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

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{x(\log x)^{2n-1}}{1-x^2}dx=-\frac{π^{2n}}{4n}|B_{2n}|\\
&(2) \displaystyle\int_0^1 \frac{(1+x^2)(\log x)^{2n}}{(1-x^2)^2}dx=\frac{2^{2n}-1}{2}π^{2n}|B_{2n}|\\
\end{alignat}ただし、全て \(n \in \mathrm{N}\)








<証明>

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{x(\log x)^{2n-1}}{1-x^2}dx\\
&=\displaystyle\int_0^1 x(\log x)^{2n-1}\displaystyle\sum_{m=0}^{\infty} x^{2m}dx=\displaystyle\sum_{m=0}^{\infty} \displaystyle\int_0^1 x^{2m+1}(\log x)^{2n-1}dx\\
&=\displaystyle\sum_{m=0}^{\infty} \frac{(2n-1)!(-1)^{2n-1}}{(2m+2)^{2n}}=-\frac{(2n-1)!}{2^{2n}}\displaystyle\sum_{m=0}^{\infty} \frac{1}{(m+1)^{2n}}\\
&=-\frac{(2n-1)!}{2^{2n}}ζ(2n)=-\frac{(2n-1)!}{2^{2n}} \cdot \frac{(2π)^{2n}}{2 \cdot (2n)!}|B_{2n}|=-\frac{π^{2n}}{4n}|B_{2n}|
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x(\log x)^{2n-1}}{1-x^2}dx=-\frac{π^{2n}}{4n}|B_{2n}|$$







\begin{alignat}{2}
&(2)  \displaystyle\int_0^1 \frac{(1+x^2)(\log x)^{2n}}{(1-x^2)^2}dx\\
&=\displaystyle\int_0^1 (1+x^2)(\log x)^{2n}\displaystyle\sum_{m=1}^{\infty} mx^{2(m-1)}dx\\
&=\displaystyle\sum_{m=1}^{\infty} m \displaystyle\int_0^1 x^{2(m-1)}(1+x^2)(\log x)^{2n}dx\\
&=\displaystyle\sum_{m=1}^{\infty} \left\{\frac{(2n)!(-1)^{2n}}{(2m-1)^{2n+1}}+\frac{(2n)!(-1)^{2n}}{(2m+1)^{2n+1}}\right\}\\
&=(2n)!\displaystyle\sum_{m=1}^{\infty} m\left\{\frac{1}{(2m-1)^{2n+1}}+\frac{1}{(2m+1)^{2n+1}}\right\}\\
&=(2n)!\left\{1 \cdot \left(1+\frac{1}{3^{2n+1}}\right)+2 \cdot \left(\frac{1}{3^{2n+1}}+\frac{1}{5^{2n+1}}\right)+ 3 \cdot \left(\frac{1}{5^{2n+1}}+\frac{1}{7^{2n+1}}\right)+ \cdots\right\}\\
&=(2n)!\left(1+\frac{1}{3^{2n}}+\frac{1}{5^{2n}}+\frac{1}{7^{2n}}+ \cdots\right)\\
&=(2n)!\displaystyle\sum_{m=0}^{\infty} \frac{1}{(2m+1)^{2n}}=(2n)!\left(1-\frac{1}{2^{2n}}\right)ζ(2n)\\
&=(2n)!\left(1-\frac{1}{2^{2n}}\right) \cdot \frac{(2π)^{2n}}{2(2n)!}|B_{2n}|=\frac{2^{2n}-1}{2}π^{2n}|B_{2n}|
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(1+x^2)(\log x)^{2n}}{(1-x^2)^2}dx=\frac{2^{2n}-1}{2}π^{2n}|B_{2n}|$$

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