(logx)^2(1-x^{n+1})/(1-x)^2[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{(\log x)^2(1-x^{n+1})}{(1-x)^2}dx=2(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{n-k+1}{k^3}\\
&(2)  \displaystyle\int_0^1 \frac{(\log x)^2\{1+(-1)^nx^{n+1}\}}{(1+x)^2}dx=\frac{3}{2}(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{(-1)^{k-1}(n-k+1)}{k^3}\\
&(3)  \displaystyle\int_0^1 \frac{(\log x)^2(1-x^{2n+2})}{(1-x^2)^2}dx=\frac{7}{4}(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{n-k+1}{(2k-1)^3}\\
\end{alignat}ただし、全て \(n \in \mathrm{N}\)












<証明>

次の定積分の結果を用います。[詳細はこちらです。(A)(B)(C)(D)(E)(F)]
\begin{alignat}{2}
&(A)  \displaystyle\int_0^1 \frac{(\log x)^2}{1-x}dx=2ζ(3)\\
&(B)  \displaystyle\int_0^1 \frac{(\log x)^2}{1+x}dx=\frac{3}{2}ζ(3)\\
&(C)  \displaystyle\int_0^1 \frac{(\log x)^2}{1-x^2}dx=\frac{7}{4}ζ(3)\\
&(D)  \displaystyle\int_0^1 \frac{x^n(\log x)^2}{1+x}dx=(-1)^n\left\{\frac{3}{2}ζ(3)+2\displaystyle\sum_{k=1}^n \frac{(-1)^k}{k^3}\right\}\\
&(E)  \displaystyle\int_0^1 \frac{x^n(\log x)^2}{1-x}dx=2\left\{ζ(3)-\displaystyle\sum_{k=1}^n \frac{1}{k^3}\right\}\\
&(F)  \displaystyle\int_0^1 \frac{x^{2n}(\log x)^2}{1-x^2}dx=\frac{7}{4}ζ(3)-2\displaystyle\sum_{k=1}^n \frac{1}{(2k-1)^3}\\
\end{alignat}ただし、全て \(n \in \mathrm{N}\)




また、次の等比数列の等式を被積分関数に代入します。
\begin{alignat}{2}
&(α)  1+x+x^2+ \cdot +x^n=\frac{1-x^{n+1}}{1-x}=\displaystyle\sum_{m=0}^n x^m\\
&(β)  1-x+x^2- \cdots +(-1)^nx^n=\frac{1+(-1)^nx^{n+1}}{1+x}=\displaystyle\sum_{m=0}^n (-1)^m x^m\\
&(γ)  1+x^2+x^4+ \cdots +x^{2n}=\frac{1-x^{2n+2}}{1-x^2}=\displaystyle\sum_{m=0}^n x^{2m}\\
\end{alignat}






\begin{alignat}{2}
(1)  \displaystyle\int_0^1 \frac{(\log x)^2(1-x^{n+1})}{(1-x)^2}dx&=\displaystyle\int_0^1 \frac{(\log x)^2}{1-x} \cdot \frac{1-x^{n+1}}{1-x}dx=\displaystyle\int_0^1 \frac{(\log x)^2}{1-x} \left(\displaystyle\sum_{m=0}^n x^m\right)dx\\
&=\displaystyle\sum_{m=0}^n \displaystyle\int_0^1 \frac{x^m (\log x)^2}{1-x}dx=\displaystyle\int_0^1 \frac{(\log x)^2}{1-x}dx+\displaystyle\sum_{m=1}^n\displaystyle\int_0^1 \frac{x^m (\log x)^2}{1-x}dx\\
&=2ζ(3)+\displaystyle\sum_{m=1}^n\left\{2ζ(3)-2 \displaystyle\sum_{k=0}^{m} \frac{1}{k^3}\right\}\\
&=2ζ(3)-2nζ(3)-2 \displaystyle\sum_{m=1}^n\left(1+\frac{1}{2^3}+\frac{1}{3^3}+ \cdots +\frac{1}{m^3}\right)\\
&=2(n+1)ζ(3)-2 \left\{1+\left(1+\frac{1}{2^3}\right)+\left(1+\frac{1}{2^3}+\frac{1}{3^3}\right)+ \cdots +\left(1+\frac{1}{2^3}+\frac{1}{3^3}+ \cdots +\frac{1}{n^3}\right)\right\}\\
&=2(n+1)ζ(3)-2\left\{\frac{n}{1^3}+\frac{n-1}{2^3}+\frac{n-2}{3^3}+ \cdots +\frac{2}{(n-1)^3}+\frac{1}{n^3}\right\}\\
&=2(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{n-k+1}{k^3}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(\log x)^2(1-x^{n+1})}{(1-x)^2}dx=2(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{n-k+1}{k^3}$$








\begin{alignat}{2}
(2)  \displaystyle\int_0^1 \frac{(\log x)^2\{1+(-1)^nx^{n+1}\}}{(1+x)^2}dx&=\displaystyle\int_0^1 \frac{(\log x)^2}{1+x} \cdot \frac{1+(-1)^nx^{n+1}}{1+x}dx=\displaystyle\int_0^1 \frac{(\log x)^2}{1+x} \left\{\displaystyle\sum_{m=0}^n (-1)^mx^m\right\}dx\\
&=\displaystyle\sum_{m=0}^n (-1)^m\displaystyle\int_0^1 \frac{x^m (\log x)^2}{1+x}dx=\displaystyle\int_0^1 \frac{(\log x)^2}{1+x}dx+\displaystyle\sum_{m=1}^n(-1)^m\displaystyle\int_0^1 \frac{x^m (\log x)^2}{1+x}dx\\
&=\frac{3}{2}ζ(3)+\displaystyle\sum_{m=1}^n(-1)^m\left[(-1)^{m}\left\{\frac{3}{2}ζ(3)+2 \displaystyle\sum_{k=1}^{m} \frac{(-1)^k}{k^3}\right\}\right]\\
&=\frac{3}{2}ζ(3)+\frac{3}{2}nζ(3)-2 \displaystyle\sum_{m=1}^n\left\{1-\frac{1}{2^3}+\frac{1}{3^3}- \cdots +\frac{(-1)^{m-1}}{m^3}\right\}\\
&=\frac{3}{2}(n+1)ζ(3)-2 \left[1+\left(1-\frac{1}{2^3}\right)+\left(1-\frac{1}{2^3}+\frac{1}{3^3}\right)+ \cdots +\left\{1-\frac{1}{2^3}+\frac{1}{3^3}- \cdots +\frac{(-1)^{n-1}}{n^3}\right\}\right]\\
&=\frac{3}{2}(n+1)ζ(3)-2\left\{\frac{n}{1^3}-\frac{n-1}{2^3}+\frac{n-2}{3^3}- \cdots +\frac{(-1)^{n-2} \cdot 2}{(n-1)^3}+\frac{(-1)^{n-1}}{n^3}\right\}\\
&=\frac{3}{2}(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{(-1)^{k-1}(n-k+1)}{k^3}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(\log x)^2\{1+(-1)^nx^{n+1}\}}{(1+x)^2}dx=\frac{3}{2}(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{(-1)^{k-1}(n-k+1)}{k^3}$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^1 \frac{(\log x)^2(1-x^{2n+2})}{(1-x^2)^2}dx&=\displaystyle\int_0^1 \frac{(\log x)^2}{1-x^2} \cdot \frac{1-x^{2n+2}}{1-x^2}dx=\displaystyle\int_0^1 \frac{(\log x)^2}{1-x^2} \left(\displaystyle\sum_{m=0}^n x^{2m}\right)dx\\
&=\displaystyle\sum_{m=0}^n \displaystyle\int_0^1 \frac{x^m (\log x)^2}{1-x^2}dx=\displaystyle\int_0^1 \frac{(\log x)^2}{1-x^2}dx+\displaystyle\sum_{m=1}^n\displaystyle\int_0^1 \frac{x^{2m} (\log x)^2}{1-x^2}dx\\
&=\frac{7}{4}ζ(3)+\displaystyle\sum_{m=1}^n\left\{\frac{7}{4}ζ(3)-2 \displaystyle\sum_{k=0}^{m-1} \frac{1}{(2k+1)^3}\right\}\\
&=\frac{7}{4}ζ(3)+\frac{7}{4}nζ(3)-2 \displaystyle\sum_{m=1}^n\left\{1+\frac{1}{3^3}+\frac{1}{5^3}+ \cdots +\frac{1}{(2m-1)^3}\right\}\\
&=\frac{7}{4}(n+1)ζ(3)-2 \left[1+\left(1+\frac{1}{3^3}\right)+\left(1+\frac{1}{3^3}+\frac{1}{5^3}\right)+ \cdots +\left\{1+\frac{1}{3^3}+\frac{1}{5^3}+ \cdots +\frac{1}{(2n-1)^3}\right\}\right]\\
&=\frac{7}{4}(n+1)ζ(3)-2\left\{\frac{n}{1^3}+\frac{n-1}{3^3}+\frac{n-2}{5^3}+ \cdots +\frac{2}{(2n-3)^3}+\frac{1}{(2n-1)^3}\right\}\\
&=\frac{7}{4}(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{n-k+1}{(2k-1)^3}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(\log x)^2(1-x^{2n+2})}{(1-x^2)^2}dx=\frac{7}{4}(n+1)ζ(3)-2 \displaystyle\sum_{k=1}^n \frac{n-k+1}{(2k-1)^3}$$

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