AIが出力した積分を解く②
前回に引き続き
Chat-GPTに「もっと難しい定積分の問題作って」と入力して出てきた 問題 を解いたときのメモです。
問題
$$\int_{0}^{1}\frac{x^2\ln^2(x)}{(1-x^2)^2}dx=\frac{1}{8}\left(\pi^2-7\zeta(3)\right)$$
下準備
今回解くのに使う道具をまとめときます。
$$\frac{x^2}{(1-x^2)^2}
=
\frac{1}{4}\left(
\frac{1}{(x-1)^2} + \frac{1}{(x+1)^2} + \frac{1}{x-1} - \frac{1}{x+1}
\right)$$
$$\int_{0}^{1}\ln^2(x)x^ndx = \frac{2}{(n+1)^3}$$
上の部分分数分解をしてから下の式でそれぞれ処理していきます。
解く
$$
\begin{eqnarray}
\int_{0}^{1}\frac{x^2\ln^2(x)}{(1-x^2)^2}dx
&=&
\frac{1}{4}\int_{0}^{1}\ln^2(x)\left(
\frac{1}{(x-1)^2} + \frac{1}{(x+1)^2} + \frac{1}{x-1} - \frac{1}{x+1}
\right)dx\\
&=&
\frac{1}{4}\left(
\int_{0}^{1}\frac{\ln^2(x)}{(x-1)^2}dx+
\int_{0}^{1}\frac{\ln^2(x)}{(x+1)^2}dx+
\int_{0}^{1}\frac{\ln^2(x)}{x-1}dx-
\int_{0}^{1}\frac{\ln^2(x)}{x+1}dx
\right)
\end{eqnarray}
$$
$$
\begin{eqnarray}
\int_{0}^{1}\frac{\ln^2(x)}{(x-1)^2}dx
&=&
\int_{0}^{1}\ln^2(x)(1+x+x^2+\cdots)(1+x+x^2+\cdots)dx\\
&=&
\int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}(n+1)x^n dx\\
&=&
\sum_{n=0}^{\infty}\int_{0}^{1}(n+1)\ln^2(x)x^n dx \\
&=&
\sum_{n=0}^{\infty}\frac{2(n+1)}{(n+1)^3}
=
2\zeta(2)
=
\frac{\pi^2}{3}
\end{eqnarray}
$$
$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{(x+1)^2}dx &=& \int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}(n+1)(-1)^{n}x^n dx\\ &=& \sum_{n=0}^{\infty}\frac{2(n+1)(-1)^{n}}{(n+1)^3}\\ &=& 2\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}-4\sum_{n=1}^{\infty}\frac{1}{(2n)^2} = 2\zeta(2)-\zeta(2) = \frac{\pi^2}{6} \end{eqnarray} $$
$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{x-1}dx &=& -\int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}x^n dx\\ &=& -\sum_{n=0}^{\infty}\frac{2}{(n+1)^3} = -2\zeta(3) \end{eqnarray} $$
$$ \begin{eqnarray} \int_{0}^{1}\frac{\ln^2(x)}{x+1}dx &=& \int_{0}^{1}\ln^2(x)\sum_{n=0}^{\infty}(-1)^nx^n dx\\ &=& \sum_{n=0}^{\infty}\frac{2(-1)^n}{(n+1)^3}\\ &=& 2\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}-4\sum_{n=1}^{\infty}\frac{1}{(2n)^3} = 2\zeta(3)-\frac{1}{2}\zeta(3) = \frac{3}{2}\zeta(3) \end{eqnarray} $$
$$ \therefore \int_{0}^{1}\frac{x^2\ln^2(x)}{(1-x^2)^2}dx = \frac{1}{4}\left( \frac{\pi^2}{3} + \frac{\pi^2}{6} - 2\zeta(3)- \frac{3}{2}\zeta(3) \right) =\frac{1}{8}\left(\pi^2-7\zeta(3)\right) $$