Improper Integrals

∫1∞dxxp\int_1^\infty \frac{dx}{x^p} converges only for p>1p > 1, plus the singular kind ∫01dxx=2\int_0^1 \frac{dx}{\sqrt{x}} = 2 — the twin of the p-series scene.

∫1∞dxxp=1p−1  (p>1)\int_1^{\infty}\frac{dx}{x^p} = \frac{1}{p-1}\ \ (p>1)
∫01dxx=2\int_0^1\frac{dx}{\sqrt{x}} = 2

Part of these learning paths

Examples & Exercises

Compute ∫1∞dxx2=1\int_1^\infty \frac{dx}{x^2} = 1 directly.

∫1Rdxx2=1−1R→1\int_1^R \frac{dx}{x^2} = 1 - \frac{1}{R} \to 1
p=2,R=20p=2, R=20: area 0.950.95 approaches 11.
Show proof
  1. ∫1Rx−2 dx=[−x−1]1R=1−1R\int_1^R x^{-2}\,dx = \left[-x^{-1}\right]_1^R = 1 - \frac{1}{R}.
  2. As R→∞R \to \infty, 1R→0\frac{1}{R} \to 0, so the area tends to 11. At R=20R = 20: 1−120=0.951 - \frac{1}{20} = 0.95 — matching the caption's figure.

Prove ∫1∞dxxp\int_1^\infty \frac{dx}{x^p} diverges for p≤1p \leq 1.

∫1Rdxx=ln⁡R→∞\int_1^R \frac{dx}{x} = \ln R \to \infty
p=1p=1: the area grows like ln⁡R\ln R without bound.
Show proof
  1. p=1p = 1: ∫1Rdxx=ln⁡R→∞\int_1^R \frac{dx}{x} = \ln R \to \infty as R→∞R \to \infty. At R=50R = 50 the area is ln⁡50≈3.912\ln 50 \approx 3.912.
  2. p<1p < 1: ∫1Rx−p dx=R1−p−11−p\int_1^R x^{-p}\,dx = \frac{R^{1-p}-1}{1-p}, and since 1−p>01 - p > 0 this diverges to ∞\infty as R→∞R \to \infty.

Compute ∫01dxx=2\int_0^1 \frac{dx}{\sqrt{x}} = 2 as a limit at the singularity.

∫ε1x−1/2 dx=2−2ε→2\int_\varepsilon^1 x^{-1/2}\,dx = 2 - 2\sqrt{\varepsilon} \to 2
p=1/2,ε=0.01p=1/2, \varepsilon=0.01: area 1.81.8 approaches 22.
Show proof
  1. ∫ε1x−1/2 dx=[2x]ε1=2−2ε\int_\varepsilon^1 x^{-1/2}\,dx = \left[2\sqrt{x}\right]_\varepsilon^1 = 2 - 2\sqrt{\varepsilon}.
  2. As ε→0+\varepsilon \to 0^+, ε→0\sqrt{\varepsilon} \to 0, so the area tends to 22. At ε=0.01\varepsilon = 0.01: 2−20.01=2−0.2=1.82 - 2\sqrt{0.01} = 2 - 0.2 = 1.8 — matching the caption's figure.

10. Integration