AIが出力した積分を解く②

$$ \begin{eqnarray} \int_{0}^{1}\frac{x^2\ln^2(x)}{(1-x^2)^2}dx &=& \frac{1}{4}\int_{0}^{1}\ln^2(x)\left( \frac{1}{(x-1)^2} + \frac{1}{(x+1)^2} + \frac{1}{x-1} - \frac{1}{x+1} \right)dx\\ &=& \frac{1}{4}\left( \int_{0}^{1}\frac{\ln^2(x)}{(x-1)^2}dx+ \int_{0}^{1}\frac{\ln^2(x)}{(x+1)^2}dx+ \int_{0}^{1}\frac{\ln^2(x)}{x-1}dx- \int_{0}^{1}\frac{\ln^2(x)}{x+1}dx \right) \end{eqnarray} $$
$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{(x-1)^2}dx &=& \int_{0}^{1}\ln^2(x)(1+x+x^2+\cdots)(1+x+x^2+\cdots)dx\\ &=& \int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}(n+1)x^n dx\\ &=& \sum_{n=0}^{\infty}\int_{0}^{1}(n+1)\ln^2(x)x^n dx \\ &=& \sum_{n=0}^{\infty}\frac{2(n+1)}{(n+1)^3} = 2\zeta(2) = \frac{\pi^2}{3} \end{eqnarray} $$

$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{(x+1)^2}dx &=& \int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}(n+1)(-1)^{n}x^n dx\\ &=& \sum_{n=0}^{\infty}\frac{2(n+1)(-1)^{n}}{(n+1)^3}\\ &=& 2\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}-4\sum_{n=1}^{\infty}\frac{1}{(2n)^2} = 2\zeta(2)-\zeta(2) = \frac{\pi^2}{6} \end{eqnarray} $$

$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{x-1}dx &=& -\int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}x^n dx\\ &=& -\sum_{n=0}^{\infty}\frac{2}{(n+1)^3} = -2\zeta(3) \end{eqnarray} $$

$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{x+1}dx &=& \int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}(-1)^nx^n dx\\ &=& \sum_{n=0}^{\infty}\frac{2(-1)^n}{(n+1)^3}\\ &=& 2\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}-4\sum_{n=1}^{\infty}\frac{1}{(2n)^3} = 2\zeta(3)-\frac{1}{2}\zeta(3) = \frac{3}{2}\zeta(3) \end{eqnarray} $$

$$ \therefore \int_{0}^{1}\frac{x^2\ln^2(x)}{(1-x^2)^2}dx = \frac{1}{4}\left( \frac{\pi^2}{3} + \frac{\pi^2}{6} - 2\zeta(3)- \frac{3}{2}\zeta(3) \right) =\frac{1}{8}\left(\pi^2-7\zeta(3)\right) $$