判别法推导

发表: 5/26/2026 更新: 5/26/2026 字数: 0 字 时长: 0 分钟

基准 $P$ 积分判别

1.无穷区间 $P$ 积分

$\begin{array}{l}{\displaystyle \mathrm{记}{I}_{+}={\int }_a^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\mathrm{d}x=\underset{b\to +\mathrm{\infty }}{lim}{\int }_a^b\frac{1}{x^{\alpha }}\mathrm{d}x(a>0)}\\ {\displaystyle I_{+}=\underset{b\to +\mathrm{\infty }}{lim}\frac{x^{1-\alpha }}{1-\alpha }{|}_a^b=\underset{b\to +\mathrm{\infty }}{lim}\frac{b^{1-\alpha }-a^{1-\alpha }}{1-\alpha }}\\ {\displaystyle \mathrm{①}\alpha >1:1-\alpha <0,\underset{b\to +\mathrm{\infty }}{lim}{b}^{1-\alpha }\to 0}\\ {\displaystyle I_{+}=\underset{b\to +\mathrm{\infty }}{lim}\frac{b^{1-\alpha }-a^{1-\alpha }}{1-\alpha }=\frac{-a^{1-\alpha }}{1-\alpha }\Rightarrow \mathrm{收}\mathrm{敛}}\\ \mathrm{②}\alpha =1:\\ {\displaystyle I_{+}=\underset{b\to +\mathrm{\infty }}{lim}{\int }_a^b\frac{1}{x}\mathrm{d}x=\underset{b\to +\mathrm{\infty }}{lim}\ln x{|}_a^b=\underset{b\to +\mathrm{\infty }}{lim}[\ln b-\ln a]=\mathrm{\infty }\Rightarrow \mathrm{发}\mathrm{散}}\\ {\displaystyle \mathrm{③}\alpha <1:1-\alpha >0,\underset{b\to +\mathrm{\infty }}{lim}{b}^{1-\alpha }\to \mathrm{\infty }}\\ {\displaystyle I_{+}=\underset{b\to +\mathrm{\infty }}{lim}\frac{b^{1-\alpha }-a^{1-\alpha }}{1-\alpha }=\mathrm{\infty }\Rightarrow \mathrm{发}\mathrm{散}}\end{array}$

Conc

$\begin{array}{rl}\alpha & >1:{\int }_a^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\mathrm{d}x\mathrm{收}\mathrm{敛}\\ \alpha & \le 1:{\int }_a^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\mathrm{d}x\mathrm{发}\mathrm{散}\end{array}$

$\begin{array}{l}\begin{array}{rl}{I}_{+}& =\underset{b\to +\mathrm{\infty }}{lim}{\int }_a^b\frac{1}{x^{\alpha }}\mathrm{d}x\stackrel{x=-t}{=}\underset{b\to +\mathrm{\infty }}{lim}{\int }_{-b}^{-a}\frac{1}{(-x{)}^{\alpha }}\mathrm{d}x\\ & =\underset{b\to +\mathrm{\infty }}{lim}{\int }_{-b}^{-a}\frac{1}{|x{|}^{\alpha }}\mathrm{d}x=\underset{b\to +\mathrm{\infty }}{lim}{\int }_{-b}^{-a}|\frac{1}{x^{\alpha }}|\mathrm{d}x\end{array}\\ {\displaystyle \therefore \underset{b\to +\mathrm{\infty }}{lim}{\int }_{-b}^{-a}|\frac{1}{x^{\alpha }}|\mathrm{d}x\mathrm{与}{I}_{+}\mathrm{有}\mathrm{相}\mathrm{同}\mathrm{的}\mathrm{敛}\mathrm{散}\mathrm{性}}\\ {\displaystyle \mathrm{又}\mathrm{根}\mathrm{据}\mathrm{绝}\mathrm{对}\mathrm{收}\mathrm{敛}:I_{-}=\underset{b\to +\mathrm{\infty }}{lim}{\int }_{-b}^{-a}\frac{1}{x^{\alpha }}\mathrm{d}x\mathrm{与}{I}_{+}\mathrm{有}\mathrm{相}\mathrm{同}\mathrm{的}\mathrm{敛}\mathrm{散}\mathrm{性}}\end{array}$

Conc

$\begin{array}{rl}\alpha & >1:{\int }_{-\mathrm{\infty }}^{-a}\frac{1}{x^{\alpha }}\mathrm{d}x\mathrm{收}\mathrm{敛}\\ \alpha & \le 1:{\int }_{-\mathrm{\infty }}^{-a}\frac{1}{x^{\alpha }}\mathrm{d}x\mathrm{发}\mathrm{散}\end{array}$

2.瑕点型 $P$ 积分

$\begin{array}{l}{\displaystyle \mathrm{记}{I}_a={\int }_{a^{+}}^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x=\underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x(a>0)}\\ {\displaystyle I_a=\underset{\epsilon \to 0^{+}}{lim}\frac{(x-a{)}^{1-\alpha }}{1-\alpha }{|}_{a+\epsilon }^b=\underset{\epsilon \to 0^{+}}{lim}\frac{(b-a{)}^{1-\alpha }-{\epsilon }^{1-\alpha }}{1-\alpha }}\\ {\displaystyle \mathrm{①}\alpha >1:1-\alpha <0,\underset{\epsilon \to 0^{+}}{lim}{\epsilon }^{1-\alpha }\to \mathrm{\infty }}\\ {\displaystyle I_a=\underset{\epsilon \to 0^{+}}{lim}\frac{(b-a{)}^{1-\alpha }-{\epsilon }^{1-\alpha }}{1-\alpha }=\mathrm{\infty }\Rightarrow \mathrm{发}\mathrm{散}}\\ \mathrm{②}\alpha =1:\\ \begin{array}{rl}{I}_a& =\underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{x-a}\mathrm{d}x=\underset{\epsilon \to 0^{+}}{lim}\ln (x-a){|}_{a+\epsilon }^b\\ & =\underset{\epsilon \to 0^{+}}{lim}[\ln (b-a)-\ln \epsilon ]=\mathrm{\infty }\Rightarrow \mathrm{发}\mathrm{散}\end{array}\\ {\displaystyle \mathrm{③}\alpha <1:1-\alpha >0,\underset{\epsilon \to 0^{+}}{lim}{\epsilon }^{1-\alpha }\to 0}\\ {\displaystyle I_a=\underset{\epsilon \to 0^{+}}{lim}\frac{(b-a{)}^{1-\alpha }-{\epsilon }^{1-\alpha }}{1-\alpha }=\frac{(b-a{)}^{1-\alpha }}{1-\alpha }\Rightarrow \mathrm{收}\mathrm{敛}}\end{array}$

Conc

$\begin{array}{rl}\alpha & <1:\underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x\mathrm{收}\mathrm{敛}\\ \alpha & \ge 1:\underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x\mathrm{发}\mathrm{散}\end{array}$

$\begin{array}{l}{\displaystyle I_a=\underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x\stackrel{x+t=a+b}{=}\underset{\epsilon \to 0^{+}}{lim}{\int }_a^{b-\epsilon }\frac{1}{(b-x{)}^{\alpha }}\mathrm{d}x}\\ {\displaystyle \therefore I_b=\underset{\epsilon \to 0^{+}}{lim}{\int }_a^{b-\epsilon }\frac{1}{(b-x{)}^{\alpha }}\mathrm{d}x\mathrm{与}{I}_a\mathrm{有}\mathrm{相}\mathrm{同}\mathrm{的}\mathrm{敛}\mathrm{散}\mathrm{性}}\end{array}$

Conc

$\begin{array}{rl}\alpha & <1:\underset{\epsilon \to 0^{+}}{lim}{\int }_a^{b-\epsilon }\frac{1}{(b-x{)}^{\alpha }}\mathrm{d}x\mathrm{收}\mathrm{敛}\\ \alpha & \ge 1:\underset{\epsilon \to 0^{+}}{lim}{\int }_a^{b-\epsilon }\frac{1}{(b-x{)}^{\alpha }}\mathrm{d}x\mathrm{发}\mathrm{散}\end{array}$

极限形式 $P$ 判别法

注: $\underset{x\to ◻}{lim}\frac{f(x)}{g(x)}$ 存在或为无穷大，瑕积分的瑕点必须是非震荡型无穷间断点

1.无穷区间

对于积分 $I_{+}={\displaystyle {\int }_a^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\mathrm{d}x(a>0)}$，我们已经知道：
${\displaystyle {\int }_a^{+\mathrm{\infty }}\frac{1}{x^{\alpha }}\mathrm{d}x\{\begin{array}{ll}\text{收敛},& \alpha >1\\ \text{发散},& \alpha \le 1\end{array}}$

设 $f(x),g(x)$ 在 $[a,+\mathrm{\infty })$ 上非负连续，基准函数 ${\displaystyle g(x)=\frac{1}{x^{\alpha }}}$
${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}\frac{f(x)}{g(x)}=\underset{x\to +\mathrm{\infty }}{lim}{x}^{\alpha }f(x)=L}$

无穷区间积分:${\displaystyle {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\triangleq \underset{b\to +\mathrm{\infty }}{lim}{\int }_a^{b}f(x)\mathrm{d}x}$
无穷区间 P 积分:${\displaystyle {\int }_a^{+\mathrm{\infty }}g(x)\mathrm{d}x\triangleq \underset{b\to +\mathrm{\infty }}{lim}{\int }_a^b\frac{1}{x^{\alpha }}\mathrm{d}x}$

1. $\alpha >1$
   • 若 $0<L<+\mathrm{\infty }:$ ${\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x$ 与 ${\int }_a^{+\mathrm{\infty }}g(x)\mathrm{d}x$ 同敛散，因此 ${\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛};$
   • 若 $L=0:$ ${\int }_a^{+\mathrm{\infty }}g(x)\mathrm{d}x\mathrm{收}\mathrm{敛}\Rightarrow {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛}.$

${\displaystyle \Rightarrow \mathrm{\exists }\alpha >1,\underset{x\to +\mathrm{\infty }}{lim}{x}^{\alpha }f(x)\mathrm{存}\mathrm{在}⟹{\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛}}$

2. $\alpha \le 1$
   • 若 $0<L<+\mathrm{\infty }:$ ${\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x$ 与 ${\int }_a^{+\mathrm{\infty }}g(x)\mathrm{d}x$ 同敛散，因此 ${\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{发}\mathrm{散};$
   • 若 $L=+\mathrm{\infty }:$ ${\int }_a^{+\mathrm{\infty }}g(x)\mathrm{d}x\mathrm{发}\mathrm{散}\Rightarrow {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}.$

${\displaystyle \Rightarrow \mathrm{\exists }\alpha \le 1,\underset{x\to +\mathrm{\infty }}{lim}{x}^{\alpha }f(x)=k(\ne 0)\mathrm{或}\mathrm{\infty }⟹{\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}}$

---

同理，因为积分 ${\displaystyle I_{-}={\int }_{-\mathrm{\infty }}^{-a}\frac{1}{x^{\alpha }}\mathrm{d}x(a>0),}$ 与 $I_{+}$ 具有相同的敛散性
所以通过 ${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}\frac{f(x)}{g(x)}=\underset{x\to -\mathrm{\infty }}{lim}{x}^{\alpha }f(x)}$ 可推出对于无穷区间积分 ${\displaystyle {\int }_{-\mathrm{\infty }}^{-a}f(x)\mathrm{d}x}$ 有着与 ${\displaystyle {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x}$ 相同的判别法.

Conc

1. $f(x)\in c[a,+\mathrm{\infty })(a>0)$
   ① ${\displaystyle \mathrm{\exists }\alpha >1:\underset{x\to +\mathrm{\infty }}{lim}{x}^{\alpha }f(x)\mathrm{\exists }\Rightarrow {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛};}$
   ② ${\displaystyle \mathrm{\exists }\alpha \le 1:\underset{x\to +\mathrm{\infty }}{lim}{x}^{\alpha }f(x)=k(\ne 0)\mathrm{或}\mathrm{\infty }\Rightarrow {\int }_a^{+\mathrm{\infty }}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}.}$
2. $f(x)\in c(-\mathrm{\infty },a](a<0)$
   ① ${\displaystyle \mathrm{\exists }\alpha >1:\underset{x\to -\mathrm{\infty }}{lim}{x}^{\alpha }f(x)\mathrm{\exists }\Rightarrow {\int }_{-\mathrm{\infty }}^{a}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛};}$
   ② ${\displaystyle \mathrm{\exists }\alpha \le 1:\underset{x\to -\mathrm{\infty }}{lim}{x}^{\alpha }f(x)=k(\ne 0)\mathrm{或}\mathrm{\infty }\Rightarrow {\int }_{-\mathrm{\infty }}^{a}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}.}$

2.瑕点型

对于积分 $I_a={\displaystyle \underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x(a>0)}$，我们已经知道：
${\displaystyle {\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x\{\begin{array}{ll}\text{收敛},& \alpha <1\\ \text{发散},& \alpha \ge 1\end{array}}$

设 $f(x),g(x)$ 在 $(a,b]$ 上非负连续，基准函数 ${\displaystyle g(x)=\frac{1}{(x-a{)}^{\alpha }}}$
${\displaystyle \underset{x\to a^{+}}{lim}\frac{f(x)}{g(x)}=\underset{x\to a^{+}}{lim}(x-a{)}^{\alpha }f(x)=L}$

瑕积分:${\displaystyle {\int }_a^{b}f(x)\mathrm{d}x\triangleq \underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^{b}f(x)\mathrm{d}x}$
瑕点型 P 积分:${\displaystyle {\int }_a^{b}g(x)\mathrm{d}x\triangleq \underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }^b\frac{1}{(x-a{)}^{\alpha }}\mathrm{d}x}$
极限存在 $⟺$ 积分收敛；极限不存在 $⟺$ 积分发散.

1. $\alpha <1$
   • 若 $0<L<+\mathrm{\infty }:$ ${\int }_a^{b}f(x)\mathrm{d}x$ 与 ${\int }_a^{b}g(x)\mathrm{d}x$ 同敛散，因此 ${\int }_a^{b}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛};$
   • 若 $L=0:$ ${\int }_a^{b}g(x)\mathrm{d}x\mathrm{收}\mathrm{敛}\Rightarrow {\int }_a^{b}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛}.$

${\displaystyle \Rightarrow \mathrm{\exists }\alpha <1,\underset{x\to a^{+}}{lim}(x-a{)}^{\alpha }f(x)\mathrm{存}\mathrm{在}⟹{\int }_a^{b}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛}}$

2. $\alpha \ge 1$
   • 若 $0<L<+\mathrm{\infty }:$ ${\int }_a^{b}f(x)\mathrm{d}x$ 与 ${\int }_a^{b}g(x)\mathrm{d}x$ 同敛散，因此 ${\int }_a^{b}f(x)\mathrm{d}x\mathrm{发}\mathrm{散};$
   • 若 $L=+\mathrm{\infty }:$ ${\int }_a^{b}g(x)\mathrm{d}x\mathrm{发}\mathrm{散}\Rightarrow {\int }_a^{b}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}.$

${\displaystyle \Rightarrow \mathrm{\exists }\alpha \ge 1,\underset{x\to a^{+}}{lim}(x-a{)}^{\alpha }f(x)=k(\ne 0)\mathrm{或}\mathrm{\infty }⟹{\int }_a^{b}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}}$

---

同理，因为 ${\displaystyle I_b={\int }_a^{b-\epsilon }\frac{1}{(b-x{)}^{\alpha }}\mathrm{d}x(a>0),}$ 与 $I_a$ 具有相同的敛散性
所以通过 ${\displaystyle \underset{x\to b^{-}}{lim}\frac{f(x)}{g(x)}=\underset{x\to b^{-}}{lim}(b-x{)}^{\alpha }f(x)}$ 可推出对于瑕积分 ${\displaystyle {\int }_a^{b}f(x)\mathrm{d}x(b\mathrm{为}\mathrm{唯}\mathrm{一}\mathrm{瑕}\mathrm{点})}$ 有着与 ${\displaystyle {\int }_a^{b}f(x)\mathrm{d}x(a\mathrm{为}\mathrm{唯}\mathrm{一}\mathrm{瑕}\mathrm{点})}$ 相同的判别法.

Conc

1. $f(x)\in c[a,b],a\mathrm{为}\mathrm{唯}\mathrm{一}\mathrm{瑕}\mathrm{点}$
   ① ${\displaystyle \mathrm{\exists }\alpha <1:\underset{x\to a+}{lim}(x-a{)}^{\alpha }f(x)\mathrm{\exists }\Rightarrow {\int }_a^{b}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛};}$
   ② ${\displaystyle \mathrm{\exists }\alpha \ge 1:\underset{x\to a+}{lim}(x-a{)}^{\alpha }f(x)=k(\ne 0)\mathrm{或}\mathrm{\infty }\Rightarrow {\int }_a^{b}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}.}$
2. $f(x)\in c[a,b],b\mathrm{为}\mathrm{唯}\mathrm{一}\mathrm{瑕}\mathrm{点}$
   ① ${\displaystyle \mathrm{\exists }\alpha <1:\underset{x\to b-}{lim}(b-x{)}^{\alpha }f(x)\mathrm{\exists }\Rightarrow {\int }_a^{b}f(x)\mathrm{d}x\mathrm{收}\mathrm{敛};}$
   ② ${\displaystyle \mathrm{\exists }\alpha \ge 1:\underset{x\to b-}{lim}(b-x{)}^{\alpha }f(x)=k(\ne 0)\mathrm{或}\mathrm{\infty }\Rightarrow {\int }_a^{b}f(x)\mathrm{d}x\mathrm{发}\mathrm{散}.}$

极限形式比较判别法

${\displaystyle \underset{x\to ◻}{lim}\frac{f(x)}{g(x)}=L,f(x)\ge 0,g(x)\ge 0}$

1. $0<L<+\mathrm{\infty }:$ ${\displaystyle \int f(x)\mathrm{d}x}$ 与 ${\displaystyle \int g(x)\mathrm{d}x}$ 同敛散
2. $L=0:$ 若 ${\displaystyle \int g(x)\mathrm{d}x}$ 收敛，则 ${\displaystyle \int f(x)\mathrm{d}x}$ 收敛
3. $L=+\mathrm{\infty }:$ 若 ${\displaystyle \int g(x)\mathrm{d}x}$ 发散，则 ${\displaystyle \int f(x)\mathrm{d}x}$ 发散

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}\to {\int }_a^{+\mathrm{\infty }}\underset{x\to -\mathrm{\infty }}{lim}\to {\int }_{-\mathrm{\infty }}^a\underset{x\to a^{+}}{lim}\to {\int }_a^b\underset{x\to b^{-}}{lim}\to {\int }_a^b}$