Section03_广义积分

正常积分

1. 积分区间有限
2. 被积函数在积分期间内连续或除有限个第一类间断点外处处连续

区间无穷的反常积分

$f(x)\in C[a,+\infty),\ \int_{a}^{+\infty}f(x)\cdot dx$

1. 定义 $\displaystyle\int_{a}^{b} f(x)\cdot dx = F(b)-F(a)$
   • $F(b)-F(a)$ 与 $\displaystyle \int_{a}^{+\infty}f(x)\cdot dx$ 不同
   • $\displaystyle \lim_{b\to \infty} [F(b)-F(a)]$ 与 $\displaystyle \int_{a}^{\infty}f(x)\cdot dx$ 相同
     1. 若 $\displaystyle \lim_{b\to \infty} [F(b)-F(a)] = A$ 则 $\displaystyle \int_{a}^{\infty}f(x)\cdot dx=A$
     2. 若 $\displaystyle \lim_{b\to \infty} [F(b)-F(a)]\ \nexists$ 则 $\displaystyle \int_{a}^{\infty}f(x)\cdot dx$ 发散
2. 判别法
   1. 若 $\displaystyle \exists\ \alpha>1; \lim_{x\to +\infty}x^{\alpha}f(x)\ \exists$，则收敛
   2. 若 $\displaystyle \exists\ \alpha\le 1; \lim_{x\to +\infty}x^{\alpha}f(x)\ = k(\neq 0)\text{或}\infty$，则发散

例题

• 例1 $\displaystyle \int_{1}^{+\infty}\frac{dx}{1+x^{2}}$ $$\begin{array}{ll} &\displaystyle \int_{1}^{b}\frac{dx}{1+x^{2}} = \arctan x\vert^{b}_{1} = \arctan b - \frac{\pi}{4}\\ \therefore & \displaystyle \lim_{b\to +\infty}(\arctan b -\frac{\pi}{4}) = \frac{\pi}{4} \\ \therefore & \displaystyle \int_{1}^{+\infty}\frac{dx}{1+x^{2}} = \frac{\pi}{4} \\ \end{array}$$
• 例2 判断 $\displaystyle \int_{0}^{+\infty}\frac{\sqrt{x}}{1+x^{2}}\cdot dx$ 的敛散性 $$\begin{array}{ll} \because & \displaystyle \lim_{x\to +\infty} x^{\frac{3}{2}}\frac{\sqrt{x}}{1+x^{2}} = 1, \alpha=\frac{3}{2}>1 \\ \therefore & \text{原式收敛} \\ \end{array}$$

$\Gamma$ 函数

$$\Gamma(\alpha) = \int_{0}^{+\infty}x^{\alpha-1}e^{-x}\cdot dx$$

1. $\Gamma(\alpha+1)= \alpha\Gamma(\alpha)$
2. $\Gamma(n+1) = n!\quad n\in\mathbb{N}$
3. $\Gamma(\frac{1}{2}) = \sqrt{\pi}$

例题

1. $\displaystyle \int_{0}^{+\infty}x^{5}e^{-x}\cdot dx$ $$\begin{split} \int_{0}^{+\infty}x^{5}e^{-x}\cdot dx = \Gamma(6) = 6! \end{split}$$
2. $\displaystyle \int_{0}^{+\infty}x\sqrt{x}e^{-x}\cdot dx$ $$\begin{split} &\int_{0}^{+\infty} x \sqrt{x}e^{-x}\cdot dx = \int_{0}^{+\infty} x^{\frac{3}{2}}e^{-x}\cdot dx\\ = & \Gamma(\frac{5}{2}) = \frac{3}{2}\cdot\frac{1}{2}\Gamma(\frac{1}{2}) = \frac{3 \sqrt{\pi}}{4} \\ \end{split}$$
3. $\displaystyle \int_{0}^{+\infty}x^{2}e^{-x^{2}}\cdot dx$ $$\begin{split} &\int_{0}^{+\infty}x^{2}e^{-x^{2}}\cdot dx \xlongequal{x = \sqrt{t}}\int_{0}^{+\infty} \frac{1}{2}t^{\frac{1}{2}}e^{-t}\cdot dt\\ =& \frac{1}{2}\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{4} \end{split}$$

$f(x)\in C(-\infty,a],\ \int_{-\infty}^{a}f(x)\cdot dx$

1. 定义 $\displaystyle\int_{b}^{a} f(x)\cdot dx = F(a)-F(b)$
   • $F(a)-F(b)$ 与 $\displaystyle \int_{-\infty}^{a}f(x)\cdot dx$ 不同
   • $\displaystyle \lim_{b\to -\infty} [F(a)-F(b)]$ 与 $\displaystyle \int_{-\infty}^{a}f(x)\cdot dx$ 相同
     1. 若 $\displaystyle \lim_{b\to -\infty} [F(a)-F(b)] = A$ 则 $\displaystyle \int_{-\infty}^{a}f(x)\cdot dx=A$
     2. 若 $\displaystyle \lim_{b\to -\infty} [F(a)-F(b)]\ \nexists$ 则 $\displaystyle \int_{-\infty}^{a}f(x)\cdot dx$ 发散
2. 判别法
   1. 若 $\displaystyle \exists\ \alpha>1; \lim_{x\to -\infty}x^{\alpha}f(x)\ \exists$，则收敛
   2. 若 $\displaystyle \exists\ \alpha\le 1; \lim_{x\to -\infty}x^{\alpha}f(x)\ = k(\neq 0)\text{或}\infty$，则发散

区间有限的反常积分

$f(x)\in C(a,b]\text{且} f(a+0) = \infty\quad \int_{a}^{b}f(x)\cdot dx$

1. 定义 $\displaystyle\forall\ \epsilon >0,\ \int_{a+\epsilon}^{b} f(x)\cdot dx$
   • $F(b)-F(a+\epsilon)$ 与 $\displaystyle \int_{a+\epsilon}^{b}f(x)\cdot dx$ 不同
   • $\displaystyle \lim_{\epsilon\to 0}[F(b)-F(a+\epsilon)]$ 与 $\displaystyle \int_{a+\epsilon}^{b}f(x)\cdot dx$ 相同
     1. 若 $\displaystyle \lim_{\epsilon\to 0}[F(b)-F(a+\epsilon)]=A$ 则 $\displaystyle \int_{a+\epsilon}^{b}f(x)\cdot dx=A$
     2. 若 $\displaystyle \lim_{\epsilon\to 0}[F(b)-F(a+\epsilon)]\ \nexists$ 则 $\displaystyle \int_{a+\epsilon}^{b}f(x)\cdot dx$ 发散
2. 判别法
   1. 若 $\displaystyle\exists\ \alpha<1, \lim_{x\to a^{+}}(x-a)^{\alpha}f(x)\ \exists$ 收敛
   2. 若 $\displaystyle\exists\ \alpha\ge 1,\lim_{x\to a^{+}}(x-a)^{\alpha}f(x) = k(\neq 0)\text{或}\infty$ 发散

例题

1. $\displaystyle \int_{1}^{2}\frac{dx}{x\sqrt{x-1}}$ $$\begin{array}{ll} &\displaystyle \forall\ \epsilon > 0 \int_{1+\epsilon}^{2}\frac{dx}{x\sqrt{x-1}} = \int_{1+\epsilon}^{2}\frac{d(x-1)}{[1+(x-1)]\sqrt{x-1}} \\ & \displaystyle = \int_{\epsilon}^{1} \frac{dx}{(1+x)\sqrt{x}} = 2\int_{\epsilon}^{1} \frac{d(\sqrt{x})}{1+(\sqrt{x})^{2}} = 2\arctan \sqrt{x}\vert_{\epsilon}^{1} \\ & \displaystyle= 2(\frac{\pi}{4} - \arctan \sqrt{\epsilon})\\ \because & \displaystyle \lim_{\epsilon\to 0^{+}}\arctan \sqrt{\epsilon} = 0 \\ \therefore & \text{原式} = \frac{\pi}{2} \\ \end{array}$$

2. $\displaystyle \int_{1}^{2}\frac{dx}{(x-1)^{2}}$ $$\begin{array}{ll} &\displaystyle \forall\ \epsilon > 0 \int_{1+\epsilon}^{2}\frac{dx}{(x-1)^{2}} = \int_{\epsilon}^{1}\frac{dx}{x^{2}} = \left.-\frac{1}{x}\right\vert_{\epsilon}^{1} \\ & \displaystyle = -1+\frac{1}{\epsilon} \\ \because & \displaystyle \lim_{\epsilon\to 0^{+}} \frac{1}{\epsilon}= +\infty \\ \therefore & \text{原式发散} \end{array}$$

3. 讨论 $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{x^{2}+x}}$ 的敛散性 $$\begin{array}{ll} \because & \displaystyle \lim_{x\to 0^{+}} (x-0)^{\frac{1}{2}}\frac{1}{\sqrt{x}\sqrt{x+1}} = 1 \text{ 且 } \alpha=\frac{1}{2}<1 \\ \therefore & \displaystyle \int_{0}^{1}\frac{dx}{\sqrt{x^{2}+x}} \text{收敛} \\ & \displaystyle \text{原式}=\int_{0}^{1}\frac{dx}{\sqrt{x(1+x)}} = 2 \int_{0}^{1}\frac{d \sqrt{x}}{\sqrt{1+(\sqrt{x})^{2}}} \\ & \displaystyle = 2 \int_{0}^{1} \frac{dx}{\sqrt{1+x^{2}}} \xlongequal{x = \tan t} 2 \int_{0}^{\frac{\pi}{4}} \cos t \cdot d(\tan t) = 2\int_{0}^{\frac{\pi}{4}}\sec t\cdot dt \\ & = (2\ln\vert \sec t +\tan t\vert )\vert _{0}^{\frac{\pi}{4}} = 2\ln(\sqrt{2} +1) \end{array}$$

$f(x)\in C[a,b)\text{且} f(b-0) = \infty\quad \int_{a}^{b}f(x)\cdot dx$

1. 定义 $\displaystyle\forall\ \epsilon >0,\ \int_{a}^{b-\epsilon} f(x)\cdot dx$
   • $F(b-\epsilon)-F(a)$ 与 $\displaystyle \int_{a}^{b-\epsilon}f(x)\cdot dx$ 不同
   • $\displaystyle \lim_{\epsilon\to 0}[F(b-\epsilon)-F(a)]$ 与 $\displaystyle \int_{a}^{b-\epsilon}f(x)\cdot dx$ 相同
     1. 若 $\displaystyle \lim_{\epsilon\to 0}[F(b-\epsilon)-F(a)]=A$ 则 $\displaystyle \int_{a}^{b-\epsilon}f(x)\cdot dx=A$
     2. 若 $\displaystyle \lim_{\epsilon\to 0}[F(b-\epsilon)-F(a)]\ \nexists$ 则 $\displaystyle \int_{a}^{b-\epsilon}f(x)\cdot dx$ 发散
2. 判别法
   1. 若 $\displaystyle\exists\ \alpha<1, \lim_{x\to b^{-}}(b-x)^{\alpha}f(x)\ \exists$ 收敛
   2. 若 $\displaystyle\exists\ \alpha\ge 1,\lim_{x\to b^{-}}(b-x)^{\alpha}f(x) = k(\neq 0)\text{或}\infty$ 发散

例题

1. $\displaystyle \int_{0}^{1}\ln(1-x)\cdot dx$ $$\begin{array}{ll} &\forall\ \epsilon > 0 \\ & \displaystyle\int_{0}^{1-\epsilon} \ln (1-x)\cdot dx = x\ln(1-x)\vert^{1-\epsilon}_{0} + \int_{0}^{1-\epsilon}\frac{x}{1-x}\cdot dx\\ & \displaystyle = x\ln(1-x)\vert^{1-\epsilon}_{0} - \int_{0}^{1-\epsilon}1\cdot dx + \int_{0}^{1-\epsilon}\frac{1}{1-x}\cdot dx \\ & = x\ln(1-x)\vert_{0}^{1-\epsilon} - x\vert_{0}^{1-\epsilon} -\ln(1-x)\vert _{0}^{1-\epsilon} \\ & = (x-1)\ln(1-x)\vert^{1-\epsilon}_{0}- x\vert_{0}^{1-\epsilon} \\ \because & \displaystyle \lim_{\epsilon\to 0^{+}}(1-\epsilon - 1)\ln(1-1+\epsilon) = -\lim_{\epsilon\to 0^{+}}\frac{\ln \epsilon}{\frac{1}{\epsilon}} \\ & = \displaystyle \lim_{\epsilon\to0^{+}}\frac{\frac{1}{\epsilon}}{\frac{1}{\epsilon^{2}}} = \lim_{\epsilon\to 0^{+}}\epsilon = 0 \\ \therefore & \text{原式} = 0 -1 = -1 \\ \end{array}$$
2. $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}$ $$\begin{array}{ll} \because & \displaystyle \lim_{x\to 1^{-}} (1-x)^{\frac{1}{2}}\frac{1}{\sqrt{(1-x)(1+x)}} = \frac{1}{\sqrt{2}} \text{ 且 } \alpha=\frac{1}{2}<1\\ \therefore & \displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}} \text{收敛} \\ \therefore & \displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}} = \arcsin x\vert^{1}_{0} = \frac{\pi}{2} \\ \end{array}$$