ConceptComplete

Improper Integrals

Improper integrals extend the Riemann integral to unbounded intervals or unbounded functions. They are defined as limits of proper integrals. Convergence tests (comparison, limit comparison, absolute convergence) determine whether an improper integral converges or diverges. Improper integrals appear throughout analysis, probability, and physics.


Definitions

Definition6.1Improper integral (Type I: infinite interval)

If ff is integrable on [a,R][a, R] for all R>aR > a, the improper integral is

โˆซaโˆžf(x)โ€‰dx=limโกRโ†’โˆžโˆซaRf(x)โ€‰dx,\int_a^\infty f(x) \, dx = \lim_{R \to \infty} \int_a^R f(x) \, dx,

provided the limit exists.

Definition6.2Improper integral (Type II: unbounded function)

If ff is integrable on [a+ฯต,b][a+\epsilon, b] for all ฯต>0\epsilon > 0 but unbounded near aa, the improper integral is

โˆซabf(x)โ€‰dx=limโกฯตโ†’0+โˆซa+ฯตbf(x)โ€‰dx.\int_a^b f(x) \, dx = \lim_{\epsilon \to 0^+} \int_{a+\epsilon}^b f(x) \, dx.

Exampleโˆซโ‚^โˆž 1/xยฒ dx converges

โˆซ1โˆž1x2โ€‰dx=limโกRโ†’โˆžโˆซ1R1x2โ€‰dx=limโกRโ†’โˆž[โˆ’1x]1R=limโกRโ†’โˆž(1โˆ’1R)=1.\int_1^\infty \frac{1}{x^2} \, dx = \lim_{R \to \infty} \int_1^R \frac{1}{x^2} \, dx = \lim_{R \to \infty} \left[-\frac{1}{x}\right]_1^R = \lim_{R \to \infty} \left(1 - \frac{1}{R}\right) = 1.

Exampleโˆซโ‚^โˆž 1/x dx diverges

โˆซ1โˆž1xโ€‰dx=limโกRโ†’โˆžlnโก(R)=โˆž.\int_1^\infty \frac{1}{x} \, dx = \lim_{R \to \infty} \ln(R) = \infty.

The integral diverges.

Exampleโˆซโ‚€ยน 1/โˆšx dx converges

โˆซ011xโ€‰dx=limโกฯตโ†’0+โˆซฯต1xโˆ’1/2โ€‰dx=limโกฯตโ†’0+[2x]ฯต1=2.\int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{\epsilon \to 0^+} \int_\epsilon^1 x^{-1/2} \, dx = \lim_{\epsilon \to 0^+} [2\sqrt{x}]_\epsilon^1 = 2.


Convergence tests

Theorem6.1Comparison test

If 0โ‰คf(x)โ‰คg(x)0 \leq f(x) \leq g(x) for xโ‰ฅax \geq a, then:

  • If โˆซaโˆžg\int_a^\infty g converges, then โˆซaโˆžf\int_a^\infty f converges.
  • If โˆซaโˆžf\int_a^\infty f diverges, then โˆซaโˆžg\int_a^\infty g diverges.
ExampleUsing comparison test

Does โˆซ1โˆžsinโก2(x)x2โ€‰dx\int_1^\infty \frac{\sin^2(x)}{x^2} \, dx converge? Since 0โ‰คsinโก2(x)โ‰ค10 \leq \sin^2(x) \leq 1, we have

sinโก2(x)x2โ‰ค1x2.\frac{\sin^2(x)}{x^2} \leq \frac{1}{x^2}.

Since โˆซ1โˆž1x2โ€‰dx=1\int_1^\infty \frac{1}{x^2} \, dx = 1 converges, by comparison, โˆซ1โˆžsinโก2(x)x2โ€‰dx\int_1^\infty \frac{\sin^2(x)}{x^2} \, dx converges.


Summary

Improper integrals extend integration to unbounded domains or functions:

  • Type I: Infinite intervals โˆซaโˆžf\int_a^\infty f.
  • Type II: Unbounded functions near a point.
  • Convergence tests: comparison, limit comparison, absolute convergence.

See Riemann Integral for the foundation.