Convergence of improper integrals: the p-test

About this lesson

Calculus BC (AP-aligned) › Improper integrals

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Convergence of improper integrals: the p-test

Warm-up

Three quick questions from earlier lessons.

Discover

∫1∞dxx\displaystyle\int_1^{\infty} \dfrac{dx}{x} diverges and ∫1∞dxx2\displaystyle\int_1^{\infty} \dfrac{dx}{x^2} converges. What about ∫1∞dxx3\displaystyle\int_1^{\infty} \dfrac{dx}{x^3}?

  1. It converges
  2. It diverges

Key idea

Powers of xx decide convergence at once.

p-test: ∫1∞dxxp\displaystyle\int_1^{\infty} \dfrac{dx}{x^p} converges exactly when p>1p > 1 (to 1p−1\tfrac{1}{p - 1}). ∫01dxxp\displaystyle\int_0^1 \dfrac{dx}{x^p} converges exactly when p<1p < 1 (to 11−p\tfrac{1}{1 - p}). For p=1p = 1 both diverge.

Worked example

Classify three integrals.

∫1∞x−3/2 dx\displaystyle\int_1^{\infty} x^{-3/2}\,dx converges, to 22
p = 3/2 > 1 at ∞.
∫01x−1/2 dx\displaystyle\int_0^1 x^{-1/2}\,dx converges, to 22
p = 1/2 < 1 near 0.
∫1∞x−1/2 dx\displaystyle\int_1^{\infty} x^{-1/2}\,dx diverges
p = 1/2 ≤ 1 at ∞.

Your turn, guided

Your turn: the value of ∫1∞dxx4\displaystyle\int_1^{\infty} \dfrac{dx}{x^4}.

13\tfrac{1}{3}
1/(p − 1) with p = 4.

Practice ★

  1. For which pp does ∫2∞dxxp\displaystyle\int_{ 2 }^{\infty} \dfrac{dx}{x^p} converge?
    1. p>1p > 1
    2. p≥1p \ge 1
    3. p<1p < 1
  1. Evaluate ∫3∞5x3 dx\displaystyle\int_{ 3 }^{\infty} \dfrac{ 5 }{x^3}\,dx.
  2. Evaluate ∫4∞1x3 dx\displaystyle\int_{ 4 }^{\infty} \dfrac{ 1 }{x^3}\,dx.

Watch out!

Practice ★★

  1. Which improper integrals converge? Tick all that do.
    1. ∫1∞3x2 dx\displaystyle\int_1^{\infty} \dfrac{ 3 }{x^2}\,dx
    2. ∫013x dx\displaystyle\int_0^1 \dfrac{ 3 }{\sqrt{x} }\,dx
    3. ∫0∞e−3x dx\displaystyle\int_0^{\infty} e^{-3x}\,dx
    4. ∫1∞3x dx\displaystyle\int_1^{\infty} \dfrac{ 3 }{\sqrt{x} }\,dx
    5. ∫013x2 dx\displaystyle\int_0^1 \dfrac{ 3 }{x^2}\,dx
  2. Which improper integrals converge? Tick all that do.
    1. ∫1∞5x2 dx\displaystyle\int_1^{\infty} \dfrac{ 5 }{x^2}\,dx
    2. ∫015x dx\displaystyle\int_0^1 \dfrac{ 5 }{\sqrt{x} }\,dx
    3. ∫0∞e−5x dx\displaystyle\int_0^{\infty} e^{-5x}\,dx
    4. ∫1∞5x dx\displaystyle\int_1^{\infty} \dfrac{ 5 }{\sqrt{x} }\,dx
    5. ∫015x2 dx\displaystyle\int_0^1 \dfrac{ 5 }{x^2}\,dx

Exit ticket

Three questions to check you have got it.

  1. Evaluate ∫3∞5x3 dx\displaystyle\int_{ 3 }^{\infty} \dfrac{ 5 }{x^3}\,dx.
  2. Evaluate ∫4∞1x3 dx\displaystyle\int_{ 4 }^{\infty} \dfrac{ 1 }{x^3}\,dx.
  3. Evaluate ∫4∞4x3 dx\displaystyle\int_{ 4 }^{\infty} \dfrac{ 4 }{x^3}\,dx.

Summary card

The p-test

  • ∫1∞x−p dx\int_1^{\infty} x^{-p}\,dx: converges for p>1p > 1.
  • ∫01x−p dx\int_0^1 x^{-p}\,dx: converges for p<1p < 1.
  • p=1p = 1: both diverge.

Watch out: the condition flips between ∞\infty and 00.

End of the lesson

Back to the chapter

Answers: Convergence of improper integrals: the p-test

A

  1. p>1p > 1

B

  1. 518\frac{5}{18}
  2. 132\frac{1}{32}

C

  1. ∫1∞3x2 dx\displaystyle\int_1^{\infty} \dfrac{ 3 }{x^2}\,dx · ∫013x dx\displaystyle\int_0^1 \dfrac{ 3 }{\sqrt{x} }\,dx · ∫0∞e−3x dx\displaystyle\int_0^{\infty} e^{-3x}\,dx
  2. ∫1∞5x2 dx\displaystyle\int_1^{\infty} \dfrac{ 5 }{x^2}\,dx · ∫015x dx\displaystyle\int_0^1 \dfrac{ 5 }{\sqrt{x} }\,dx · ∫0∞e−5x dx\displaystyle\int_0^{\infty} e^{-5x}\,dx

D

  1. 518\frac{5}{18}
  2. 132\frac{1}{32}
  3. 18\frac{1}{8}

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