中册 5.1 反常积分计算 第3题

\section*{解题过程：} （1） $\displaystyle \int_{2}^{+\infty} \frac{1}{x^{2}+x-2} \mathrm{~d} x=\frac{1}{3} \int_{2}^{+\infty}\left(\frac{1}{x-1}-\frac{1}{x+2}\right) \mathrm{d} x=\left.\frac{1}{3} \ln \frac{x-1}{x+2}\right|_{2} ^{+\infty}=\frac{2}{3} \ln 2$ ． （2）方法 1：令 $x=a \tan \theta$ ，则 $$ \begin{aligned} \int_{0}^{+\infty} \frac{1}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x & =\frac{1}{a^{2(n+1)}} \int_{0}^{\frac{\pi}{2}} \frac{a}{\left(\sec ^{2} \theta\right)^{n+1}} \sec ^{2} \theta \mathrm{~d} \theta=\frac{1}{a^{2 n+1}} \int_{0}^{\frac{\pi}{2}} \cos ^{2 n} \theta \mathrm{~d} \theta \\ & =\frac{1}{2 a^{2 n+1}} \mathrm{~B}\left(\frac{1}{2}, n+\frac{1}{2}\right)=\frac{1}{a^{2 n+1}} \cdot \frac{2 n-1}{2 n} \cdot \frac{2 n-3}{2(n-1)} \cdots \frac{1}{2} \cdot \frac{\pi}{2} . \end{aligned} $$ 方法 2：$\displaystyle I_{n}=\int_{0}^{+\infty} \frac{1}{\left(x^{2}+a^{2}\right)^{n}} \mathrm{~d} x=\left.\frac{x}{\left(x^{2}+a^{2}\right)^{n}}\right|_{0} ^{+\infty}+2 n \int_{0}^{+\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x$ $$ =2 n \int_{0}^{+\infty} \frac{x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x=2 n \int_{0}^{+\infty} \frac{x^{2}+a^{2}-a^{2}}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x=2 n I_{n}-2 n a^{2} I_{n+1} $$ 由此可得 $\displaystyle I_{n}=a^{2} \frac{2 n}{2 n-1} I_{n+1}$ ．于是 $$ I_{n+1}=\int_{0}^{+\infty} \frac{1}{\left(x^{2}+a^{2}\right)^{n+1}} \mathrm{~d} x=\frac{1}{a^{2 n+1}} \frac{2 n-1}{2 n} \cdot \frac{2 n-3}{2(n-1)} \cdots \frac{1}{2} \frac{\pi}{2} $$ （3）令 $\displaystyle t=\frac{1}{1+x^{n}}$ ，则 $$ \int_{0}^{+\infty} \frac{1}{x^{n}+1} \mathrm{~d} x=\frac{1}{n} \int_{0}^{1} t^{-\frac{1}{n}}(1-t)^{-\frac{n-1}{n}} \mathrm{~d} t=\frac{1}{n} \mathrm{~B}\left(\frac{1}{n}, \frac{n-1}{n}\right)=\frac{1}{n} \Gamma\left(\frac{1}{n}\right) \Gamma\left(\frac{n-1}{n}\right)=\frac{\pi}{n \sin \frac{\pi}{n}} $$ 特别地， $\displaystyle \int_{0}^{+\infty} \frac{1}{1+x^{6}} \mathrm{~d} x=\frac{1}{6} \Gamma\left(\frac{1}{6}\right) \Gamma\left(\frac{5}{6}\right)=\frac{1}{6} \frac{\pi}{\sin \frac{\pi}{6}}=\frac{1}{3} \pi$ ． （4） $\displaystyle \int_{0}^{+\infty} \frac{x-\sin x}{x^{3}} \mathrm{~d} x=-\frac{1}{2} \int_{0}^{+\infty}(x-\sin x) \mathrm{d}\left(\frac{1}{x^{2}}\right)=-\frac{1}{2}\left(\left.\frac{x-\sin x}{x^{2}}\right|_{0} ^{+\infty}-\int_{0}^{+\infty} \frac{1-\cos x}{x^{2}} \mathrm{~d} x\right)$ $$ =\frac{1}{2} \int_{0}^{+\infty} \frac{1-\cos x}{x^{2}} \mathrm{~d} x=-\frac{1}{2}\left(\left.\frac{1-\cos x}{x}\right|_{0} ^{+\infty}-\int_{0}^{+\infty} \frac{\sin x}{x} \mathrm{~d} x\right)=\frac{1}{2} \int_{0}^{+\infty} \frac{\sin x}{x} \mathrm{~d} x=\frac{\pi}{4} $$ （5）记 $\displaystyle t=\sqrt{a} x+\frac{b}{\sqrt{a}}, \lambda=\frac{\sqrt{c a-b^{2}}}{a}$ ，则 $$ \int_{-\infty}^{+\infty} \frac{1}{\left(a x^{2}+2 b x+c\right)^{\alpha}} \mathrm{d} x=\frac{2}{a^{\alpha}} \int_{0}^{+\infty} \frac{1}{\left(t^{2}+\lambda^{2}\right)^{\alpha}} \mathrm{d} t $$ 令 $\mathrm{d} t=\lambda \tan \theta$ ，则 $$ \int_{0}^{+\infty} \frac{1}{\left(t^{2}+\lambda^{2}\right)^{\alpha}} \mathrm{d} t=\frac{1}{\lambda^{2 \alpha}} \int_{0}^{\frac{\pi}{2}} \frac{\lambda}{\left(\sec ^{2} \theta\right)^{\alpha}} \sec ^{2} \theta \mathrm{~d} \theta=\frac{1}{\lambda^{2 \alpha-1}} \int_{0}^{\frac{\pi}{2}} \cos ^{2 \alpha-2} \theta \mathrm{~d} \theta=\frac{1}{2 \lambda^{2 \alpha-1}} \mathrm{~B}\left(\frac{1}{2}, \alpha-\frac{1}{2}\right) $$ 于是 $\displaystyle \int_{-\infty}^{+\infty} \frac{1}{\left(a x^{2}+2 b x+c\right)^{\alpha}} \mathrm{d} x=\frac{2}{a^{\alpha}} \int_{0}^{+\infty} \frac{1}{\left(t^{2}+\lambda^{2}\right)^{\alpha}} \mathrm{d} t=\frac{1}{a^{\alpha} \lambda^{2 \alpha-1}} \mathrm{~B}\left(\frac{1}{2}, \alpha-\frac{1}{2}\right)$ ． （6）由（2）得 $\displaystyle \int_{-\infty}^{+\infty} \frac{1}{\left(x^{2}+2 x+2\right)^{n}} \mathrm{~d} x=\int_{-\infty}^{+\infty} \frac{1}{\left[(x+1)^{2}+1\right]^{n}} \mathrm{~d}(x+1)=\int_{-\infty}^{+\infty} \frac{1}{\left(t^{2}+1\right)^{n}} \mathrm{~d} t=2 \int_{0}^{+\infty} \frac{1}{\left(t^{2}+1\right)^{n}} \mathrm{~d} t$ $$ =2 \cdot \frac{2 n-3}{2 n-2} \cdot \frac{2 n-5}{2 n-4} \cdots \frac{1}{2} \cdot \frac{\pi}{2}=\frac{(2 n-3)!!}{(2 n-2)!!} \cdot \pi $$ （7）当 $p=q$ 时， $$ \int \frac{\mathrm{d} x}{\left(x^{2}+p\right)\left(x^{2}+q\right)}=\int \frac{\mathrm{d} x}{\left(x^{2}+p\right)^{2}}=\frac{1}{2 p} \cdot \frac{x}{x^{2}+p}+\frac{1}{2 p} \cdot \frac{\mathrm{~d} x}{x^{2}+p}=\frac{1}{2 p} \cdot \frac{x}{x^{2}+p}+\frac{1}{2 p \sqrt{p}} \arctan \frac{x}{\sqrt{p}}+C $$ $$ \int_{0}^{+\infty} \frac{\mathrm{d} x}{\left(p+x^{2}\right)^{2}}=\lim _{A \rightarrow+\infty} \int_{0}^{A} \frac{\mathrm{~d} x}{\left(x^{2}+p\right)^{2}}=\lim _{A \rightarrow+\infty}\left(\frac{1}{2 p} \frac{A}{A^{2}+p}+\frac{1}{2 p \sqrt{p}} \arctan \frac{A}{\sqrt{p}}\right)=\frac{\pi}{4 p \sqrt{p}} $$ 当 $p \neq q$ 时， $$ \begin{aligned} & \int \frac{\mathrm{d} x}{\left(x^{2}+p\right)\left(x^{2}+q\right)}=\frac{1}{p-q} \int\left(\frac{1}{x^{2}+q}-\frac{1}{x^{2}+p}\right) \mathrm{d} x=\frac{1}{p-q}\left(\frac{1}{\sqrt{q}} \arctan \frac{x}{\sqrt{q}}-\frac{1}{\sqrt{p}} \arctan \frac{x}{\sqrt{p}}\right)+C . \\ & \int_{0}^{+\infty} \frac{\mathrm{d} x}{\left(x^{2}+p\right)\left(x^{2}+q\right)}=\lim _{A \rightarrow+\infty} \frac{1}{p-q}\left(\frac{1}{\sqrt{q}} \arctan \frac{A}{\sqrt{q}}-\frac{1}{\sqrt{p}} \arctan \frac{A}{\sqrt{p}}\right)=\frac{\pi}{2 \sqrt{p q}(\sqrt{p}+\sqrt{q})} . \end{aligned} $$ 综上，只要 $p, q>0$ 就有 $\displaystyle \int_{0}^{+\infty} \frac{\mathrm{d} x}{\left(x^{2}+\dot{p}\right)\left(x^{2}+q\right)}=\frac{\pi}{2 \sqrt{p q}(\sqrt{p}+\sqrt{q})}$ ．