(logx)^2log(1+x)[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 (\log x)^2 \log(1-x)dx=\frac{π^2}{3}-6+2ζ(3)\\
&(2) \displaystyle\int_0^1 (\log x)^2 \log(1+x)dx=\frac{π^2}{6}-6+4 \log 2 +\frac{3}{2}ζ(3)\\
&(3) \displaystyle\int_0^1 (\log x)^3 \log(1-x)dx=-π^2-\frac{π^4}{15}+24-6ζ(3)\\
&(4) \displaystyle\int_0^1 (\log x)^3 \log(1+x)dx=-\frac{π^2}{2}-\frac{7}{120}π^4+24-12 \log 2 -\frac{9}{2}ζ(3)\\
&(5) \displaystyle\int_0^1 \log \left(\frac{1-ax}{1-a}\right) \cdot \frac{1}{\log x}dx=-\displaystyle\sum_{n=1}^{\infty} \frac{a^n \log (n+1)}{n}  (0 \lt a \lt 1)
\end{alignat}







<証明>

予め、途中で用いる次の定積分を計算しておきます。
\begin{alignat}{2}
&(A) \displaystyle\int_0^1 x^n (\log x)^2dx=\frac{2}{(n+1)^3}\\
&(B) \displaystyle\int_0^1 x^n (\log x)^3dx=-\frac{6}{(n+1)^4}
\end{alignat}

\begin{alignat}{2}
&(A) \displaystyle\int_0^1 x^n (\log x)^2dx=\left[\frac{x^{n+1}}{n+1}(\log x)^2\right]_0^1-\displaystyle\int_0^1 \frac{x^{n+1}}{n+1}\cdot 2(\log x) \cdot \frac{1}{x}dx\\
&                  =-\frac{2}{n+1}\displaystyle\int_0^1 x^n \log xdx\\
&                  =-\frac{2}{n+1}\left(\left[\frac{x^{n+1}}{n+1}\log x\right]_0^1-\displaystyle\int_0^1 \frac{x^{n+1}}{n+1} \cdot \frac{1}{x}dt\right)\\
&                  =\frac{2}{(n+1)^2}\displaystyle\int_0^1 x^ndx=\frac{2}{(n+1)^2}\left[\frac{x^{n+1}}{n+1}\right]_0^1=\frac{2}{(n+1)^3}
&\\
&\\
&(B) \displaystyle\int_0^1 x^n (\log x)^3dx=\left[\frac{x^{n+1}}{n+1}(\log x)^3\right]_0^1-\displaystyle\int_0^1 \frac{x^{n+1}}{n+1}\cdot 3(\log x)^2 \cdot \frac{1}{x}dx\\
&                  =-\frac{3}{n+1}\displaystyle\int_0^1 x^n (\log x)^2dx\\
&                  =-\frac{3}{n+1} \cdot \frac{2}{(n+1)^3}=-\frac{6}{(n+1)^4}
\end{alignat}




\((1)\) から \((4)\) まで \(\log (1 \pm x)\) を級数で表します。

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 (\log x)^2 \log(1-x)dx=-\displaystyle\int_0^1 (\log x)^2 \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}dx\\
&                        =-\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\displaystyle\int_0^1 x^n (\log x)^2dx\\
&                        =-\displaystyle\sum_{n=1}^{\infty} \frac{2}{n(n+1)^3}\\
\end{alignat}部分分数分解を行います。$$=-2\displaystyle\sum_{n=1}^{\infty} \left\{\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)^3}\right\}$$それぞれの級数は
\begin{alignat}{2}
&\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1\\
&\displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1)^2}=ζ(2)-1,  \displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1)^3}=ζ(3)-1
\end{alignat}となるので$$=-2\{1-ζ(2)+1-ζ(3)+1\}=2ζ(2)+2ζ(3)-6=\frac{π^2}{3}+2ζ(3)-6$$以上より$$\displaystyle\int_0^1 (\log x)^2 \log(1-x)dx=\frac{π^2}{3}-6+2ζ(3)$$







\begin{alignat}{2}
&(2) \displaystyle\int_0^1 (\log x)^2 \log(1+x)dx=-\displaystyle\int_0^1 (\log x)^2 \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}dx\\
&                        =\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\displaystyle\int_0^1 x^n (\log x)^2dx\\
&                        =\displaystyle\sum_{n=1}^{\infty} \frac{2(-1)^{n-1}}{n(n+1)^3}\\
\end{alignat}部分分数分解を行います。$$=2\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\left\{\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)^3}\right\}$$それぞれの級数は
\begin{alignat}{2}
&\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)- \cdots\\
&                      =1+2\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \cdots \right)\\
&                      =-1+2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \cdots \right)\\
&                      =-1+2 \log 2\\
&\\
&\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{(n+1)^2}=η(2)-1=\frac{1}{2}ζ(2)-1=\frac{π^2}{12}-1\\
&\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{(n+1)^3}=η(3)-1=\frac{3}{4}ζ(3)-1
\end{alignat}となるので$$=2\left\{-1+2 \log2 +\frac{π^2}{12}-1+\frac{3}{4}ζ(3)-1\right\}=\frac{π^2}{6}-6+4 \log 2 +\frac{3}{2}ζ(3)$$以上より$$\displaystyle\int_0^1 (\log x)^2 \log(1+x)dx=\frac{π^2}{6}-6+4 \log 2 +\frac{3}{2}ζ(3)$$







\begin{alignat}{2}
&(3) \displaystyle\int_0^1 (\log x)^3 \log(1-x)dx=-\displaystyle\int_0^1 (\log x)^3 \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}dx\\
&                        =-\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\displaystyle\int_0^1 x^n (\log x)^3dx\\
&                        =\displaystyle\sum_{n=1}^{\infty} \frac{6}{n(n+1)^4}\\
\end{alignat}部分分数分解を行います。$$=-2\displaystyle\sum_{n=1}^{\infty} \left\{\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)^3}-\frac{1}{(n+1)^4}\right\}$$それぞれの級数は
\begin{alignat}{2}
&\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1,  \displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1)^2}=ζ(2)-1\\
&\displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1)^3}=ζ(3)-1,  \displaystyle\sum_{n=1}^{\infty} \frac{1}{(n+1)^4}=ζ(4)-1
\end{alignat}となるので$$=6\{1-ζ(2)+1-ζ(3)+1-ζ(4)+1\}=6\left\{4-\frac{π^2}{6}-ζ(3)-\frac{π^4}{90}\right\}=24-π^2-6ζ(3)-\frac{π^4}{15}$$以上より$$\displaystyle\int_0^1 (\log x)^3 \log(1-x)dx=-π^2-\frac{π^4}{15}+24-6ζ(3)$$







\begin{alignat}{2}
&(4) \displaystyle\int_0^1 (\log x)^3 \log(1+x)dx=\displaystyle\int_0^1 (\log x)^3 \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}dx\\
&                        =\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\displaystyle\int_0^1 x^n (\log x)^3dx\\
&                        =\displaystyle\sum_{n=1}^{\infty} \frac{6(-1)^n}{n(n+1)^4}\\
\end{alignat}部分分数分解を行います。$$=6\displaystyle\sum_{n=1}^{\infty}(-1)^n \left\{\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)^3}-\frac{1}{(n+1)^4}\right\}$$それぞれの級数は
\begin{alignat}{2}
&\displaystyle\sum_{n=1}^{\infty}(-1)^n\left(\frac{1}{n}-\frac{1}{n+1}\right)=-\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)+ \cdots\\
&                      =1-2\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \cdots \right)\\
&                      =1-2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \cdots \right)\\
&                      =1-2 \log 2\\
&\\
&\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{(n+1)^2}=η(2)-1=\frac{1}{2}ζ(2)-1=\frac{π^2}{12}-1\\
&\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{(n+1)^3}=η(3)-1=\frac{3}{4}ζ(3)-1\\
&\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{(n+1)^4}=η(4)-1=\frac{7}{8}ζ(4)-1=\frac{7}{720}π^4-1
\end{alignat}
となるので$$=6\left\{1-2 \log 2-\frac{π^2}{12}+1-\frac{3}{4}ζ(3)+1-\frac{7}{720}π^4+1\right\}=24-12 \log 2 -\frac{π^2}{2}-\frac{9}{2}ζ(3)-\frac{7}{120}π^4$$以上より$$\displaystyle\int_0^1 (\log x)^3 \log(1+x)dx=-\frac{π^2}{2}-\frac{7}{120}π^4+24-12 \log 2 -\frac{9}{2}ζ(3)$$







\((5)\) \(\log\) を切り離してから級数で表します。
\begin{alignat}{2}
&\displaystyle\int_0^1 \log \left(\frac{1-ax}{1-a}\right) \cdot \frac{1}{\log x}dx=\displaystyle\int_0^1 \frac{\log(1-ax)-\log (1-a)}{\log x}dx\\
&                       =-\displaystyle\int_0^1 \frac{1}{\log x} \left\{\displaystyle\sum_{n=1}^{\infty} \frac{(ax)^n}{n}-\displaystyle\sum_{n=1}^{\infty} \frac{a^n}{n}\right\}dx\\
&                       =-\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{n}\displaystyle\int_0^1 \frac{x^n-1}{\log x}dx
\end{alignat}ここで$$\displaystyle\int_0^n x^a da=\left[\frac{x^a}{\log x}\right]_0^n=\frac{x^n-1}{\log x}$$であるから
\begin{alignat}{2}
&=-\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{n}\displaystyle\int_0^1 \left(\displaystyle\int_0^n x^a da\right)dx=-\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{n}\displaystyle\int_0^n \displaystyle\int_0^1 x^a dxda\\
&=-\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{n}\displaystyle\int_0^n\left[\frac{x^{a+1}}{a+1}\right]_0^1da=-\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{n}\displaystyle\int_0^n\frac{1}{a+1}da\\
&=-\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{n}[\log(a+1)]_0^n=-\displaystyle\sum_{n=1}^{\infty} \frac{a^n \log (n+1)}{n}
\end{alignat}以上より$$\displaystyle\int_0^1 \log \left(\frac{1-ax}{1-a}\right) \cdot \frac{1}{\log x}dx=-\displaystyle\sum_{n=1}^{\infty} \frac{a^n \log (n+1)}{n}$$

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