Improper Integrals
Integrals over intervals that run forever or blow up. The exponent p decides convergence against divergence.
An endless region can still have finite area. Its tail just has to thin out fast enough. Too thick a tail piles area up forever.
For example, a π§ water leak that slows down quickly spills only one bucketful, even leaking forever. A leak that slows too little overflows the bucket in the end.
β« 1 β d x x p = 1 p β 1 Β Β ( p > 1 ) \int_1^{\infty}\frac{dx}{x^p} = \frac{1}{p-1}\ \ (p>1) β« 1 β β x p d x β = p β 1 1 β Β Β ( p > 1 ) β« 0 1 d x x = 2 \int_0^1\frac{dx}{\sqrt{x}} = 2 β« 0 1 β x β d x β = 2 Part of these learning paths
Examples & Exercises
Compute β« 1 β d x x 2 = 1 \int_1^\infty \frac{dx}{x^2} = 1 β« 1 β β x 2 d x β = 1
β« 1 R d x x 2 = 1 β 1 R β 1 \int_1^R \frac{dx}{x^2} = 1 - \frac{1}{R} \to 1 β« 1 R β x 2 d x β = 1 β R 1 β β 1 p = 2 , R = 20 p=2, R=20 p = 2 , R = 20 0.95 0.95 0.95 1 1 1 Show proof
β« 1 R x β 2 β d x = [ β x β 1 ] 1 R = 1 β 1 R \int_1^R x^{-2}\,dx = \left[-x^{-1}\right]_1^R = 1 - \frac{1}{R} β« 1 R β x β 2 d x = [ β x β 1 ] 1 R β = 1 β R 1 β
As R β β R \to \infty R β β 1 R β 0 \frac{1}{R} \to 0 R 1 β β 0 1 1 1 R = 20 R = 20 R = 20 1 β 1 20 = 0.95 1 - \frac{1}{20} = 0.95 1 β 20 1 β = 0.95
Prove β« 1 β d x x p \int_1^\infty \frac{dx}{x^p} β« 1 β β x p d x β p β€ 1 p \leq 1 p β€ 1
β« 1 R d x x = ln β‘ R β β \int_1^R \frac{dx}{x} = \ln R \to \infty β« 1 R β x d x β = ln R β β p = 1 p=1 p = 1 ln β‘ R \ln R ln R Show proof
p = 1 p = 1 p = 1 β« 1 R d x x = ln β‘ R β β \int_1^R \frac{dx}{x} = \ln R \to \infty β« 1 R β x d x β = ln R β β R β β R \to \infty R β β R = 50 R = 50 R = 50 ln β‘ 50 β 3.912 \ln 50 \approx 3.912 ln 50 β 3.912
p < 1 p < 1 p < 1 β« 1 R x β p β d x = R 1 β p β 1 1 β p \int_1^R x^{-p}\,dx = \frac{R^{1-p}-1}{1-p} β« 1 R β x β p d x = 1 β p R 1 β p β 1 β 1 β p > 0 1 - p > 0 1 β p > 0 β \infty β R β β R \to \infty R β β
Compute β« 0 1 d x x = 2 \int_0^1 \frac{dx}{\sqrt{x}} = 2 β« 0 1 β x β d x β = 2
β« Ξ΅ 1 x β 1 / 2 β d x = 2 β 2 Ξ΅ β 2 \int_\varepsilon^1 x^{-1/2}\,dx = 2 - 2\sqrt{\varepsilon} \to 2 β« Ξ΅ 1 β x β 1/2 d x = 2 β 2 Ξ΅ β β 2 p = 1 / 2 , Ξ΅ = 0.01 p=1/2, \varepsilon=0.01 p = 1/2 , Ξ΅ = 0.01 1.8 1.8 1.8 2 2 2 Show proof
β« Ξ΅ 1 x β 1 / 2 β d x = [ 2 x ] Ξ΅ 1 = 2 β 2 Ξ΅ \int_\varepsilon^1 x^{-1/2}\,dx = \left[2\sqrt{x}\right]_\varepsilon^1 = 2 - 2\sqrt{\varepsilon} β« Ξ΅ 1 β x β 1/2 d x = [ 2 x β ] Ξ΅ 1 β = 2 β 2 Ξ΅ β
As Ξ΅ β 0 + \varepsilon \to 0^+ Ξ΅ β 0 + Ξ΅ β 0 \sqrt{\varepsilon} \to 0 Ξ΅ β β 0 2 2 2 Ξ΅ = 0.01 \varepsilon = 0.01 Ξ΅ = 0.01 2 β 2 0.01 = 2 β 0.2 = 1.8 2 - 2\sqrt{0.01} = 2 - 0.2 = 1.8 2 β 2 0.01 β = 2 β 0.2 = 1.8
11. Integration