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Wize University Calculus 2 Textbook > Improper Integrals

Comparison Theorem for Improper Integrals

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Comparison Theorem for Improper Integrals

Common Improper Integrals

1. ∫a∞1xp  ⁣dx{\displaystyle\int^\infty_a}\frac{1}{x^p}\de{x}
  • If p≤1p\le1: diverges
  • If p>1p>1: converges

2. ∫0a1xp  ⁣dx{\displaystyle\int^a_0}\frac{1}{x^p}\de{x}
  • If p<1p<1: converges
  • If p≥1p\ge1: diverges

Comparison Theorem

Suppose that ff and gg are continuous and f(x)≥g(x)≥0f(x)\ge g(x)\ge 0 for x≥a.x \ge a.
a. If ∫a∞f(x)  ⁣dx{\displaystyle\int^\infty_a}f(x)\de{x} converges, then ∫a∞g(x)  ⁣dx{\displaystyle\int^\infty_a}g(x)\de{x} converges.
b. If ∫a∞g(x)  ⁣dx{\displaystyle\int^\infty_a}g(x)\de{x} diverges, then ∫a∞f(x)  ⁣dx{\displaystyle\int^\infty_a}f(x)\de{x} diverges.

Notes:

  • The theorem only tells you if an improper integral diverges or converges, not the value it converges to
  • Often, we will use the common improper integrals and the Comparison Theorem

Practice: Comparison Test

Determine whether the following improper integral is convergent or divergent:
∫1∞xx2+x  ⁣dx{\displaystyle\int}^\infty_1\frac{\sqrt{x}}{x^2+x}\de{x}

Practice: Comparison Theorem

Determine whether ∫02arctan⁡x5x dx{\displaystyle \int^2_0}\frac{\arctan x^5}{\sqrt{x}}\ dx converges or diverges.

Practice: Comparison Theorem

Determine if ∫01/41x2+x5  ⁣dx{\displaystyle\int^{1/4}_0}\dfrac{1}{x^2+x^5}\de{x} is convergent or divergent.