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Use Limit Comparison Test or Direct Comparison Test?

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

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

  • πŸ“Definition: click the footnote to review the concept if needed.

    • direct comparison test1
    • limit comparison test2
    • improper integral3
  • πŸ—ƒExample: Investigate the convergence of ∫1∞1βˆ’eβˆ’xxdx\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)=1βˆ’eβˆ’xx\displaystyle \textcolor{purple}{f(x)=\frac{1-e^{-x}}{x}} and
      • g(x)=1x\displaystyle \textcolor{pink}{g(x)=\frac{1}{x}}
      • because 1βˆ’eβˆ’xx<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
        • ∫a∞g(x) dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx}Β converges, or
        • ∫a∞f(x) dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx}Β diverges
      • however, in this case, the ∫a∞g(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 ∫a∞f(x) dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx} diverges is antecedent (ε› ) and ∫a∞g(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
        • lim⁑xβ†’βˆžf(x)g(x)=lim⁑xβ†’βˆž(1βˆ’eβˆ’xx)(x1)=lim⁑xβ†’βˆž(1βˆ’eβˆ’x)=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, ∫1∞1βˆ’eβˆ’xxdx\displaystyle \int_1^{\infty}\frac{1-e^{-x}}{x}dx diverges because ∫1∞1xdx\displaystyle \int_1^{\infty}\frac{1}{x}dx diverges.
  • πŸ“œKey Takeaway:

    • Use limit comparison testβœ… when you can’t use direct comparison testπŸ™ˆ.
    • Identify what is antecedent and what is consequent in the mathematical definition.
  • πŸ”—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 0≀f(x)≀g(x)0\leq f(x)\leq g(x) for all xβ‰₯ax\geq a. Then ∫a∞f(x) dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx}Β converges if ∫a∞g(x) dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx}Β converges, and ∫a∞g(x) dx\color{pink}{\displaystyle \int _a^{\infty } g(x) \, dx}Β diverges if ∫a∞f(x) dx\color{purple}{\displaystyle \int _a^{\infty } f(x) \, dx}Β diverges. ↩ ↩2

  2. Limit Comparison Test: If the positive function ff and gg are continuous on [a,∞)[a,\infty), and if lim⁑xβ†’βˆžf(x)g(x)=L,0<L<∞, \lim_{x\to\infty}\frac{f(x)}{g(x)}=L, \quad 0<L<\infty, then ∫a∞f(x)dxand∫a∞g(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. AnΒ improper integralΒ is defined by ∫a∞f(x) dx=lim⁑Nβ†’βˆžβˆ«aNf(x) dx.\int _a^{\infty } f(x) \, dx = \lim _{N\rightarrow \infty } \int _a^{N} f(x) \, dx. ↩ ↩2