term-by-term integration
It is tempting to integrate an infinite sum by integrating each piece and adding the answers: integral of (sum f_n) = sum of (integral of f_n). This is exactly swapping the order of a sum (a limit of partial sums) and an integral (also a limit). The swap is legal — but only under a hypothesis, and uniform convergence is a clean one that licenses it.
Theorem (sequence form): if f_n are Riemann integrable on [a, b] and f_n -> f uniformly on [a, b], then f is Riemann integrable and the integral of f_n over [a,b] converges to the integral of f. Series form: if the partial sums of sum f_n converge uniformly on [a, b] with each f_n integrable, then integral over [a,b] of sum f_n equals sum of (integral over [a,b] of f_n).
Why uniform convergence suffices: |integral of f_n - integral of f| <= integral of |f_n - f| <= (b - a) times sup over [a,b] of |f_n - f|, and the right-hand side vanishes as n -> infinity. The honest warning: the hypothesis cannot simply be dropped. Under mere pointwise convergence the integrals can fail to converge to the integral of the limit — a famous failure, fixed in the more powerful Lebesgue theory by the dominated and monotone convergence theorems.
On [0, x] with |x| < 1, integrate the geometric series 1/(1 - t) = sum_{n>=0} t^n term by term. The series converges uniformly on [-r, r] for any r < 1, so integration gives -log(1 - x) = sum_{n>=0} x^{n+1}/(n+1) = x + x^2/2 + x^3/3 + ... — the power series for the logarithm, derived rigorously.
Integrating a geometric series term by term yields the log series.