中册 5.1 反常积分计算 第7题

\section*{解题过程：} （1）由于 $$ \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \mathrm{~d} x \xlongequal{t=x^{2}} \int_{0}^{1} \frac{t}{\sqrt{1-t}} \frac{1}{2} \frac{1}{\sqrt{t}} \mathrm{~d} t=\frac{1}{2} \int_{0}^{1} t^{\frac{1}{2}}(1-t)^{-\frac{1}{2}} \mathrm{~d} t=\frac{1}{2} \mathrm{~B}\left(\frac{1}{2}, \frac{3}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{3}{2}\right)}{2 \Gamma(2)}=\frac{\pi}{4}, $$ 所以 $\displaystyle \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \mathrm{~d} x$ 收玫，其值为 $\displaystyle \frac{\pi}{4}$ ． （2）由于 $$ \int_{1}^{+\infty} \frac{2-x^{2}}{x^{3} \sqrt{x^{2}-1}} \mathrm{~d} x \xlongequal{t=x^{2}} \frac{1}{2} \int_{1}^{+\infty} \frac{2-t}{t^{2} \sqrt{t-1}} \mathrm{~d} t \xlongequal{u=\sqrt{t-1}} \int_{0}^{+\infty} \frac{1-u^{2}}{\left(1+u^{2}\right)^{2}} \mathrm{~d} u=\int_{0}^{+\infty} \mathrm{d} \frac{u}{1+u^{2}}=\left.\frac{u}{1+u^{2}}\right|_{0} ^{+\infty}=0 . $$ 所以 $\displaystyle \int_{1}^{+\infty} \frac{2-x^{2}}{x^{3} \sqrt{x^{2}-1}} \mathrm{~d} x$ 收玫，其值为 0 。 （3）令 $t=-\ln x$ ，则 $\displaystyle \int_{0}^{1} \frac{\ln x}{x^{\alpha}} \mathrm{d} x=-\int_{0}^{+\infty} t \mathrm{e}^{(\alpha-1) t} \mathrm{~d} t=-\frac{1}{(1-\alpha)^{2}}$ ． （4）作三角变换 $x=\tan t$ ，则 $\displaystyle \int_{0}^{+\infty} \frac{\arctan x}{\left(1+x^{2}\right)^{\frac{3}{2}}} \mathrm{~d} x=\int_{0}^{\frac{\pi}{2}} \frac{t \sec ^{2} t}{\sec ^{3} t} \mathrm{~d} t=\int_{0}^{\frac{\pi}{2}} t \cos t \mathrm{~d} t=\frac{\pi}{2}-1$ ． （5） $\displaystyle \int_{0}^{+\infty} \frac{x \mathrm{e}^{-x}}{\left(1+\mathrm{e}^{-x}\right)^{2}} \mathrm{~d} x=\int_{0}^{+\infty} \frac{x \mathrm{e}^{x}}{\left(1+\mathrm{e}^{x}\right)^{2}} \mathrm{~d} x=-\int_{0}^{+\infty} x \mathrm{~d} \frac{1}{1+\mathrm{e}^{x}}=-\left.\frac{x}{1+\mathrm{e}^{x}}\right|_{0} ^{+\infty}+\int_{0}^{+\infty} \frac{1}{1+\mathrm{e}^{x}} \mathrm{~d} x$ $$ =\int_{0}^{+\infty} \frac{\mathrm{e}^{x}}{\mathrm{e}^{2 x}+\mathrm{e}^{x}} \mathrm{~d} x=\int_{0}^{+\infty} \frac{1}{\mathrm{e}^{2 x}+\mathrm{e}^{x}} \mathrm{de}^{x} \xlongequal{\mathrm{e}^{x}=1} \int_{1}^{+\infty} \frac{1}{t+t^{2}} \mathrm{~d} t=\left.\ln \frac{t}{1+t}\right|_{1} ^{+\infty}=\ln 2 $$ （6） $\displaystyle \int_{1}^{+\infty} \frac{\mathrm{d} x}{\mathrm{e}^{1+x}+\mathrm{e}^{3-x}}=\int_{1}^{+\infty} \frac{\mathrm{d} x}{\mathrm{e}^{3-x}\left(\mathrm{e}^{-2+2 x}+1\right)}=\frac{1}{\mathrm{e}^{2}} \int_{1}^{+\infty} \frac{\mathrm{de}^{x-1}}{\left(\mathrm{e}^{x-1}\right)^{2}+1}=\frac{1}{\mathrm{e}^{2}} \cdot \frac{\pi}{4}=\frac{\pi}{4 \mathrm{e}^{2}}$ ． （7）方法 1 ：令 $\mathrm{e}^{-x}=t$ ，则 $$ \int_{0}^{+\infty} \frac{1}{\left(1+\mathrm{e}^{x}\right)^{2}} \mathrm{~d} x=\int_{1}^{0} \frac{1}{\left(1+t^{-1}\right)^{2}}\left(-\frac{1}{t}\right) \mathrm{d} t=\int_{0}^{1} \frac{t}{(1+t)^{2}} \mathrm{~d} t=\left.\left(\ln (1+t)+\frac{1}{1+t}\right)\right|_{0} ^{1}=\ln 2-\frac{1}{2} . $$ 方法 2：令 $\mathrm{e}^{x}=t$ ，则 $$ \int_{0}^{+\infty} \frac{1}{\left(1+\mathrm{e}^{x}\right)^{2}} \mathrm{~d} x=\int_{1}^{+\infty} \frac{1}{t(1+t)^{2}} \mathrm{~d} t=\int_{1}^{+\infty}\left(\frac{1}{t}-\frac{1}{1+t}-\frac{1}{(1+t)^{2}}\right) \mathrm{d} t=\left.\left(\ln \frac{t}{1+t}+\frac{1}{1+t}\right)\right|_{1} ^{+\infty}=\ln 2-\frac{1}{2} $$ （8） $\displaystyle \int_{1}^{2}\left[\frac{1}{x \ln ^{2} x}-\frac{1}{(x-1)^{2}}\right] \mathrm{d} x=\lim _{u \rightarrow 1^{+}} \int_{u}^{2}\left[\frac{1}{x \ln ^{2} x}-\frac{1}{(x-1)^{2}}\right] \mathrm{d} x=\left.\lim _{u \rightarrow 1^{+}}\left(-\frac{1}{\ln x}+\frac{1}{x-1}\right)\right|_{u} ^{2}$ $$ \begin{aligned} & =\lim _{u \rightarrow 1^{+}}\left(-\frac{1}{\ln 2}+\frac{1}{2-1}+\frac{1}{\ln u}-\frac{1}{u-1}\right) \\ & =-\frac{1}{\ln 2}+1+\lim _{u \rightarrow 1^{+}}\left(\frac{1}{\ln u}-\frac{1}{u-1}\right)=\frac{3}{2}-\frac{1}{\ln 2} . \end{aligned} $$ （9）方法 1 ：令 $\displaystyle x=\frac{1}{t}$ ，则 $\displaystyle \int_{1}^{+\infty} \frac{\ln \left(1+x^{2}\right)}{x^{3}} \mathrm{~d} x=\int_{0}^{1} t\left(\ln \left(1+t^{2}\right)-2 \ln t\right) \mathrm{d} t=\left.\frac{1}{2} t^{2}\left(\ln \left(1+t^{2}\right)-2 \ln t\right)\right|_{0} ^{1}-\frac{1}{2} \int_{0}^{1} t^{2}\left(\frac{2 t}{1+t^{2}}-\frac{2}{t}\right) \mathrm{d} t=\ln 2$ ． 方法 2： $\displaystyle \int_{1}^{+\infty} \frac{\ln \left(1+x^{2}\right)}{x^{3}} \mathrm{~d} x=\frac{1}{2} \int_{1}^{+\infty} \frac{\ln \left(1+x^{2}\right)}{x^{4}} \mathrm{~d} x^{2} \xlongequal{t=x^{2}} \frac{1}{2} \int_{1}^{+\infty} \frac{\ln (1+t)}{t^{2}} \mathrm{~d} t=-\frac{1}{2} \int_{1}^{+\infty} \cdot \ln (1+t) \mathrm{d}\left(\frac{1}{t}\right)$ $$ =-\frac{1}{2}\left[\left.\frac{\ln (1+t)}{t}\right|_{1} ^{+\infty}-\int_{1}^{+\infty} \frac{1}{t(1+t)} \mathrm{d} t\right]=\frac{1}{2}\left(\ln 2-\left.\ln \frac{1+t}{t}\right|_{1} ^{+\infty}\right)=\ln 2 . $$ （10）令 $\displaystyle x=\frac{1}{t}$ ，则 $\displaystyle \int_{1}^{+\infty} \frac{\ln (1+x)-\ln x}{1+x^{2}} \mathrm{~d} x=\int_{0}^{1} \frac{\ln (t+1)}{1+t^{2}} \mathrm{~d} t=\frac{\pi}{8} \ln 2$ ． 注： $\displaystyle \int_{0}^{1} \frac{\ln (t+1)}{1+t^{2}} \mathrm{~d} t$ 的计算：见定积分题 22 ．