中册 5.1 反常积分计算 第8题

\section*{解题过程：} （1）由于 $\displaystyle \lim _{x \rightarrow 1^{-}} \frac{\ln x}{1-x}=\lim _{x \rightarrow 1^{-}} \frac{\frac{1}{x}}{-1}=-1$ ，故 $x=1$ 不是取点，因而只有 $x=0$ 为㻓点． 又 $\displaystyle \lim _{x \rightarrow 0^{+}} x^{\frac{1}{2}} \frac{\ln x}{x-1}=0$ ，由于 $\displaystyle p=\frac{1}{2}, \lambda=0$ ，故积分 $\displaystyle \int_{0}^{1} \frac{\ln x}{1-x} \mathrm{~d} x$ 收玫。 令 $1-x=t$ ，则 $\displaystyle \int_{0}^{1} \frac{\ln x}{1-x} \mathrm{~d} x=\int_{0}^{1} \frac{\ln (1-t)}{t} \mathrm{~d} t=-\int_{0}^{1} \sum_{n=1}^{\infty} \frac{t^{n-1}}{n} \mathrm{~d} t=-\sum_{n=1}^{\infty} \frac{1}{n^{2}}=-\frac{\pi^{2}}{6}$ ． （2）由于 $\displaystyle \ln (1+x)=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{n}}{n}$ ，所以 $$ \int_{0}^{1} \frac{\ln (1+x)}{x} \mathrm{~d} x=\int_{0}^{1} \sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n} x^{n-1} \mathrm{~d} x=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n^{2}}=\sum_{k=1}^{\infty} \frac{1}{(2 k-1)^{2}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}}=\frac{\pi^{2}}{8}-\frac{1}{4} \cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{12} . $$ （3）由于 $\displaystyle \ln (1-x)=-\sum_{n=1}^{\infty} \frac{x^{n}}{n}$ ，所以 $\displaystyle I=\int_{0}^{1} \frac{\ln (1-x)}{x} \mathrm{~d} x=-\int_{0}^{1} \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} \mathrm{~d} x=-\sum_{n=1}^{\infty} \frac{1}{n^{2}}=-\frac{\pi^{2}}{6}$ ． （4） $\displaystyle \int_{0}^{1} \frac{\ln x}{1-x^{2}} \mathrm{~d} x=\frac{1}{2}\left(\int_{0}^{1} \frac{\ln x}{1-x} \mathrm{~d} x+\int_{0}^{1} \frac{\ln x}{1+x} \mathrm{~d} x\right)$ ． 而 $$ \begin{aligned} \int_{0}^{1} \frac{\ln x}{1+x} \mathrm{~d} x & =\int_{0}^{1} \sum_{n=0}^{\infty}(-1)^{n} x^{n} \ln x \mathrm{~d} x=-\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{(n+1)^{2}}=\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n^{2}} \\ & =-\left(\sum_{k=1}^{\infty} \frac{1}{(2 k-1)^{2}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}}\right)=-\frac{\pi^{2}}{12} \end{aligned} $$ 所以 $\displaystyle \quad \int_{0}^{1} \frac{\ln x}{1-x^{2}} \mathrm{~d} x=\frac{1}{2}\left(-\frac{\pi^{2}}{6}-\frac{\pi^{2}}{12}\right)=-\frac{3 \pi^{2}}{24}$ ． （5）令 $\mathrm{e}^{-x}=t$ ，则 $$ \begin{aligned} \int_{0}^{+\infty} \frac{x}{\mathrm{e}^{x}+1} \mathrm{~d} x & =-\int_{0}^{+\infty} \frac{x}{\mathrm{e}^{-x}+1} \mathrm{de}^{-x}=-\int_{0}^{1} \frac{\ln t}{t+1} \mathrm{~d} t=\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n^{2}} \\ & =\sum_{k=1}^{\infty} \frac{1}{(2 k-1)^{2}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}}=\frac{\pi^{2}}{8}-\frac{1}{4} \cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{12} \end{aligned} $$ （6）令 $t=\mathrm{e}^{-x}$ ，利用 $\displaystyle \ln (1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{n}$ 及 $\displaystyle \sum \frac{1}{(2 k-1)^{2}}=\frac{\pi^{2}}{8}$ ，逐项积分有 $$ \begin{aligned} \int_{0}^{+\infty} \ln \left(1+\mathrm{e}^{-x}\right) \mathrm{d} x & =\int_{0}^{1} \frac{\ln (1+t)}{t} \mathrm{~d} t=\int_{0}^{1} \sum(-1)^{n+1} \frac{t^{n-1}}{n} \mathrm{~d} t=\sum \frac{(-1)^{n+1}}{n^{2}} \\ & =\sum_{k=1}^{\infty} \frac{1}{(2 k-1)^{2}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}}=\frac{\pi^{2}}{8}-\frac{1}{4} \cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{12} \end{aligned} $$