Riemann integration the fundamental theorem and improper integrals

An integral can exist even when its integrand has jumps, and an antiderivative can recover that integral under hypotheses weaker than continuity. An improper integral introduces a different limit, at an infinite endpoint or a singularity. We keep these constructions separate so that the fundamental theorem is applied only where it is justified.

1 Darboux integrability on a finite interval

Let $f$ be bounded on $[a,b]$, with $a<b$. For a partition $P:a=x_0<\cdots<x_n=b$, define

$$U(f,P)=\sum_i\sup_{[x_{i-1},x_i]}f\,\Delta x_i,\qquad L(f,P)=\sum_i\inf_{[x_{i-1},x_i]}f\,\Delta x_i.$$

Refinement lowers $U$ and raises $L$. Common refinement proves every lower sum is below every upper sum. The upper integral is $\inf_PU$ and the lower integral is $\sup_PL$. Equality defines Riemann integrability. Equivalently, for every $\varepsilon>0$ some partition has $U-L<\varepsilon$: squeezing proves sufficiency, and near-optimal upper/lower partitions plus common refinement prove necessity.

Uniform continuity proves every continuous function on a closed bounded interval is integrable. Monotonicity is another sufficient condition even with discontinuities. For nondecreasing $f$, an equal-length $n$-part partition gives

$$U-L=\frac{b-a}{n}\sum_{i=1}^n[f(x_i)-f(x_{i-1})] =\frac{b-a}{n}[f(b)-f(a)]\longrightarrow0.$$

The integral is linear and order preserving. For example, $|\int_a^bf|\le\int_a^b|f|\le(b-a)\sup|f|$. These follow first from cell sums and then from limits. Additivity across an interior split point follows by combining partitions on the two pieces.

2 The fundamental theorem in two directions

FTC part one. If bounded $f$ is Riemann integrable on $[a,b]$ and $F(x)=\int_a^xf(t)dt$, then $F$ is Lipschitz. At every interior point where $f$ is continuous, $F'(x)=f(x)$.

Proof. If $|f|\le M$, additivity gives $|F(y)-F(x)|\le M|y-x|$. For a continuity point and nonzero small $h$,

$$\frac{F(x+h)-F(x)}h-f(x)=\frac1h\int_x^{x+h}[f(t)-f(x)]dt.$$

Its absolute value is bounded by the supremum of $|f(t)-f(x)|$ between the two endpoints, tending to zero. The oriented integral handles negative $h$ as well. $\square$

At a jump, differentiability can fail: for $f=0$ below $1/2$ and $f=1$ above it, the accumulated function has left slope 0 and right slope 1 at $1/2$, regardless of the assigned integrand value at the jump.

FTC part two. If $G$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'=f$ there with bounded Riemann integrable $f$ on the closed interval, then

$$\int_a^bf(x)dx=G(b)-G(a).$$

Proof. On every partition cell the mean value theorem supplies an interior point $\xi_i$ with $G(x_i)-G(x_{i-1})=f(\xi_i)\Delta x_i$. Summing telescopes to the endpoint difference. As partition mesh tends to zero, these Riemann sums converge to the integral. Endpoint values of $f$ do not affect it. $\square$

Thus continuity of the integrand is sufficient but not necessary for the second part. One cannot apply this argument to an unbounded singular integrand on a closed interval; that requires an improper limit.

The convergence of all sufficiently fine tagged sums, used in FTC part two, follows directly from the Darboux definition. If $|f|\le M$ and $f$ is integrable with value $I$, fix $\varepsilon>0$ and choose $P_0$ with $U(f,P_0)-L(f,P_0)<\varepsilon$. Let $k$ be the number of interior division points of $P_0$. For an arbitrary tagged partition $P$ of mesh $\eta$, refine it by these fixed points to $Q$. Only intervals of $P$ containing a division point of $P_0$ in their interiors are split; their total length is at most $k\eta$. Choose the same tags on all unsplit intervals and arbitrary permitted tags on the split pieces. If $S_P,S_Q$ are the tagged sums, boundedness gives $|S_P-S_Q|\le2Mk\eta$. Since $Q$ refines $P_0$, both $S_Q$ and $I$ lie between $L(f,P_0)$ and $U(f,P_0)$. Thus

$$|S_P-I|<\varepsilon+2Mk\eta.$$

For fixed $P_0$ this proves convergence as $\eta\to0$, uniformly over the tags; if $k=0$ the second term is zero. This supplies the missing link from the Darboux criterion to the telescoping mean-value sums, while allowing discontinuous bounded integrands.

It also closes the absolute-value facts used in estimates. For each interval $J$, $\operatorname{osc}(|f|,J)\le\operatorname{osc}(f,J)$ because $||u|-|v||\le|u-v|$. The Darboux criterion therefore makes $|f|$ integrable whenever $f$ is. Finite tagged sums satisfy the triangle inequality; passing to their just-proved limits yields $|\int f|\le\int|f|$. The bound $\int|f|\le M(b-a)$ follows from order preservation. No continuity assumption has been added to FTC part two.

3 Improper endpoints and the Cauchy criterion

For locally Riemann integrable $f$ on $[a,\infty)$, define $\int_a^\infty f=\lim_{B\to\infty}\int_a^Bf$ when the finite limit exists. At a singular endpoint $a$, define $\int_a^bf=\lim_{\delta\downarrow0}\int_{a+\delta}^bf$. If both endpoints are improper, split at a fixed interior point and require both limits separately. A symmetric cancellation is not the definition of convergence.

The infinite-endpoint integral converges exactly when its tails satisfy: for every $\varepsilon>0$, some $A$ makes $|\int_C^Bf|<\varepsilon$ for all $B>C\ge A$. Necessity follows by subtracting two near-limit accumulated integrals. Sufficiency follows because the accumulated values are Cauchy as the endpoint tends to infinity; completeness of $\mathbb R$ gives a finite limit.

For $f\ge0$, accumulated integrals are increasing, so convergence is equivalent to their boundedness. Consequently, $0\le f\le g$ eventually and convergence of $\int g$ imply convergence of $\int f$. If $\int|f|$ converges, the absolute tail bound and Cauchy criterion prove convergence of $\int f$. This is absolute convergence; the converse fails.

For the model powers, the two integrals converge exactly under the respective conditions shown:

$$\int_1^\infty x^{-p}dx\quad(p>1),\qquad \int_0^1x^{-p}dx\quad(p<1).$$

Evaluate the antiderivative on a proper finite interval first, then take the endpoint limit; the logarithm handles $p=1$.

4 A conditionally convergent oscillatory integral

For $B>A\ge1$, integration by parts gives

$$\int_A^B\frac{\sin x}{x}dx=\left[-\frac{\cos x}{x}\right]_A^B-\int_A^B\frac{\cos x}{x^2}dx.$$

The absolute tail is at most $1/A+1/B+(1/A-1/B)=2/A$. The Cauchy criterion proves convergence of $\int_1^\infty\sin x/x\,dx$.

It is not absolutely convergent. On each interval $[n\pi+\pi/6,n\pi+5\pi/6]$, $|\sin x|\ge1/2$ and $x\le(n+1)\pi$. Its contribution to $\int|\sin x|/x$ is at least $1/[3(n+1)]$. Summing the disjoint intervals produces a divergent harmonic lower bound. Cancellation is essential, so absolute-value operations or rearrangements need their own justification.

The trigonometric identities used here are established analytically in the course's elementary-functions chapter, not treated as evidence that any improper interchange is permissible.

Exercises with complete solutions

E1 Foundation. Let $f(x)=0$ on $[0,1/2)$ and $f(x)=2$ on $[1/2,1]$. Find $F(x)=\int_0^xf$ and identify where its derivative fails.

S1. $F(x)=0$ for $0\le x\le1/2$, and $F(x)=2x-1$ for $1/2\le x\le1$. It is continuous and Lipschitz but its left/right derivatives at $1/2$ are 0 and 2, so no derivative exists there. At every point of $(0,1)$ other than $1/2$, FTC part one applies. At the endpoints only one-sided derivatives are available: $F'_+(0)=0$ and $F'_-(1)=2$.

E2 Convergence. Determine convergence and value, when finite, of $\int_1^\infty x^{-3/2}dx$ and $\int_0^1x^{-3/2}dx$.

S2. On finite endpoints the antiderivative is $-2x^{-1/2}$. The integral from 1 to infinity is 2, whereas the lower-endpoint integral diverges to positive infinity. The same power can behave differently at zero and infinity.

E3 Technique. Evaluate $\int_1^\infty (\log x)/x^2\,dx$ with a justified endpoint limit.

S3. Integrating by parts on $[1,B]$ gives $-(\log B)/B+1-1/B$. Since $(\log B)/B\to0$, the limit is 1. For example, the logarithmic/exponential construction gives $\log B=o(B)$, or a justified elementary limit supplies the same endpoint fact. All finite-interval integrations occur away from a singularity.

E4 Cancellation. Explain why $\int_{-1}^1dx/x$ is divergent as an improper integral despite zero symmetric truncated integrals.

S4. Splitting at zero requires separate integrals over $(-1,0)$ and $(0,1)$; each diverges logarithmically with opposite signs. A zero symmetric principal value is a different limit, not convergence of either required improper part. Writing infinity minus infinity does not produce an integral value.

Summary

Darboux sums define finite-interval integration independently of antiderivatives. The two parts of FTC have distinct hypotheses and conclusions. Improper integrals add endpoint limits, where Cauchy tails, comparison and absolute convergence govern which operations remain legitimate.