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Use Limit Comparison Test or Direct Comparison Test?

December 7, 2024
3 min read
  • 💬Question: We start with the question that when should I use for improper integral?

  • 🗣Answer: When you can’t do a direct comparison, use the limit comparison.

  • 📝Definition: click the footnote to review the concept if needed.

    • direct comparison test1
    • limit comparison test2
    • improper integral3
  • 🗃Example: Investigate the convergence of 11exxdx\displaystyle \int_1^{\infty}\frac{1-e^{-x}}{x}dx

  • ✍Step by Step Reasoning:

    • By the improper integral3, we know that the integrand suggests a comparison of
      • f(x)=1exx\displaystyle \textcolor{purple}{f(x)=\frac{1-e^{-x}}{x}} and
      • g(x)=1x\displaystyle \textcolor{pink}{g(x)=\frac{1}{x}}
      • because 1exx<1x\displaystyle \textcolor{purple}{\frac{1-e^{-x}}{x}} < \textcolor{pink}{\frac{1}{x}} in [1,)\displaystyle [1,\infty).
    • However, we cannot use the Direct Comparison Test1.
      • from its definition we know that when f(x)g(x)f(x) \leq g(x), the antecedents are
        • ag(x)dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx} converges, or
        • af(x)dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx} diverges
      • however, in this case, the ag(x)dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx} diverges… This doesn’t obey the rule.
      • Meaning, the rule of direct comparison suggests that af(x)dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx} diverges is antecedent (因) and ag(x)dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx} diverges is consequent (果).
      • Now it’s the opposite.
    • Therefore, we should use the Limit Comparison Test2.
      • using the Limit Comparison Test we find that
        • limxf(x)g(x)=limx(1exx)(x1)=limx(1ex)=1\displaystyle \lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\left(\frac{1-e^{-x}}{x}\right)\left(\frac{x}{1}\right)=\lim_{x\to\infty}(1-e^{-x})=1
      • which is a positive finite limit.
    • Result.
      • Therefore, 11exxdx\displaystyle \int_1^{\infty}\frac{1-e^{-x}}{x}dx diverges because 11xdx\displaystyle \int_1^{\infty}\frac{1}{x}dx diverges.
  • 📜Key Takeaway:

    • Use limit comparison test✅ when you can’t use direct comparison test🙈.
    • Identify what is antecedent and what is consequent in the mathematical definition.
  • 🔗Reference: Thomas Calculus > Chapter 8 Techniques of Integration > 8.8 Improper Integrals

Footnotes

  1. Direct Comparison Test: Let ff and gg be continuous on [a,)[a,\infty) with 0f(x)g(x)0\leq f(x)\leq g(x) for all xax\geq a. Then af(x)dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx} converges if ag(x)dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx} converges, and ag(x)dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx} diverges if af(x)dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx} diverges. 2

  2. Limit Comparison Test: If the positive function ff and gg are continuous on [a,)[a,\infty), and if limxf(x)g(x)=L,0<L<, \lim_{x\to\infty}\frac{f(x)}{g(x)}=L, \quad 0<L<\infty, then af(x)dxandag(x)dx\int_{a}^{\infty}f(x)dx\quad\text{and}\quad \int_{a}^{\infty}g(x)dx both converge or both diverge. 2

  3. An improper integral is defined by af(x)dx=limNaNf(x)dx.\int _a^{\infty } f(x) \, dx = \lim _{N\rightarrow \infty } \int _a^{N} f(x) \, dx. 2

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