4北海道大学 2001年 数学 第4問解答・解説

$$\displaystyle \begin{aligned}\frac{1}{1-x^2}&=\frac12\left(\frac{1}{1+x}+\frac{1}{1-x}\right)\quad(-1<x<1),\\ \int_0^a\frac{dx}{1-x^2}&=\frac12\left[\log(1+x)-\log(1-x)\right]_0^a=\frac12\log\frac{1+a}{1-a}.\qquad\text{(1)}\\[4pt]\frac{x^{2n+2}}{1-x^2}&=\frac{1}{1-x^2}-\sum\limits _{k=0}^{n}x^{2k},\\ \int_0^a\frac{x^{2n+2}}{1-x^2}\,dx&=\int_0^a\frac{dx}{1-x^2}-\sum\limits _{k=0}^{n}\frac{a^{2k+1}}{2k+1}\\ &=\frac12\log\frac{1+a}{1-a}-\sum\limits _{k=0}^{n}\frac{a^{2k+1}}{2k+1}.\qquad\text{(2)}\\ I_n(a)&=\int_0^a\frac{x^{2n+2}}{1-x^2}\,dx,\qquad b=|a|<1,\\ |I_n(a)|&=\int_0^b\frac{t^{2n+2}}{1-t^2}\,dt\leq\frac{b^{2n+3}}{1-b^2}\longrightarrow0.\\ \text{(2)より}\quad \sum\limits _{k=0}^{n}\frac{a^{2k+1}}{2k+1}&=\frac12\log\frac{1+a}{1-a}-I_n(a)\longrightarrow\frac12\log\frac{1+a}{1-a}.\\ \therefore\quad \log\frac{1+a}{1-a}&=2\sum\limits _{k=0}^{\infty}\frac{a^{2k+1}}{2k+1}.\qquad\text{(3)}\end{aligned}$$