考虑 ${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1+x}{1+x^2}\mathrm{d}x}$. 注意到 $$\begin{array}{rl}& {\int }_c^A\frac{1+x}{1+x^2}\mathrm{d}x=\arctan x{|}_c^A+\frac{1}{2}\ln (1+x^2){|}_c^A\\ \stackrel{A\to +\mathrm{\infty }}{\to }& \frac{\pi }{2}-\arctan c+(+\mathrm{\infty })-\frac{1}{2}\ln (1+c^2)=+\mathrm{\infty },\end{array}$$ 故发散. 但 $${\int }_{-A}^A\frac{1+x}{1+x^2}\mathrm{d}x=\arctan x{|}_{-A}^A+\frac{1}{2}\ln (1+x^2){|}_{-A}^A\stackrel{A\to +\mathrm{\infty }}{\to }\pi ,$$ 故这个积分在 Cauchy 主值意义下收敛, 在一般形态下不收敛.

构造变上限函数 $$F(A)={\int }_1^A\frac{1}{x^p}\mathrm{d}x=\{\begin{array}{rl}& \ln x{|}_1^A=\ln A,p=1,\\ & {\frac{x^{1-p}}{1-p}|}_1^A=\frac{A^{1-p}}{1-p}+\frac{1}{p-1},p\ne 1\end{array}(A\ge 1),$$ 则 $$\underset{A\to +\mathrm{\infty }}{lim}F(A)=\{\begin{array}{rl}& +\mathrm{\infty },p\le 1,\\ & \frac{1}{p-1},p>1.\end{array}$$

1. ${\displaystyle I_n={\int }_0^{\mathrm{\infty }}{x}^n{\mathrm{e}}^{-x}\mathrm{d}x}$.
   得 $$I_n=-{\int }_0^{\mathrm{\infty }}{x}^n\mathrm{d}{\mathrm{e}}^{-x}=-x^n{\mathrm{e}}^{-x}{|}_0^{+\mathrm{\infty }}+n{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-x}{x}^{n-1}\mathrm{d}x=0+n{I}_{n-1},$$ 而 ${\displaystyle I_0={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-x}\mathrm{d}x=1}$, 因此可以递推得到结果.
2. ${\displaystyle I={\int }_0^{\mathrm{\infty }}\frac{x\mathrm{d}x}{(x^2+a^2{)}^{\frac{3}{2}}}}$.
   作代换 ${\displaystyle x=a\tan t}$, 有 $$I={\int }_0^{\frac{\pi }{2}}\frac{a^2\tan t{\sec }^{2}t}{(a\sec t{)}^3}=\frac{1}{a}{\int }_0^{\frac{\pi }{2}}\sin t\mathrm{d}t.$$

由已知, 结合 定理, ${\displaystyle \mathrm{\forall }\epsilon >0}$, ${\displaystyle \mathrm{\exists }{A}_0\ge a}$, ${\displaystyle \mathrm{\forall }{A}^{\prime },A^{″}>A_0}$: ${\displaystyle {\int }_{A^{\prime }}^{A^{″}}|f(x)|\mathrm{d}x<\epsilon }$, 于是 $$|{\int }_{A^{\prime }}^{A^{″}}f(x)\mathrm{d}x|\le {\int }_{A^{\prime }}^{A^{″}}|f(x)|\mathrm{d}x<\epsilon ,$$ 于是 ${\displaystyle {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x}$ 满足 Cauchy 收敛准则, 它收敛.

记 $$F(A)={\int }_a^{A}f(x)\mathrm{d}x,G(A)={\int }_a^{A}g(x)\mathrm{d}x,$$ 则 ${\displaystyle 0\le F(A)\le G(A)}$. 因此 ${\displaystyle G(A)}$ 有界时 ${\displaystyle F(A)}$ 有界, ${\displaystyle F(A)}$ 无界时 ${\displaystyle G(A)}$ 无界. 再结合 定理, 得证.

4. 由已知, 对 ${\displaystyle \epsilon =\frac{l}{2}}$, ${\displaystyle \mathrm{\exists }{A}_0>0,\mathrm{\forall }x>A_0}$: $$|\frac{f(x)}{g(x)}-l|<\frac{l}{2}\Rightarrow \frac{l}{2}g(x)<f(x)<\frac{3l}{2}g(x),$$ 结合 比较判别法得 ${\displaystyle {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x,{\int }_a^{+\mathrm{\infty }}g(x)\mathrm{d}x}$ 同敛散.
5. 易得 ${\displaystyle \mathrm{\exists }{A}_0>0,\mathrm{\forall }x>A_0}$: ${\displaystyle \frac{f(x)}{g(x)}>1\Rightarrow f(x)>g(x)}$.
6. 易得 ${\displaystyle \mathrm{\exists }{A}_0>0,\mathrm{\forall }x>A_0}$: ${\displaystyle \frac{f(x)}{g(x)}<1\Rightarrow f(x)<g(x)}$.

${\displaystyle p\le 0}$ 时发散.
${\displaystyle p>1}$ 时, 取 ${\displaystyle \epsilon >0}$, ${\displaystyle p-\epsilon >1}$, 则 $$\frac{\ln x}{1+x^p}\cdot x^{p-\epsilon }=\ln x\cdot x^{-\epsilon }\cdot \frac{x^p}{1+x^p}\to 0.$$ 而 ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{1}{x^{p-\epsilon }}\mathrm{d}x}$ 收敛, 因此 ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\ln x}{1+x^p}\mathrm{d}x}$ 收敛.
${\displaystyle 0<p\le 1}$ 时, $$\frac{\ln x}{1+x^p}\ge \frac{\ln x}{1+x}\ge \frac{1}{1+x},$$ 发散.

1. ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\cos x\sin \frac{1}{x}}{x}\mathrm{d}x}$
   考察 ${\displaystyle {\int }_1^{+\mathrm{\infty }}\left|\frac{\cos x\sin \frac{1}{x}}{x}\right|\mathrm{d}x}$. 注意到 $$\left|\frac{\cos x\sin \frac{1}{x}}{x}\right|\le \frac{\sin \frac{1}{x}}{x},\underset{x\to +\mathrm{\infty }}{lim}\frac{\sin \frac{1}{x}}{x}\cdot x^2=1,$$ 则 ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\sin \frac{1}{x}}{x}\mathrm{d}x}$ 收敛, 从而原式绝对收敛.
2. ${\displaystyle {\int }_1^{+\mathrm{\infty }}[\sin \frac{1}{x}-\ln (\frac{1}{x}+\sqrt{1+\frac{1}{x^2}})]\mathrm{d}x}$
   注意到 $$\begin{array}{rl}& \sin \frac{1}{x}-\ln (\frac{1}{x}+\sqrt{1+\frac{1}{x^2}})\\ =& \frac{1}{x}-\frac{1}{6x^3}+o\left(\frac{1}{x^3}\right)-\ln (1+\frac{1}{x}+\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right))\\ =& \frac{1}{x}-\frac{1}{6x^3}+o\left(\frac{1}{x^3}\right)\\ & -(\frac{1}{x}+\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right))+\frac{1}{2}{[\frac{1}{x}+\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)]}^2+o\left(\frac{1}{x^2}\right)\\ =& o\left(\frac{1}{x^2}\right),\end{array}$$ 从而 $$\underset{x\to +\mathrm{\infty }}{lim}\frac{|\sin \frac{1}{x}-\ln (\frac{1}{x}+\sqrt{1+\frac{1}{x^2}})|}{\frac{1}{x^2}}=0,l=0,p=2,$$ 故原式绝对收敛.

• Abel 判别法: ${\displaystyle \mathrm{\forall }\epsilon >0}$, ${\displaystyle \mathrm{\exists }{A}_0>a}$, ${\displaystyle \mathrm{\forall }{A}^{\prime },A^{″}>A_0}$: ${\displaystyle |{\int }_{A^{\prime }}^{A^{″}}f(x)\mathrm{d}x|<\epsilon }$. 根据 ${\displaystyle g(x)}$ 在 ${\displaystyle [a,+\mathrm{\infty })}$ 上单调有界和 积分第二中值定理, ${\displaystyle \mathrm{\forall }{A}^{\prime },A^{″}>A_0}$: $$|{\int }_{A^{\prime }}^{A^{″}}f(x)g(x)\mathrm{d}x|\le |g(A^{\prime })|\cdot |{\int }_{A^{\prime }}^{\xi }f(x)\mathrm{d}x|+|g(A^{″})|\cdot |{\int }_{\xi }^{A^{″}}f(x)\mathrm{d}x|<M\epsilon +M\epsilon .$$
• Dirichlet 判别法: $$|{\int }_{\xi }^{A^{″}}f(x)\mathrm{d}x|=|{\int }_a^{\xi }f(x)\mathrm{d}x-{\int }_a^{A^{″}}f(x)\mathrm{d}x|=|F(\xi )-F(A^{″})|\le M+M=2M.$$

1. ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\sin x}{x^p}\mathrm{d}x}$, ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\cos x}{x^p}\mathrm{d}x}$
   ${\displaystyle p\le 0}$ 发散已经证明了.
   ${\displaystyle p>1}$ 时, ${\displaystyle \left|\frac{\sin x}{x^p}\right|\le \frac{1}{x^p}}$, 结合 ${\displaystyle p}$ 积分, 绝对收敛.
   ${\displaystyle 0<p\le 1}$ 时,

   • ${\displaystyle f(x)=\sin x}$, ${\displaystyle F(A)={\int }_1^A\sin x\mathrm{d}x=\cos 1-\cos A}$ 有界,
   • ${\displaystyle g(x)=\frac{1}{x^p}}$ 单减且 ${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}g(x)=0}$,

   从而由 Dirichlet 判别法, 收敛. 但是只是条件收敛. 因为 $$\frac{|\sin x|}{x^p}\ge \frac{(\sin x{)}^2}{x^p}=\frac{1}{2x^p}-\frac{\cos 2x}{2x^p},$$ 而 ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{1}{2x^p}\mathrm{d}x}$ 发散, ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\cos 2x}{2x^p}}$ 收敛, 从而绝对值积分发散.

2. ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\sin x}{x^p}\arctan x\mathrm{d}x}$, ${\displaystyle p>0}$
   刚刚已经证明 ${\displaystyle {\int }_1^{+\mathrm{\infty }}\frac{\sin x}{x^p}\mathrm{d}x}$ 收敛. 而 ${\displaystyle \arctan x}$ 在 ${\displaystyle [1,+\mathrm{\infty })}$ 上单增, ${\displaystyle 0\le \arctan x<\frac{\pi }{2}}$, 从而根据 Abel 判别法, 收敛.
   ${\displaystyle p>1}$ 时, ${\displaystyle |\frac{\sin x}{x^p}\arctan x|\le \frac{\frac{\pi }{2}}{x^p}}$, 所以绝对收敛.
   ${\displaystyle 0<p\le 1}$ 时, $$|\frac{\sin x}{x^p}\arctan x|\ge \frac{{\sin }^{2}x}{x^p}\arctan 1=\arctan 1(\frac{1}{2x^p}-\frac{\cos 2x}{2x^p}),$$ 所以条件收敛.

3. ${\displaystyle {\int }_1^{+\mathrm{\infty }}\ln (1+\frac{\sin x}{x^p})\mathrm{d}x,p>0}$
   注意到 $$\begin{array}{rl}\ln (1+\frac{\sin x}{x^p})& =\frac{\sin x}{x^p}-\frac{{\sin }^{2}x}{2x^{2p}}+\frac{{\sin }^{3}x}{3x^{3p}}\cdot {\ln }^{‴}(1+y){|}_{\frac{\sin \xi }{{\xi }^p}}\\ & =\frac{\sin x}{x^p}-\frac{{\sin }^{2}x}{2x^{2p}}+o\left(\frac{1}{x^{2p}}\right)\\ & =\frac{\sin x}{x^p}+\frac{\cos 2x}{4x^{2p}}-\frac{1}{4x^{2p}}+o\left(\frac{1}{x^{2p}}\right).\end{array}$$
   ${\displaystyle p>1}$ 时, 绝对收敛.
   ${\displaystyle \frac{1}{2}<p\le 1}$ 时, 条件收敛.
   ${\displaystyle 0<p\le \frac{1}{2}}$ 时, 发散.