Riemann and Improper Integration
This domain tests the Darboux formulation of the Riemann integral, integrability criteria, the fundamental theorem, and the careful limiting process that defines improper integrals at infinite limits or vertical asymptotes.
Integration theory
For a partition P with mesh widths Δx_i, define upper sums U(f, P) = Σ M_i Δx_i and lower sums L(f, P) = Σ m_i Δx_i. A bounded f is Riemann integrable iff sup_P L = inf_P U, equivalently iff for every ε > 0 some partition has U − L < ε. Continuous functions on [a, b] are integrable, as are monotone functions and bounded functions with finitely many discontinuities. The integral is linear and additive over adjacent intervals, and |∫f| ≤ ∫|f|.
Fundamental theorem
If f is continuous on [a, b], then F(x) = ∫_a^x f is C¹ with F' = f. If G' = f on [a, b], then ∫_a^b f = G(b) − G(a). Substitution and integration by parts follow from the chain and product rules.
Improper integrals
Set ∫a^∞ f = lim{R→∞} ∫_a^R f, and handle an endpoint singularity by a one-sided limit. For 0 ≤ f ≤ g, convergence of ∫g forces convergence of ∫f. The p-test: ∫_1^∞ x^{−p} converges iff p > 1, while ∫_0^1 x^{−p} converges iff p < 1. Absolute convergence implies convergence, but ∫_1^∞ (sin x)/x converges only conditionally.
Worked example
Compute ∫_0^1 ln x dx, improper at 0. For a > 0, an antiderivative is x ln x − x, so ∫_a^1 ln x dx = (0 − 1) − (a ln a − a) = −1 − a ln a + a. As a → 0+, a ln a → 0, since writing a ln a = ln a/(1/a) and applying L'Hôpital gives limit 0. Hence the improper integral converges to −1.
Traps
Splitting an integral across a singularity without separate one-sided limits; applying the comparison test to sign-changing integrands; assuming convergence implies absolute convergence; using the fundamental theorem across a discontinuity; forgetting that Riemann integrability presupposes boundedness, so unbounded integrands need the improper theory.