xlog(1-x)log(1+x)[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 x \log (1-x) \log (1+x)dx=\frac{1}{4}-\log 2\\
&(2) \displaystyle\int_0^1 x^2 \log (1-x) \log (1+x)dx=\frac{17}{27}-\frac{π^2}{18}-\frac{8}{9}\log 2+\frac{1}{3}(\log 2)^2\\
&(3) \displaystyle\int_0^1 x^3 \log (1-x) \log (1+x)dx=\frac{13}{96}-\frac{2 }{3}\log 2\\
\end{alignat}






次の定積分の結果を用います。(詳細はこちらです。)
\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \log x \log(2-x)dx=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2\\
&(2) \displaystyle\int_0^1 x\log x \log(2-x)dx=\frac{7}{4}-\frac{π^2}{6}- \log 2+(\log 2)^2\\
&(3) \displaystyle\int_0^1 x^2\log x \log(2-x)dx=\frac{115}{54}-\frac{2}{9}π^2- \frac{8}{9}\log 2+\frac{4}{3}(\log 2)^2\\
&(4) \displaystyle\int_0^1 x^3\log x \log(2-x)dx=\frac{865}{288}-\frac{π^2}{3}- \log 2+2(\log 2)^2\\
\end{alignat}





<証明>

\((1)\) から \((3)\) まで \(1-x=t\) と置きます。\((dx=-dt)\)

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 x \log (1-x) \log (1+x)dx\\
&=\displaystyle\int_1^0 (1-t) \log t \log (2-t)(-dt)\\
&=\displaystyle\int_0^1 (1-x) \log x \log (2-x)dx\\
&=\displaystyle\int_0^1 \log x \log (2-x)dx-\displaystyle\int_0^1 x \log x \log (2-x)dx\\
&=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2-\left\{\frac{7}{4}-\frac{π^2}{6}- \log 2+(\log 2)^2\right\}\\
&=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2-\frac{7}{4}+\frac{π^2}{6}+ \log 2-(\log 2)^2\\
&=\frac{1}{4}-\log 2
\end{alignat}以上より$$\displaystyle\int_0^1 x \log (1-x) \log (1+x)dx=\frac{1}{4}-\log 2$$






\begin{alignat}{2}
&(2) \displaystyle\int_0^1 x^2 \log (1-x) \log (1+x)dx\\
&=\displaystyle\int_1^0 (1-t)^2 \log t \log (2-t)(-dt)\\
&=\displaystyle\int_0^1 (1-x)^2 \log x \log (2-x)dx\\
&=\displaystyle\int_0^1 (1-2x+x^2) \log x \log (2-x)dx\\
&=\displaystyle\int_0^1 \log x \log (2-x)dx-2\displaystyle\int_0^1 x \log x \log (2-x)dx+\displaystyle\int_0^1 x^2 \log x \log (2-x)dx\\
&=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2-2\left\{\frac{7}{4}-\frac{π^2}{6}- \log 2+(\log 2)^2\right\}+\frac{115}{54}-\frac{2}{9}π^2- \frac{8}{9}\log 2+\frac{4}{3}(\log 2)^2\\
&=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2-\frac{7}{2}+\frac{π^2}{3}+2 \log 2-2(\log 2)^2+\frac{115}{54}-\frac{2}{9}π^2- \frac{8}{9}\log 2+\frac{4}{3}(\log 2)^2\\
&=\frac{17}{27}-\frac{π^2}{18}-\frac{8}{9}\log 2+\frac{1}{3}(\log 2)^2
\end{alignat}以上より$$\displaystyle\int_0^1 x^2 \log (1-x) \log (1+x)dx=\frac{17}{27}-\frac{π^2}{18}-\frac{8}{9}\log 2+\frac{1}{3}(\log 2)^2$$






\begin{alignat}{2}
&(3) \displaystyle\int_0^1 x^3 \log (1-x) \log (1+x)dx\\
&=\displaystyle\int_1^0 (1-t)^3 \log t \log (2-t)(-dt)\\
&=\displaystyle\int_0^1 (1-x)^3 \log x \log (2-x)dx\\
&=\displaystyle\int_0^1 (1-3x+3x^2-x^3) \log x \log (2-x)dx\\
&=\displaystyle\int_0^1 \log x \log (2-x)dx-3\displaystyle\int_0^1 x \log x \log (2-x)dx+3\displaystyle\int_0^1 x^2 \log x \log (2-x)dx-\displaystyle\int_0^1 x^3 \log x \log (2-x)dx\\
&=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2-3\left\{\frac{7}{4}-\frac{π^2}{6}- \log 2+(\log 2)^2\right\}+3\left\{\frac{115}{54}-\frac{2}{9}π^2- \frac{8}{9}\log 2+\frac{4}{3}(\log 2)^2\right\}-\left\{\frac{865}{288}-\frac{π^2}{3}- \log 2+2(\log 2)^2\right\}\\
&=2-\frac{π^2}{6}-2 \log 2+(\log 2)^2-\frac{21}{4}+\frac{π^2}{2}+3 \log 2-3(\log 2)^2+\frac{115}{18}-\frac{2}{3}π^2- \frac{8}{3}\log 2+4(\log 2)^2-\frac{865}{288}+\frac{π^2}{3}+ \log 2-2(\log 2)^2\\
&=2-\frac{21}{4}+\frac{115}{18}-\frac{865}{288}+\left(-2+3-\frac{8}{3}+1\right) \log 2\\
&=\frac{576-1512+1840-865}{288}+\frac{-6+9-8+3}{3}\log 2\\
&=\frac{13}{96}-\frac{2 }{3}\log 2
\end{alignat}以上より$$\displaystyle\int_0^1 x^3 \log (1-x) \log (1+x)dx=\frac{13}{96}-\frac{2 }{3}\log 2$$

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