(1-x^{2n+2})logx/(1-x^2)^2[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{(1-x^{2n+2})\log x}{(1-x^2)^2}dx=-\frac{(n+1)π^2}{8}+\displaystyle\sum_{k=1}^n \frac{n-k+1}{(2k-1)^2}\\
&(2) \displaystyle\int_0^1 \frac{\{1+(-1)^nx^{n+1}\}\log x}{(1+x)^2}dx=-\frac{(n+1)π^2}{12}+\displaystyle\sum_{k=1}^n \frac{(-1)^{k-1}(n-k+1)}{k^2}\\
&(3) \displaystyle\int_0^1 \frac{(1-x^{n+1})\log x}{(1-x)^2}dx=-\frac{(n+1)π^2}{6}+\displaystyle\sum_{k=1}^n \frac{n-k+1}{k^2}\\
\end{alignat}ただし、全て \(n \in \mathrm{N}\)







<証明>

途中、次の定積分の結果を用います。(詳細はこちらです。(A)(B)(C))
\begin{alignat}{2}
&(A) \displaystyle\int_0^1 \frac{\log x}{1-x^2}dx=-\frac{π^2}{8}\\
&(B) \displaystyle\int_0^1 \frac{\log x}{1+x}dx=-\frac{π^2}{12}\\
&(C) \displaystyle\int_0^1 \frac{\log x}{1-x}dx=-\frac{π^2}{6}\\
\end{alignat}




\((1)\) 次の等比数列の和の等式を用います。$$1+x^2+x^4+ \cdots +x^{2n}=1+\displaystyle\sum_{k=1}^n x^{2k}=\frac{1-x^{2n+2}}{1-x^2}$$この式を代入すると
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{(1-x^{2n+2})\log x}{(1-x^2)^2}dx=\displaystyle\int_0^1 \frac{\log x}{1-x^2}\left(1+\displaystyle\sum_{k=1}^n x^{2k}\right)dx\\
&                   =\displaystyle\int_0^1 \frac{\log x}{1-x^2}dx+\displaystyle\sum_{k=1}^n \displaystyle\int_0^1 \frac{x^{2k}\log x}{1-x^2}dx\\
\end{alignat}\((A)\) を代入すると$$=-\frac{π^2}{8}+\displaystyle\sum_{k=1}^n \displaystyle\int_0^1 \frac{x^{2k}\log x}{1-x^2}dx$$右側の積分を計算します。
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{x^{2k}\log x}{1-x^2}dx=\displaystyle\int_0^1 x^{2k} \log x \displaystyle\sum_{m=0}^{\infty} x^{2m} dx=\displaystyle\sum_{m=0}^{\infty} \displaystyle\int_0^1 x^{2k+2m}\log xdx\\
&             =-\displaystyle\sum_{m=0}^{\infty} \frac{1}{(2k+2m+1)^2}=-\left\{\frac{1}{(2k+1)^2}+\frac{1}{(2k+3)^2}+\frac{1}{(2k+5)^2}+ \cdots\right\}\\
&             =-\left\{\displaystyle\sum_{l=1}^{\infty} \frac{1}{(2l-1)^2}-\displaystyle\sum_{m=1}^k \frac{1}{(2m-1)^2}\right\}=-\frac{π^2}{8}+\displaystyle\sum_{m=1}^k \frac{1}{(2m-1)^2}
\end{alignat}となるので、元の積分計算は
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{(1-x^{2n+2})\log x}{(1-x^2)^2}dx=-\frac{π^2}{8}+\displaystyle\sum_{k=1}^n \left\{-\frac{π^2}{8}+\displaystyle\sum_{m=1}^k \frac{1}{(2m-1)^2}\right\}\\
&                   =-\frac{π^2}{8}-\frac{nπ^2}{8}+\displaystyle\sum_{k=1}^n \left\{\displaystyle\sum_{m=1}^k \frac{1}{(2m-1)^2}\right\}\\
&                   =-\frac{(n+1)π^2}{8}+\displaystyle\sum_{k=1}^n \left\{\displaystyle\sum_{m=1}^k \frac{1}{(2m-1)^2}\right\}\\
\end{alignat}右側のシグマを計算します。
\begin{alignat}{2}
&\displaystyle\sum_{k=1}^n \left\{\displaystyle\sum_{m=1}^k \frac{1}{(2m-1)^2}\right\}=\displaystyle\sum_{k=1}^n \left\{1+\frac{1}{3^2}+\frac{1}{5^2}+ \cdots +\frac{1}{(2k-1)^2}\right\}\\
&                  =1+\left(1+\frac{1}{3^2}\right)+\left(1+\frac{1}{3^2}+\frac{1}{5^2}\right)+ \cdots +\left\{1+\frac{1}{3^2}+\frac{1}{5^2}+ \cdots +\frac{1}{(2n-1)^2}\right\}\\
&                  =1 \cdot n+\frac{1}{3^2}(n-1)+\frac{1}{5^2}(n-2)+ \cdots +\frac{1}{(2n-3)^2} \cdot 2 +\frac{1}{(2n-1)^2}\cdot 1=\displaystyle\sum_{k=1}^n \frac{n-k+1}{(2k-1)^2}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(1-x^{2n+2})\log x}{(1-x^2)^2}dx=-\frac{(n+1)π^2}{8}+\displaystyle\sum_{k=1}^n \frac{n-k+1}{(2k-1)^2}$$







\((2)\) 次の等比数列の和の等式を用います。$$1-x+x^2+ \cdots +(-1)^nx^{n}=1+\displaystyle\sum_{k=1}^n (-1)^k x^k=\frac{1-(-x)^{n+1}}{1+x}=\frac{1+(-1)^n x^{n+1}}{1+x}$$この式を代入すると
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{\{1+(-1)^nx^{n+1}\}\log x}{(1+x)^2}dx=\displaystyle\int_0^1 \frac{\log x}{1+x}\left\{1+\displaystyle\sum_{k=1}^n (-1)^kx^k\right\}dx\\
&                        =\displaystyle\int_0^1 \frac{\log x}{1+x}dx+\displaystyle\sum_{k=1}^n (-1)^k\displaystyle\int_0^1 \frac{x^k\log x}{1+x}dx\\
\end{alignat}\((B)\) を代入すると$$=-\frac{π^2}{12}+\displaystyle\sum_{k=1}^n (-1)^k\displaystyle\int_0^1 \frac{x^k\log x}{1+x}dx$$右側の積分を計算します。
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{x^k\log x}{1+x}dx=\displaystyle\int_0^1 x^k \log x \displaystyle\sum_{m=0}^{\infty} (-1)^m x^m dx=\displaystyle\sum_{m=0}^{\infty}(-1)^m \displaystyle\int_0^1 x^{k+2}\log xdx\\
&            =-\displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m}{(k+m+1)^2}=-\left\{\frac{1}{(k+1)^2}-\frac{1}{(k+2)^2}+\frac{1}{(k+3)^2}- \cdots\right\}\\
&            =-(-1)^{2k}\left\{\frac{1}{(k+1)^2}-\frac{1}{(k+2)^2}+\frac{1}{(k+3)^2}- \cdots\right\}\\
&            =-(-1)^{k}\left\{\frac{(-1)^k}{(k+1)^2}+\frac{(-1)^{k+1}}{(k+2)^2}+\frac{(-1)^{k+2}}{(k+3)^2}+\cdots\right\}\\
&            =(-1)^{k+1}\left[\left\{1-\frac{1}{2^2}+\frac{1}{3^2}- \cdots +\frac{(-1)^{k-1}}{k^2}+\frac{(-1)^k}{(k+1)^2}+\frac{(-1)^{k+1}}{(k+2)^2}+ \cdots\right\} -\left\{1-\frac{1}{2^2}+\frac{1}{3^2}- \cdots +\frac{(-1)^{k-1}}{k^2}\right\}\right]\\
&            =(-1)^{k+1}\left\{η(2)-\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\}=(-1)^{k+1}\left\{\frac{π^2}{12}-\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\}
\end{alignat}となるので、元の積分計算は
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{\{1+(-1)^nx^{n+1})\log x}{(1+x)^2}dx=-\frac{π^2}{12}+\displaystyle\sum_{k=1}^n (-1)^k (-1)^{k+1}\left\{\frac{π^2}{12}-\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\} \\
&                        =-\frac{π^2}{12}+\displaystyle\sum_{k=1}^n \left\{-\frac{π^2}{12}+\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\} \\
&                        =-\frac{π^2}{12}-\frac{nπ^2}{12}+\displaystyle\sum_{k=1}^n \left\{\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\}\\
&                        =-\frac{(n+1)π^2}{12}+\displaystyle\sum_{k=1}^n \left\{\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\}\\
\end{alignat}右側のシグマを計算します。
\begin{alignat}{2}
&\displaystyle\sum_{k=1}^n \left\{\displaystyle\sum_{m=1}^k \frac{(-1)^{m-1}}{m^2}\right\}=\displaystyle\sum_{k=1}^n \left\{1-\frac{1}{2^2}+\frac{1}{3^2}- \cdots +\frac{(-1)^{k-1}}{k^2}\right\}\\
&                 =1+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^2}+\frac{1}{3^2}\right)+ \cdots +\left\{1-\frac{1}{2^2}+\frac{1}{3^2}- \cdots +\frac{(-1)^{n-1}}{n^2}\right\}\\
&                 =1 \cdot n-\frac{1}{2^2}(n-1)+\frac{1}{3^2}(n-2)- \cdots +\frac{(-1)^{n-2}}{(n-1)^2} \cdot 2 +\frac{(-1)^{n-1}}{n^2}\cdot 1=\displaystyle\sum_{k=1}^n \frac{(-1)^{k-1}(n-k+1)}{k^2}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\{1+(-1)^nx^{n+1})\log x}{(1+x)^2}dx=-\frac{(n+1)π^2}{12}+\displaystyle\sum_{k=1}^n \frac{(-1)^{k-1}(n-k+1)}{k^2}$$







\((3)\) 次の等比数列の和の等式を用います。$$1+x+x^2+ \cdots +x^n=1+\displaystyle\sum_{k=1}^n x^k=\frac{1-x^{n+1}}{1-x}$$この式を代入すると
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{(1-x^{n+1})\log x}{(1-x)^2}dx=\displaystyle\int_0^1 \frac{\log x}{1-x}\left(1+\displaystyle\sum_{k=1}^n x^k\right)dx\\
&                   =\displaystyle\int_0^1 \frac{\log x}{1-x}dx+\displaystyle\sum_{k=1}^n \displaystyle\int_0^1 \frac{x^k\log x}{1-x}dx\\
\end{alignat}\((C)\) を代入すると$$=-\frac{π^2}{6}+\displaystyle\sum_{k=1}^n \displaystyle\int_0^1 \frac{x^k\log x}{1-x}dx$$右側の積分を計算します。
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{x^k\log x}{1-x}dx=\displaystyle\int_0^1 x^k \log x \displaystyle\sum_{m=0}^{\infty} x^m dx=\displaystyle\sum_{m=0}^{\infty} \displaystyle\int_0^1 x^{k+m}\log xdx\\
&            =-\displaystyle\sum_{m=0}^{\infty} \frac{1}{(k+m+1)^2}=-\left\{\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}+\frac{1}{(k+3)^2}+ \cdots\right\}\\
&            =-\left(\displaystyle\sum_{l=1}^{\infty} \frac{1}{l^2}-\displaystyle\sum_{m=1}^k \frac{1}{m^2}\right)=-\frac{π^2}{6}+\displaystyle\sum_{m=1}^k \frac{1}{m^2}
\end{alignat}となるので、元の積分計算は
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{(1-x^{n+1})\log x}{(1-x)^2}dx=-\frac{π^2}{6}+\displaystyle\sum_{k=1}^n \left(-\frac{π^2}{6}+\displaystyle\sum_{m=1}^k \frac{1}{m^2}\right)\\
&                   =-\frac{π^2}{6}-\frac{nπ^2}{6}+\displaystyle\sum_{k=1}^n \left(\displaystyle\sum_{m=1}^k \frac{1}{m^2}\right)\\
&                   =-\frac{(n+1)π^2}{6}+\displaystyle\sum_{k=1}^n \left(\displaystyle\sum_{m=1}^k \frac{1}{m^2}\right)\\
\end{alignat}右側のシグマを計算します。
\begin{alignat}{2}
&\displaystyle\sum_{k=1}^n \left(\displaystyle\sum_{m=1}^k \frac{1}{m^2}\right)=\displaystyle\sum_{k=1}^n \left(1+\frac{1}{2^2}+\frac{1}{3^2}+ \cdots +\frac{1}{k^2}\right)\\
&            =1+\left(1+\frac{1}{2^2}\right)+\left(1+\frac{1}{2^2}+\frac{1}{3^2}\right)+ \cdots +\left\{1+\frac{1}{2^2}+\frac{1}{3^2}+ \cdots +\frac{1}{n^2}\right\}\\
&            =1 \cdot n+\frac{1}{2^2}(n-1)+\frac{1}{3^2}(n-2)+ \cdots +\frac{1}{(n-1)^2} \cdot 2 +\frac{1}{n^2}\cdot 1=\displaystyle\sum_{k=1}^n \frac{n-k+1}{k^2}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(1-x^{n+1})\log x}{(1-x)^2}dx=-\frac{(n+1)π^2}{6}+\displaystyle\sum_{k=1}^n \frac{n-k+1}{k^2}$$

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