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Integration Techniques, L’Hopital’s Rule, and Improper Integrals

8.8 Improper Integrals

Focus on infinite intervals, the p-integral test, unbounded integrands.

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Ideas in this lesson

01Infinite intervalsWorked steps · five questions02The p-integral testWorked steps · five questions03Unbounded integrandsWorked steps · five questions
Idea 1 of 3

Infinite intervals

Replace an infinite bound by a variable and take its limit. A finite antiderivative expression alone does not prove convergence.

Worked example 1

Evaluate ∫1∞2x2 dx\int_{1}^{\infty } \dfrac{2}{x^{2}}\,dx.
  1. ∫1R2x2 dx=2−1(R−1−1)\int_{1}^{R} \dfrac{2}{x^{2}}\,dx = \dfrac{2}{-1}(R^{-1} - 1)
  2. R−1→0R^{-1} \to 0 as R→∞R \to \infty
  3. 21=2\dfrac{2}{1} = 2

Answer: 22

Worked example 2

Evaluate ∫1∞3x3 dx\int_{1}^{\infty } \dfrac{3}{x^{3}}\,dx.
  1. ∫1R3x3 dx=3−2(R−2−1)\int_{1}^{R} \dfrac{3}{x^{3}}\,dx = \dfrac{3}{-2}(R^{-2} - 1)
  2. R−2→0R^{-2} \to 0 as R→∞R \to \infty
  3. 32=32≈1.5\dfrac{3}{2} = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Idea 2 of 3

The p-integral test

At infinity, ∫1∞x−pdx\int _{1}\infty x^{-p} dx converges exactly for p>1p > 1. At zero, ∫01x−pdx\int _{0}^{1}x^{-p} dx converges exactly for p<1p < 1.

Worked example 1

Does ∫1∞2x0.5 dx\int_{1}^{\infty } \dfrac{2}{x^{0.5}}\,dx converge?
  1. ∫1\int _{1}ᴿ 2x−0.5dx=2(R0.5−1)0.52x^{-0}.5dx = \dfrac{2(R^{0}.5 - 1)}{0.5}
  2. p=0.5≤1  ⟹  p = 0.5 \le 1 \implies diverges

Answer: Diverges

Worked example 2

Does ∫1∞3x dx\int_{1}^{\infty } \dfrac{3}{x}\,dx converge?
  1. ∫1\int _{1}ᴿ 3xdx=3ln⁡R→∞\dfrac{3}{x}dx = 3 \ln R \to \infty
  2. p=1≤1  ⟹  p = 1 \le 1 \implies diverges

Answer: Diverges

Idea 3 of 3

Unbounded integrands

At an endpoint singularity, use a one-sided limit. At an interior singularity, both separate integrals must converge.

Worked example 1

Evaluate ∫012x dx\int_{0}^{1} \dfrac{2}{\sqrt{x}}\,dx.
  1. ∫ϵ12x dx=4(1−\int_{\epsilon }^{1} \dfrac{2}{\sqrt{x}}\,dx = 4(1 - √ϵ)\epsilon )
  2. lim⁡(ϵ→0+)4(1−\displaystyle \lim (\epsilon \to 0^{+}) 4(1 - √ϵ)=4\epsilon ) = 4

Answer: 44

Worked example 2

Does ∫013x dx\int_{0}^{1} \dfrac{3}{x}\,dx converge?
  1. ∫ϵ13x dx=−3ln⁡ϵ\int_{\epsilon }^{1} \dfrac{3}{x}\,dx = -3 \ln \epsilon
  2. ϵ→0+  ⟹  \epsilon \to 0^{+} \implies integral →∞  ⟹  \to \infty \implies divergent

Answer: Diverges

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.8,p8.8, p. 568568. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.