\frac{4^{2n+1}(2m-2n)!(m+n+1)!}{(2m+2n+2)!(m-n)!}\\ &=&4^{n}(4n+1)\pi\sum_{m=n}^{\infty}\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\frac{(2m)!(m+n+1)!}{(2m+2n+2)!(m-n)!} \end{eqnarray}
$t_{m}\coloneqq\frac{\begin{pmatrix}2m\\m\end{pmatrix}^{2}}{16^{m}}\frac{(2m)!(m+n+1)!}{(2m+2n+2)!(m-n)!}$とおく
\begin{eqnarray} \frac{t_{m}}{t_{m-1}}&=&\frac{(2m)^{2}(2m-1)^{2}}{m^{4}}\frac{1}{16}\frac{2m(2m-1)(m+n+1)}{(2m+2n+2)(2m+2n+1)(m-n)}\\ &=&\frac{(m-\frac{1}{2})^{2}}{m^{2}}\frac{m(2m-1)}{(2m+2n+1)(m-n)}\\ &=&\frac{(m-\frac{1}{2})^{2}}{m}\frac{2m-1}{(2m+2n+1)(m-n)} \end{eqnarray}
$\frac{t^{'}_{m+n}}{t^{'}_{m+n-1}}\coloneqq \frac{t_{m+n}}{t_{m+n-1}}=\frac{(m+n-\frac{1}{2})^{2}}{m+n}\frac{2m+2n-1}{(2m+2n+1)m}$
\begin{eqnarray} t^{'}_{m}&=&\frac{(m+n-\frac{1}{2})^{2}}{m+n}\frac{m+n-\frac{1}{2}}{(m+2n+\frac{1}{2})m}t^{'}_{m-1}\\ &=&\frac{(n+\frac{1}{2})_{m}^{3}}{(n+1)_{m}(2n+\frac{3}{2})_{m}m!}t^{'}_{0}\\ &=&\frac{1}{2\cdot16^{n}(4n+1)}\frac{\begin{pmatrix}2n\\n\end{pmatrix}^{2}}{\begin{pmatrix}4n\\2n\end{pmatrix}}\frac{(n+\frac{1}{2})_{m}^{3}}{(n+1)_{m}(2n+\frac{3}{2})_{m}m!} \end{eqnarray}
\begin{eqnarray} c_{2n}&=&\frac{\pi}{2^{2n+1}}\frac{\begin{pmatrix}2n\\n\end{pmatrix}^{2}}{\begin{pmatrix}4n\\2n\end{pmatrix}}\sum_{m=0}^{\infty}\frac{(n+\frac{1}{2})_{m}^{3}}{(n+1)_{m}(2n+\frac{3}{2})_{m}m!}\\ &=&\frac{\pi}{2^{2n+1}}\frac{\begin{pmatrix}2n\\n\end{pmatrix}^{2}}{\begin{pmatrix}4n\\2n\end{pmatrix}}{}_{3}F_{2}(n+\frac{1}{2},n+\frac{1}{2},n+\frac{1}{2};n+1,2n+\frac{3}{2};1) \end{eqnarray}

\begin{eqnarray} \Gamma(a)\Gamma(b)&=&\int_{0}^{\infty}dx\int_{0}^{\infty}dyx^{a-1}y^{b-1}e^{-(x+y)}\\ &=&\int_{0}^{\infty}dx\int_{x}^{\infty}dyx^{a-1}(y-x)^{b-1}e^{-y}\\ &=&\int_{0}^{\infty}dyy^{b-1}e^{-y}\int_{0}^{y}dxx^{a-1}(1-\frac{x}{y})^{b-1}\\ &=&\int_{0}^{\infty}dyy^{a+b-1}e^{-y}\int_{0}^{1}dxx^{a-1}(1-x)^{b-1}\\ &=&\Gamma(a+b)B(a,b) \end{eqnarray}

\begin{eqnarray} E(\frac{1}{\sqrt{2}})&=&\int_{0}^{\frac{\pi}{2}}\sqrt{1-\frac{1}{2}\sin^{2}{x}}dx\\ &=&\int_{0}^{1}\sqrt{\frac{1-\frac{1}{2}x^{2}}{1-x^{2}}}dx\\ &=&\frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-\frac{1}{2}x}{x(1-x)}}dx\\ &=&\frac{1}{2\sqrt{2}}\int_{0}^{1}\sqrt{\frac{1+x}{x(1-x)}}dx\\ &=&\frac{1}{2\sqrt{2}}\int_{0}^{1}\frac{1+x}{\sqrt{x(1-x^{2})}}dx\\ &=&\frac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\frac{1+\sin{x}}{\sqrt{\sin{x}}}dx\\ &=&\frac{\sqrt{\pi}}{4\sqrt{2}}\{\frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}+\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{5}{4})}\}\\ &=&\frac{\sqrt{\pi}}{4\sqrt{2}}(\frac{\sin{\frac{\pi}{4}}}{\pi}\Gamma^{2}(\frac{1}{4})+4\frac{\pi}{\sin{\frac{\pi}{4}}}\frac{1}{\Gamma^{2}(\frac{1}{4})})\\ &=&\frac{\sqrt{\pi}}{4\sqrt{2}}(\frac{1}{\pi\sqrt{2}}\Gamma^{2}(\frac{1}{4})+\frac{4\pi\sqrt{2}}{\Gamma^{2}(\frac{1}{4})})\\ &=&\frac{1}{8\sqrt{\pi}}\Gamma^{2}(\frac{1}{4})+\frac{\pi\sqrt{\pi}}{\Gamma^{2}(\frac{1}{4})} \end{eqnarray}

\begin{eqnarray} K(\frac{1}{\sqrt{2}})&=&\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sqrt{1-\frac{1}{2}\sin^{2}{x}}}\\ &=&\int_{0}^{1}\frac{dx}{\sqrt{(1-x^{2})(1-\frac{1}{2}x^{2})}}\\ &=&\frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)(1-\frac{1}{2}x)}}\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{dx}{\sqrt{x(1-x^{2})}}dx\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\sin^{-\frac{1}{2}}{x}dx\\ &=&\frac{1}{2\sqrt{2}}B(\frac{1}{4},\frac{1}{2})\\ &=&\frac{1}{2\sqrt{2}}\frac{\Gamma(\frac{1}{4})\sqrt{\pi}}{\Gamma(\frac{3}{4})}\\ &=&\frac{\sqrt{\pi}}{2\sqrt{2}}\frac{\Gamma^{2}(\frac{1}{4})}{\Gamma(\frac{1}{4})\Gamma(1-\frac{1}{4})}\\ &=&\frac{\Gamma^{2}(\frac{1}{4})}{4\sqrt{\pi}} \end{eqnarray}

\begin{eqnarray} \int_{-1}^{1}\{K(x)\}^{2}dx&=&2\sum_{n=0}^{\infty}\frac{c_{2n}^{2}}{4n+1}\\ &=&\frac{\pi^{2}}{2}\sum_{n=0}^{\infty}\frac{1}{2^{4n}(4n+1)}\frac{\begin{pmatrix}2n\\n\end{pmatrix}^{4}}{\begin{pmatrix}4n\\2n\end{pmatrix}^{2}}\{{}_{3}F_{2}(n+\frac{1}{2},n+\frac{1}{2},n+\frac{1}{2};n+1,2n+\frac{3}{2};1)\}^{2} \end{eqnarray}

[1]
\begin{eqnarray} \int_{0}^{1}K(x)dx&=&\int_{0}^{1}dx\int_{0}^{\frac{\pi}{2}}dy\frac{1}{\sqrt{1-x^{2}\sin^{2}{y}}}\\ &=&\int_{0}^{\frac{\pi}{2}}dy\int_{0}^{1}dx\frac{1}{\sqrt{1-x^{2}\sin^{2}{y}}}\\ &=&\int_{0}^{\frac{\pi}{2}}dy\frac{1}{\sin{y}}\int_{0}^{\sin{y}}\frac{dx}{\sqrt{1-x^{2}}}\\ &=&\int_{0}^{\frac{\pi}{2}}dy\frac{y}{\sin{y}} \end{eqnarray}
[2]
\begin{eqnarray} \frac{1}{\sin{y}}&=&\frac{2i}{e^{iy}-e^{-iy}}\\ &=&e^{-iy}\frac{2i}{1-e^{-i2y}}\\ &=&i2e^{-iy}\sum_{n=0}^{\infty}e^{-i2ny}\\ &=&i2\sum_{n=0}^{\infty}\{\cos{(2n+1)y}-i\sin{(2n+1)y}\}\\ &=&2\sum_{n=0}^{\infty}\sin{(2n+1)y} \end{eqnarray}
[3]
\begin{eqnarray} \int_{0}^{1}K(x)dx&=&2\sum_{n=0}^{\infty}\int_{0}^{\frac{\pi}{2}}y\sin{(2n+1)y}dy\\ &=&2\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{2}}\\ &=&2G \end{eqnarray}

\begin{eqnarray} \int_{0}^{1}(4x^{2}-1)K(x)dx&=&[xE(x)-2x(1-x^{2})K(x)]_{0}^{1}\\ &=&1 \end{eqnarray}
ゆえに
\begin{equation} \int_{0}^{1}x^{2}K(x)dx=\frac{1+2G}{4} \end{equation}

\begin{eqnarray} \int_{0}^{1}xK(x)dx&=&[E(x)-(1-x^{2})K(x)]_{0}^{1}\\ &=&1 \end{eqnarray}

[1]
\begin{eqnarray} \frac{\Gamma^{4}(\frac{1}{4})}{16\pi}&=&\frac{1}{8}\int_{0}^{1}dx\int_{0}^{1}dy(xy)^{-\frac{3}{4}}\{(1-x)(1-y)\}^{-\frac{1}{2}}\\ &=&\frac{1}{8}\int_{0}^{1}dx\int_{0}^{1}dy(xy)^{-\frac{3}{4}}\{1-(x+y)+xy\}^{-\frac{1}{2}}\quad(u=xy,v=x+y)\\ &=&\frac{1}{8}\int_{0}^{1}du\int_{2\sqrt{u}}^{1+u}dv\frac{u^{-\frac{3}{4}}(1-v+u)^{-\frac{1}{2}}}{\sqrt{v^{2}-4u}}\\ &=&\frac{1}{8}\int_{0}^{1}du\int_{\sqrt{u}}^{\frac{1+u}{2}}dv\frac{u^{-\frac{3}{4}}(1-2v+u)^{-\frac{1}{2}}}{\sqrt{v^{2}-u}}\\ &=&\frac{1}{8\sqrt{2}}\int_{0}^{1}duu^{-\frac{3}{4}}\int_{\sqrt{u}}^{\frac{1+u}{2}}\frac{dv}{\sqrt{(v+\sqrt{u})(v-\sqrt{u})(\frac{1+u}{2}-v)}}\\ &=&\frac{1}{8\sqrt{2}}\int_{0}^{1}duu^{-\frac{3}{4}}\frac{2}{\sqrt{\frac{1+u}{2}+\sqrt{u}}}K(\sqrt{\frac{\frac{1+u}{2}-\sqrt{u}}{\frac{1+u}{2}+\sqrt{u}}})\\ &=&\frac{1}{4}\int_{0}^{1}du\frac{u^{-\frac{3}{4}}}{1+\sqrt{u}}K(\frac{1-\sqrt{u}}{1+\sqrt{u}})\quad(t\coloneqq\frac{1-\sqrt{u}}{1+\sqrt{u}})\\ &=&\int_{0}^{1}dt\frac{1-t}{(1+t)^{3}}(\frac{1+t}{1-t})^{\frac{3}{2}}\frac{1}{1+\frac{1-t}{1+t}}K(t)dt\\ &=&\frac{1}{2}\int_{0}^{1}dt\frac{1}{(1-t)^{\frac{1}{2}}(1+t)^{\frac{1}{2}}}K(t)\\ &=&\frac{1}{2}\int_{0}^{1}dt\frac{K(t)}{\sqrt{1-t^{2}}} \end{eqnarray}

\begin{eqnarray} \int_{0}^{1}K(\sqrt{x})dx&=&\int_{0}^{1}dx\int_{0}^{1}dy\\ &=&\frac{1}{\sqrt{(1-y^{2})(1-xy^{2})}}\\ &=&\int_{0}^{1}\frac{dv}{v^{2}\sqrt{1-v^{2}}}\int_{0}^{v^{2}}\frac{du}{\sqrt{1-u}}\quad(xy^{2}=u,y=v)\\ &=&\int_{0}^{1}\frac{dv}{v^{2}\sqrt{1-v^{2}}}[-2\sqrt{1-u}]_{0}^{v^{2}}\\ &=&2\int_{0}^{1}\frac{1-\sqrt{1-v^{2}}}{v^{2}\sqrt{1-v^{2}}}dv\\ &=&2\int_{0}^{\frac{\pi}{2}}\frac{1-\cos{x}}{\sin^{2}{x}}dx\\ &=&2\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\cos{x}}\\ &=&\int_{0}^{\frac{\pi}{2}}\frac{dx}{\cos^{2}{\frac{x}{2}}}\\ &=&2\tan{\frac{x}{2}}|_{0}^{\frac{\pi}{2}}\\ &=&2 \end{eqnarray}

[1]
\begin{eqnarray} \int_{0}^{1}K(\overline{x})dx&=&\int_{0}^{1}K(\sqrt{1-x^{2}})dx\\ &=&\int_{0}^{1}dx\int_{0}^{\frac{\pi}{2}}d\theta\frac{1}{\sqrt{1-(1-x^{2})\sin^{2}{\theta}}}\\ &=&\int_{0}^{1}dx\int_{0}^{\frac{\pi}{2}}d\theta\frac{1}{\sqrt{\cos^{2}{\theta}+x^{2}\sin^{2}{\theta}}}\\ &=&\int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{1}dx\frac{1}{\sqrt{\cos^{2}{\theta}+x^{2}\sin^{2}{\theta}}}\\ &=&\int_{0}^{\frac{\pi}{2}}d\theta\frac{1}{\sin{\theta}}\int_{0}^{\sin{\theta}}dx\frac{1}{\sqrt{\cos^{2}{\theta}+x^{2}}}\\ &=&\int_{0}^{\frac{\pi}{2}}d\theta\frac{1}{\sin{\theta}}\int_{0}^{\tan{\theta}}dx\frac{1}{\sqrt{1+x^{2}}}\\ &=&\int_{0}^{\frac{\pi}{2}}d\theta\frac{1}{\sin{\theta}}[\log{(x+\sqrt{1+x^{2}})}]_{0}^{\tan{\theta}}\\ &=&\int_{0}^{\frac{\pi}{2}}d\theta\frac{\log{(\tan{\theta}+\frac{1}{\cos{\theta}})}}{\sin{\theta}} \end{eqnarray}
[2]
\begin{eqnarray} \frac{d}{d\theta}\log{(\tan{\theta}+\frac{1}{\cos{\theta}})}&=& \frac{\frac{1}{\cos^{2}{\theta}}+\frac{\sin{\theta}}{\cos^{2}{\theta}}}{\tan{\theta}+\frac{1}{\cos{\theta}}}d\theta\\ &=&\frac{d\theta}{\cos{\theta}} \end{eqnarray}
\begin{eqnarray} \int\frac{d\theta}{\sin{\theta}}&=&\frac{1}{2}\int\frac{d\theta}{\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}}\quad(x=\sin{\frac{\theta}{2}},dx=\frac{1}{2}\cos{\frac{\theta}{2}}d\theta)\\ &=&\int\frac{dx}{x(1-x^{2})}\\ &=&\int dx(\frac{1}{2}\frac{1}{1-x}+\frac{1}{x}-\frac{1}{2}\frac{1}{1+x})\\ &=&\frac{1}{2}\log{\frac{x^{2}}{1-x^{2}}}\\ &=&\log{\tan{\frac{\theta}{2}}} \end{eqnarray}
[3]
\begin{eqnarray} \int_{0}^{\frac{\pi}{2}}d\theta\frac{\log{(\tan{\theta}+\frac{1}{\cos{\theta}})}}{\sin{\theta}}&=&[\log{\tan{\frac{\theta}{2}}}\log{(\tan{\theta}+\frac{1}{\cos{\theta}})}]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\frac{\log{\tan{\frac{\theta}{2}}}}{\cos{\theta}}d\theta\\ &=&\lim_{\epsilon\rightarrow0}\log{\tan{(\frac{\pi}{4}-\frac{\epsilon}{2})}\log{\{\tan{\{\frac{\pi}{2}-\epsilon)+\frac{1}{\cos{(\frac{\pi}{2}-\epsilon)}}\}}}}-\lim_{\epsilon\rightarrow0}\log{\tan{\frac{\epsilon}{2}}}\log{(\tan{\epsilon}+\frac{1}{\cos{\epsilon}})}-\int_{0}^{\frac{\pi}{2}}\frac{\log{\tan{\frac{\theta}{2}}}}{\cos{\theta}}d\theta \end{eqnarray}
[4]
\begin{align} &\log{\tan{\frac{x}{2}}}\log{(\tan{x}+\frac{1}{\cos{x}})}\\ &=\log{\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}}\log{\frac{1+\sin{x}}{\cos{x}}}\\ &=\frac{1}{2}\log{\frac{1-\cos{x}}{1+\cos{x}}}\log{\frac{1+\sin{x}}{\cos{x}}}\\ &=\frac{1}{2}\sum_{n=1}^{\infty}\frac{\cos^{2n}{x}}{n}\log{\frac{1+\sin{x}}{\cos{x}}}\\ &\overset{x\rightarrow0\vee x\rightarrow\frac{\pi}{2}}\rightarrow0 \end{align}
[5]
\begin{eqnarray} \int_{0}^{\frac{\pi}{2}}d\theta\frac{\log{(\tan{\theta}+\frac{1}{\cos{\theta}})}}{\sin{\theta}}&=&-\int_{0}^{\frac{\pi}{2}}\frac{\log{\tan{\frac{\theta}{2}}}}{\cos{\theta}}d\theta\quad(t\coloneqq\tan{\frac{\theta}{2}},dt=\frac{1}{2}(1+t^{2})d\theta,\frac{1}{\cos{\theta}}=\frac{1+t^{2}}{t(1-t^{2})})\\ &=&-2\int_{0}^{1}\frac{\log{t}}{(1+t^{2})}\frac{1+t^{2}}{1-t^{2}}\\ &=&-2\int_{0}^{1}\frac{\log{t}}{1-t^{2}}dt\\ &=&-2\sum_{n=0}^{\infty}\int_{0}^{1}t^{2n}\log{t}dt\\ &=&2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}\\ &=&2\{\zeta(2)-\frac{1}{4}\zeta(2)\}\\ &=&\frac{3}{2}\zeta(2)\\ &=&\frac{\pi^{2}}{4} \end{eqnarray}