反常积分
(1)
∞∞{f(x)=∞g(x)=∞f′(x)/g′(x)=L(x→a)⇒f(x)g(x)→L\frac{\infty}{\infty} \left\{\begin{matrix}
f(x) = \infty\\
g(x) = \infty\\
f'(x)/g'(x) = L
\end{matrix}\right.(x\to a) \Rightarrow \frac{f(x)}{g(x)} \to L∞∞⎩⎨⎧f(x)=∞g(x)=∞f′(x)/g′(x)=L(x→a)⇒g(x)f(x)→L
(2)f(x)<<g(x)⇒(x→∞)f(x)g(x)→0f(x) << g(x) \Rightarrow(x \to \infty) \frac{f(x)}{g(x)} \to 0f(x)<<g(x)⇒(x→∞)g(x)f(x)→0.
Ex1.
lnx≪xp≪ex≪ex2(x→∞)lnx \ll x^p \ll e^x \ll e^{x^2}(x \to \infty)lnx≪xp≪ex≪ex2(x→∞)
1lnx≫1xp≫1ex≫1ex2\frac{1}{lnx} \gg \frac{1}{x^p} \gg \frac{1}{e^x} \gg \frac{1}{e^{x^2}}lnx1≫xp1≫ex1≫ex21
(3) ∫a∞f(x)dx=limn→∞∫anf(x)dx\int_a^{\infty} f(x)dx=lim_{n \to \infty}\int_a^n f(x)dx∫a∞f(x)dx=limn→∞∫anf(x)dx
Ex 2.∫0∞e−kxdx=1k\int_0^{\infty} e^{-kx}dx= \frac{1}{k}∫0∞e−kxdx=k1
∫0∞e−kxdx=∫0ne−kxdx=−1ke−kx∣0n=k→∞−1ke−kn−(−1k)=1k\int_0^{\infty} e^{-kx}dx= \int_0^ne^{-kx}dx=-\frac{1}{k}e^{-kx}|_0^n=_{k\to \infty}-\frac{1}{k}e^{-kn}-(-\frac{1}{k})=\frac{1}{k}∫0∞e−kxdx=∫0ne−kxdx=−k1e−kx∣0n=k→∞−k1e−kn−(−k1)=k1
即：
∫0∞e−kxdx=−1ke−kx∣0∞=k→∞−1ke−∞−(−1k)=1k\int_0^{\infty} e^{-kx}dx=-\frac{1}{k}e^{-kx}|_0^{\infty}=_{k\to \infty}-\frac{1}{k}e^{-\infty}-(-\frac{1}{k})=\frac{1}{k}∫0∞e−kxdx=−k1e−kx∣0∞=k→∞−k1e−∞−(−k1)=k1
Ex 3.∫−∞∞e−x2dx=π\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}∫−∞∞e−x2dx=π(博彩模型)
Ex 4.∫1∞1xdx(发散)\int_1^{\infty}\frac{1}{x}dx(发散)∫1∞x1dx(发散)
∫1∞dxx=lnx∣1∞=∞\int_1^{\infty}\frac{dx}{x}=lnx|_1^{\infty}=\infty∫1∞xdx=lnx∣1∞=∞
(4)∫1∞dxxp=x−p+1−p+1∣1∞=(∞)−p+1−p+1−1−p+1={0,p>1∞,p<1\int_1^{\infty} \frac{dx}{x^p}=\frac{x^{-p+1}}{-p+1}|_1^{\infty}=\frac{(\infty)^{-p+1}}{-p+1}-\frac{1}{-p+1} = \left\{\begin{matrix}
0,& p>1 \\
\infty,& p<1
\end{matrix}\right.∫1∞xpdx=−p+1x−p+1∣1∞=−p+1(∞)−p+1−−p+11={0,∞,p>1p<1
总结：∫1∞dxxp,发散，p<=1∫1∞dxxp，收敛，p>1,（=1p−1）\int_1^{\infty}\frac{dx}{x^p} ,发散，p<=1\\
\int_1^{\infty}\frac{dx}{x^p} ，收敛，p > 1,（=\frac{1}{p-1}）∫1∞xpdx,发散，p<=1∫1∞xpdx，收敛，p>1,（=p−11）
极限的比较
f(x) g(x)⇒x→∞∫a∞f(x)dx,∫a∞g(x)dx(两者同时收敛或发散)f(x)~g(x) \Rightarrow^{x \to \infty} \int_a^{\infty}f(x)dx,\int_a^{\infty}g(x)dx(两者同时收敛或发散)f(x) g(x)⇒x→∞∫a∞f(x)dx,∫a∞g(x)dx(两者同时收敛或发散)
Ex 5. ∫0∞dxx2+10≈∫1∞dxx=lnx∣1∞=∞x2+10≈10\int_0^{\infty} \frac{dx}{\sqrt{x^2+10}} \approx \int_1^{\infty}\frac{dx}{x}=lnx|_1^{\infty}=\infty\\
\sqrt{x^2+10} \approx \sqrt{10}∫0∞x2+10dx≈∫1∞xdx=lnx∣1∞=∞x2+10≈10
Ex 6. ∫10∞dxx3+3≈∫10∞dxx321x3+3≈1x3=1x32\int_{10}^{\infty}\frac{dx}{\sqrt{x^3+3}} \approx \int_{10}^{\infty}\frac{dx}{x^{\frac{3}{2}}}\\
\frac{1}{\sqrt{x^3+3}} \approx \frac{1}{\sqrt{x^3}}=\frac{1}{x^{\frac{3}{2}}}∫10∞x3+3dx≈∫10∞x23dxx3+31≈x31=x231
Ex 7.∫−∞∞e−x2dx=2∫0∞exdx=2∫01e−x2dx+2∫1∞e−xdx\int_{-\infty}^{\infty}e^{-x^2}dx=2\int_0^{\infty}e^xdx=2\int_0^1e^{-x^2}dx+2\int_1^{\infty}e^{-x}dx∫−∞∞e−x2dx=2∫0∞exdx=2∫01e−x2dx+2∫1∞e−xdx
***注意:***∫dxx2=−x−1∣−11=−2(是错误的)\int \frac{dx}{x^2} = -x^{-1}|_{-1}^{1}=-2(是错误的)∫x2dx=−x−1∣−11=−2(是错误的)
总结： 如果极限存在，函数就收敛
Ex 8.∫01dxx=∫01x−12dx=2x12∣01=2(收敛)\int_0^1\frac{dx}{\sqrt{x}}=\int_0^1x^{-\frac{1}{2}}dx=2x^{\frac{1}{2}}|_0^1=2(收敛)∫01xdx=∫01x−21dx=2x21∣01=2(收敛)
Ex 9.∫01dxx=lnx∣01=ln1−ln0=+∞(发散)\int_0^1\frac{dx}{x} = lnx|_0^1=ln1 - ln0 = +\infty(发散)∫01xdx=lnx∣01=ln1−ln0=+∞(发散)
总结：1x12≪1x≪1x2(x→0+)1x12≫1x≫1x2(x→∞)\frac{1}{x^{\frac{1}{2}}} \ll \frac{1}{x} \ll \frac{1}{x^2}(x \to 0^+)\\
\frac{1}{x^{\frac{1}{2}}} \gg \frac{1}{x} \gg \frac{1}{x^2}(x \to \infty)x211≪x1≪x21(x→0+)x211≫x1≫x21(x→∞)
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