8.9 不当综合器件:结合无限限制

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  不当综合器件: 结合无限限制

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  A probability density function $\begin{array}{r}f(x)\end{array}$  must have the property that integrating it over its entire domain $\begin{array}{r}a\le x\le b\end{array}$  equals 1, i.e., $\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx=1\end{array}$ . Suppose a random variable has an exponential probability density function $\begin{array}{r}f(x)=A{e}^{-|x|}\end{array}$  over all real values of $\begin{array}{r}x\end{array}$ . Then $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}A{e}^{-|x|}dx=1\end{array}$ . How do you evaluate this integral with the of integration to find the value of $\begin{array}{r}A\end{array}$ ? 
  ::概率密度函数 f( x) 必须具有将其整合到整个域的属性, 即 {abf( x) dx=1. 随机变量的指数概率密度函数 f( x) = Ae* = x 的所有实际值。 然后 {f( x) dx * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * a* = 1。 随机变量的指数概率密度函数 f( x) = Ae* * * * * * * * * * * * * * * * * * *。 * * * * * f( x) f( x) dx) * * * * * * * * * * * * * * * * * * * * * * a A 值, 如何与集成集成值一起评价此特性以找到 A ?

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  Improper Integrals
  ::不当综合器

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  The concept of  improper integrals  is an extension to the concept of definite integrals. The reason for the term improper is because those integrals either
  ::不当综合体概念是明确综合体概念的延伸,其原因是这些综合体或

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  1. include integration over infinite limits or
     ::包括无无限限制或无限制的融合,或
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  2. the integrand may become infinite within the limits of integration.
     ::在融合的限度内,整数可能变得无限。

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  Recall that in the definition of definite integral  $\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx\end{array}$  we assume that the interval of integration  $\begin{array}{r}[a,b]\end{array}$  is finite and the function  $\begin{array}{r}f\end{array}$  is continuous on this interval. In this concept, we will look at case 1, where the integration is over infinite limits, but the integrand is continuous over the limits. The table below gives the guide for evaluating these improper integrals.
  ::回顾在确定集成的abf(x)dx的定义中,我们假定集成[a,b]的间隔是有限的,而函数f在这个间隔是连续的。在这个概念中,我们将研究情况1, 集成的界限是无限的, 但整数是连续的。下表提供了评估这些不适当的集成的指南。

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  Definition of Improper Integral for continuous $\begin{array}{r}f(x)\end{array}$
  ::连续f(x)不适当综合整体定义

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  • If $\begin{array}{r}f\end{array}$  is continuous over the interval  $\begin{array}{r}[a,\mathrm{\infty }]\end{array}$  then the improper integral $\begin{array}{r}\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx\end{array}$  is defined as:
    ::如果 f 是 连续间隔[a, ] , 那么不适当的 {a}} f (x) dx 被定义为 :

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  $$\begin{array}{r}\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{l\to \mathrm{\infty }}{lim}\underset{a}{\overset{l}{\int }}f(x)dx.\end{array}$$

  
  ::======================================================================================================================================================================================================================================= =====================================================================================================================================================================================================================================================================================

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  • If $\begin{array}{r}f\end{array}$  is continuous over the interval  $\begin{array}{r}[-\mathrm{\infty },a]\end{array}$ , then the improper integral $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{a}{\int }}f(x)dx\end{array}$  is defined as:
    ::如果 f 是连续间隔[,a] ,则不适当的构件 af(x)dx 的定义如下:

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  $$\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{a}{\int }}f(x)dx=\underset{l\to -\mathrm{\infty }}{lim}\underset{l}{\overset{a}{\int }}f(x)dx.\end{array}$$

  
  ::af(x)dx=limllaf(x)dx。

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  • If the integrand  $\begin{array}{r}f\end{array}$ is continuous over the interval  $\begin{array}{r}[-\mathrm{\infty },\mathrm{\infty }]\end{array}$  then the improper integral  $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx\end{array}$ is defined as:
    ::如果整数 f 是连续间隔[,], 那么不适当的内装 f(x)dx 被定义为 :

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  $$\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{l\to -\mathrm{\infty }}{lim}\underset{l}{\overset{c}{\int }}f(x)dx+\underset{l\to \mathrm{\infty }}{lim}\underset{c}{\overset{l}{\int }}f(x)dx,\end{array}$$

  where $\begin{array}{r}c\end{array}$  is any real number.
  ::*f(x)dx=limllcf(x)dx+limlclf(x)dx,其中c为任何实际数字。

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  If the integration of the improper integral exists, then we say that it  converges . But if the limit of integration fails to exist, then the improper integral is said to diverge .  The integral above has an important geometric interpretation that you need to keep in mind. Recall that, geometrically, the definite integral $\begin{array}{r}\underset{a}{\overset{b}{\int }}f(x)dx\end{array}$ represents the . Similarly, the integral $\begin{array}{r}\underset{a}{\overset{l}{\int }}f(x)dx\end{array}$ is a definite integral that represents the area under the curve  $\begin{array}{r}f\end{array}$  over the interval  $\begin{array}{r}[a,l]\end{array}$  as the figure below shows. However, as  $\begin{array}{r}l\end{array}$  approaches $\begin{array}{r}\mathrm{\infty }\end{array}$ , this area will expand to the area under the curve of  $\begin{array}{r}f\end{array}$ and over the entire interval $\begin{array}{r}[a,\mathrm{\infty }]\end{array}$ . Therefore, the improper integral  $\begin{array}{r}\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx\end{array}$  can be thought of as the area under the function  $\begin{array}{r}f\end{array}$  over the interval  $\begin{array}{r}[a,\mathrm{\infty }]\end{array}$ .
  ::如果存在不适当整体体的整合, 那么我们就会说它会汇合。 但是, 如果融合的极限不存在, 那么不适当的整体体就会被说成是不同的。 以上整体体有一个重要的几何解释, 您需要记住。 回想一下, 从几何角度来说, 确定的整体整体 {abf( x) dx 代表着 。 同样地, 如下图所示 {alf( x) dx 是一个确定的整体体, 代表曲线下的区域, f 的间隔( a, l ) 。 但是, 如 {} / , 则此区域会扩展至 f 曲线下的区域, 以及整个间隔 {a, {f( x) x 。 因此, 不适当的整体体可以被看作 {a*f( x) x) x 函数下的区域 。

  

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  For example, let's evaluate $\begin{array}{r}\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x}\end{array}$ .
  ::例如,让我们来评估1dxxx。

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  We notice immediately that the integral is an improper integral because the upper limit of integration approaches infinity. First, replace the infinite upper limit by the finite limit  $\begin{array}{r}l\end{array}$  and take the limit of  $\begin{array}{r}l\end{array}$ to approach infinity:
  ::我们立即注意到,整体体是一个不适当的整体体,因为一体化方针的上限是无限的。 首先,将无限上限替换为有限限 I, 并用l 的上限来表示无限性 :

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  $$\begin{array}{rl}\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x}& =\underset{l\to \mathrm{\infty }}{lim}\underset{1}{\overset{l}{\int }}\frac{dx}{x}\\ & =\underset{l\to \mathrm{\infty }}{lim}[\ln x{]}_1^l\\ & =\underset{l\to \mathrm{\infty }}{lim}(\ln l-\ln 1)\\ & =\underset{l\to \mathrm{\infty }}{lim}(\ln l)\\ \underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x}& =\mathrm{\infty }\end{array}$$

  
  ::================================================================================================================================================ ===================================================================================================================================================================================================================================================================================================================

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  Thus the integral diverges .
  ::因此,整体的分歧。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to evaluate the improper integral $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}A{e}^{-|x|}dx=1\end{array}$  to find the value of $\begin{array}{r}A\end{array}$ .
  ::先前,您被要求评估不适当的整体 {( x) dx}}}}}Aexdx=1 以找到 A 的值 。

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  Using the limit definition of the improper integral,

  $$\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}A{e}^{-|x|}dx=1=\underset{p\to -\mathrm{\infty }}{lim}\underset{p}{\overset{0}{\int }}A{e}^{x}dx+\underset{p\to \mathrm{\infty }}{lim}\underset{0}{\overset{p}{\int }}A{e}^{-x}dx=2A\underset{p\to \mathrm{\infty }}{lim}{\left[-e^{-x}\right]}_0^p=2A[0+1].\end{array}$$

  
  ::使用不适当的内装件的限值定义, @Aexdx=1=limpp0Aexdx+limp0pAe-xdx=2Alimp[-e-x]0p=2A[0+1]。

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  Therefore $\begin{array}{r}A=\frac{1}{2}\end{array}$ , so that $\begin{array}{r}f(x)=\frac{1}{2}{e}^{-|x|}\end{array}$ .
  ::因此A=12, f(x)=12ex。

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  Example 2
  ::例2

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  Evaluate $\begin{array}{r}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2}\end{array}$ .
  ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:

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  $$\begin{array}{rl}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2}& =\underset{l\to \mathrm{\infty }}{lim}\underset{2}{\overset{l}{\int }}\frac{dx}{x^2}\\ & =\underset{l\to \mathrm{\infty }}{lim}{\left[\frac{-1}{x}\right]}_2^l\\ & =\underset{l\to \mathrm{\infty }}{lim}\left(\frac{-1}{l}+\frac{1}{2}\right)\\ \underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2}& =\frac{1}{2}\end{array}$$

  
  ::============================================================================================================================================================================================================================================================================================================

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  Thus the integration converges to  $\begin{array}{r}\frac{1}{2}\end{array}$ .
  ::因此,一体化已趋同到12个。

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  Example 3
  ::例3

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  Evaluate  $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}\end{array}$ .
  ::评估 dx1+x2。

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  What we need to do first is to split the integral into two intervals  $\begin{array}{r}(-\mathrm{\infty },0]\end{array}$ and  $\begin{array}{r}[0,\mathrm{\infty })\end{array}$ .
  ::我们首先需要做的是将有机体分成两个间隔(,0)和[0]。

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  So the integral becomes:
  ::因此,整体体变成:

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  $$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}& =\underset{-\mathrm{\infty }}{\overset{0}{\int }}\frac{dx}{1+x^2}+\underset{0}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}.\\ & =\underset{l\to -\mathrm{\infty }}{lim}\underset{l}{\overset{0}{\int }}\frac{dx}{1+x^2}+\underset{l\to \mathrm{\infty }}{lim}\underset{0}{\overset{l}{\int }}\frac{dx}{1+x^2}\\ & =\underset{l\to -\mathrm{\infty }}{lim}{\left[{\tan }^{-1}x\right]}_l^0+\underset{l\to \mathrm{\infty }}{lim}{\left[{\tan }^{-1}x\right]}_0^l\\ & =\underset{l\to -\mathrm{\infty }}{lim}\left[{\tan }^{-1}0-{\tan }^{-1}l\right]+\underset{l\to \mathrm{\infty }}{lim}\left[{\tan }^{-1}l-{\tan }^{-1}0\right]\\ & =\left[0-\left(-\frac{\pi }{2}\right)\right]+\left[\left(\frac{\pi }{2}\right)-0\right]\\ \underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}& =\pi \end{array}$$

  
  ::=======================================================================================================================================================+=====================================================================================================================================================================================================================================================================================================================================================================

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  Remark: In this example, we split the integral at  $\begin{array}{r}x=0\end{array}$ . However, we could have split the integral at any value of  $\begin{array}{r}x=c\end{array}$  without affecting the convergence or divergence of the integral. The choice is completely arbitrary.
  ::备注:在此示例中, 我们将整体体在 x=0 上拆分。 但是, 我们本可以将整体体在 x=c 的任何值上拆分, 而不影响整体体的趋同或差异。 选择是完全任意的 。

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  Example 4
  ::例4

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  Evaluate the improper integral  $\begin{array}{r}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-5x}dx\end{array}$ .
  ::评估不适当的整体 0e -5xdx。

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  $$\begin{array}{rl}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-5x}dx& =\underset{a\to \mathrm{\infty }}{lim}\underset{0}{\overset{a}{\int }}{e}^{-5x}dx\\ & =\underset{a\to \mathrm{\infty }}{lim}{\left[-\frac{e^{-5x}}{5}\right]}_0^a\\ & =\underset{a\to \mathrm{\infty }}{lim}\left[-\frac{e^{-5a}}{5}+\frac{1}{5}\right]\\ \underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-5x}dx& =\frac{1}{5}\\ \underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\frac{dx}{1+x^2}& =\pi \end{array}$$

  
  ::=========================================================================================================================================================================================================================================================================================================================================================================

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  Review
  ::回顾

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  1. Determine whether the following integrals are improper. If so, explain why.
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     1. $\begin{array}{r}\underset{1}{\overset{7}{\int }}\frac{x+2}{x+3}dx\end{array}$
        ::17x+2x+3dx
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     2. $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{0}{\int }}\frac{x^3+2}{x-7}dx\end{array}$
        ::0x3+2x-7dx
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     3. $\begin{array}{r}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{x-2}}dx\end{array}$
        ::01x-2dx
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     4. $\begin{array}{r}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx\end{array}$
        ::0e - xxxx
     
     ::确定以下的积分是否不适当 。 如果是的话, 请解释原因 。 @ 17x+2xx+3dx @ 0x3+2x- 7dx @ 01x- 2dx @ 01x- 2dx_x_xx_xx_xxx_2xxx_xxx_xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx1x1x- 2dxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  For #2-7, evaluate the integral or state whether it diverges.
  ::2 -7, 评估整体或状态是否不同 。

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  2. $\begin{array}{r}\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^{2.001}}dx\end{array}$
     ::==============================================================================================================================================
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  3. $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{-2}{\int }}\left[\frac{1}{x-1}-\frac{1}{x+1}\right]dx\end{array}$
     ::2[1x-1-1x+1x]dx
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  4. $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{0}{\int }}{e}^{5x}dx\end{array}$
     ::0e5xdx
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  5. $\begin{array}{r}\underset{1}{\overset{\mathrm{\infty }}{\int }}{x}^{2}dx\end{array}$
     ::===================================================================================================================================
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  6. $\begin{array}{r}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{5x}dx\end{array}$
     ::e5xxx
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  7. $\begin{array}{r}\underset{2}{\overset{\mathrm{\infty }}{\int }}\left[\frac{1}{x-1}-\frac{1}{x+1}\right]dx\end{array}$
     ::2[1x -1 -1 -1x+1x]dx
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  8. The region between the $\begin{array}{r}x\end{array}$ -axis and the curve  $\begin{array}{r}y=e^{-x}\end{array}$ for  $\begin{array}{r}x\ge 0\end{array}$ is revolved about the $\begin{array}{r}x\end{array}$ -axis.
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     1. Find the volume of revolution, $\begin{array}{r}V\end{array}$ .
        ::找出革命的卷土重来, V.
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     2. Find the surface area of the volume generated, $\begin{array}{r}S\end{array}$ .
        ::查找生成体积的表面区域, S 。
     
     ::xx轴与 y=e-x 之间的区域为 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  For #9-12, evaluate the integral.
  ::9 -12, 评估整体体.

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  9. $\begin{array}{r}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{10}{x^3}dx\end{array}$
     ::======================================================================================================================================================
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  10. $\begin{array}{r}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{10}{x^{\frac{1}{3}}}dx\end{array}$
      ::=====================================================================================================================================================
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  11. $\begin{array}{r}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{10}{x^5}dx\end{array}$
      ::210x5dx
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  12. $\begin{array}{r}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{10}{x^{\frac{1}{5}}}dx\end{array}$
      ::=====================================================================================================================================================
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  13. Based on the results of #9-12 above, tell whether the following statement is true or false:  $\begin{array}{r}\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^p}\end{array}$ converges if $\begin{array}{r}p>1\end{array}$ , and diverges otherwise.
      ::根据上面#9-12的结果,请说明以下声明是真实的还是虚假的:如果p>1, 则 1dxxp会趋同, 否则会不同 。
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  14. The Gamma Function, $\begin{array}{r}\mathrm{\Gamma }(x)\end{array}$ , is an improper integral that appears frequently in quantum physics. It is defined as $\begin{array}{r}\mathrm{\Gamma }(x)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{x-1}{e}^{-t}dt\end{array}$ . The integral converges for all  $\begin{array}{r}x\ge 0\end{array}$ . Find $\begin{array}{r}\mathrm{\Gamma }(1)\end{array}$ .
      ::Gamma 函数 \ \ (x) 是量子物理中经常出现的不适当的组合。 它被定义为 \ (x)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  15. Refer to the Gamma Function defined in the previous exercise to prove that: $\begin{array}{r}\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }(x)\end{array}$ , for all $\begin{array}{r}x\ge 0\end{array}$ .
      ::参考上一个练习中定义的 Gamma 函数以证明 : @ (x+1) =xn(x) , 代表全部 x%0 。

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。