Improper integration, absolute convergence and oscillatory tails

Proof

Prerequisite proofs used below

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Learning goals and definitions

Test every endpoint separately, prove absolute convergence implies convergence, and analyze an oscillatory integral whose absolute value is not integrable. For a function Riemann integrable on every finite $[a,R]$, define $\int_a^\infty f=\lim_{R\to\infty}\int_a^R f$ only if the limit is finite. At a finite singular endpoint use a one-sided limit. With multiple singularities, split the interval and require all pieces to converge separately.

The Cauchy criterion says the infinite tail converges exactly when every $\varepsilon>0$ has $A$ such that $|\int_S^Tf|<\varepsilon$ whenever $T>S\ge A$. This follows by applying the real Cauchy criterion to the finite-integral function as the endpoint tends to infinity; values on a finite initial segment do not affect this tail condition.

Theorem: comparison and absolute convergence with local proper integrability

Fix a real base point $A$. For comparison assume $f,g$ are Riemann integrable on every finite $[A,R]$ and $0\le f\le g$ on $[A,\infty)$. If this dominance starts only at some $B\ge A$, first replace the base point by $B$: the two proper integrals over $[A,B]$ are finite constants and cannot change convergence. If $\int_A^\infty g$ converges, then $\int_A^\infty f$ converges. For the signed assertion, assume explicitly that $f$ is Riemann integrable on every finite $[A,R]$. If $\int_A^\infty|f|$ converges, then $\int_A^\infty f$ converges. The local proper-Riemann-integrability assumption is part of the absolute-integrability convention here.

Proof. In the comparison case the finite integrals of $f$ are nonnegative, nondecreasing and bounded above by the total integral of $g$. Their supremum $L$ is finite. For every $\varepsilon>0$ a finite endpoint has integral greater than $L-\varepsilon$; every later integral lies between that value and $L$. Hence the limit is $L$.

For signed $f$, local Riemann integrability also gives local integrability of $|f|$: on every cell its oscillation is at most that of $f$, so the Darboux criterion applies. Proper integral bounds yield $|\int_S^T f|\le\int_S^T|f|$. Convergence of the nonnegative absolute integral therefore makes every signed tail arbitrarily small. Let $F(R)=\int_A^R f$. The sequence $F(A+n)$ is real Cauchy and has a finite limit $L$ by real completeness. Given $\varepsilon>0$, choose a common tail on which $|F(T)-F(S)|<\varepsilon/2$. Fix any $R$ in that tail and compare $F(R)$ with all sufficiently large $F(A+n)$. Passage to the real limit gives $|F(R)-L|\le\varepsilon/2<\varepsilon$. Thus the limit holds for every real endpoint, not only along the integer sequence. The same argument proves sufficiency of the introductory signed Cauchy-tail criterion. Its necessity follows from $|F(T)-F(S)|\le|F(T)-L|+|F(S)-L|$.

Why local integrability is essential. On $[1,\infty)$ let $f(x)=1/x^2$ at rational points and $f(x)=-1/x^2$ at irrational points. Its absolute value has improper integral one. But on any $[S,T]$ with $1\le S<T$, density gives every cell supremum at least $1/T^2$ and infimum at most $-1/T^2$. Every Darboux gap is therefore at least $2(T-S)/T^2>0$. The signed function has no proper finite integrals. Mere integrability of $|f|$ must not be used to infer the missing local hypothesis. Another singular endpoint still requires its separate test.

For $p\ne1$, integrate $x^{-p}$ exactly as $x^{1-p}/(1-p)$. The endpoint limits give $\int_1^\infty x^{-p}$ convergent exactly when $p>1$, and $\int_0^1x^{-p}$ convergent exactly when $p<1$. At $p=1$, the primitive $\ln x$ diverges at both problematic endpoints. These are different thresholds for different ends.

Worked example: conditional convergence of a sine tail

On $[1,\infty)$, integration by parts gives for $T>S\ge1$,

$$\int_S^T\frac{\sin x}{x}\,dx=\left[-\frac{\cos x}{x}\right]_S^T-\int_S^T\frac{\cos x}{x^2}\,dx.$$

The magnitude is at most $1/S+1/T+\int_S^Tx^{-2}dx=2/S$. The Cauchy criterion proves convergence, with an absolute tail bound $2/S$ for the signed integral. At zero, $\sin x/x$ has removable limit one, so $\int_0^\infty\sin x/x$ also exists after continuous extension there. We establish existence, not a particular value of this integral.

Its absolute integral diverges. On each interval $[k\pi+\pi/6,k\pi+5\pi/6]$ for integers $k\ge1$, $|\sin x|\ge1/2$ and $x\le(k+1)\pi$. The integral over that interval is therefore at least $[2\pi/3]/[2(k+1)\pi]=1/[3(k+1)]$. Summing the disjoint intervals gives a divergent harmonic lower bound. Thus the sine integral is conditionally, not absolutely, convergent.

Worked example: cancellation is not ordinary convergence

For $1/x$ on $(-1,1)$, the negative and positive one-sided integrals at zero diverge separately. Symmetric truncation gives $\int_{-1}^{-\varepsilon}dx/x+\int_\varepsilon^1dx/x=0$, but this defines only a Cauchy principal value. Independent cutoffs instead give $\ln\alpha-\ln\beta$ for exclusions $\alpha$ on the negative side and $\beta$ on the positive side, whose value depends on their ratio. The ordinary integral therefore does not exist.

Graded practice

1. Foundation. For which $p$ does $\int_0^1x^{-p}dx$ converge, and what is its value when it does?
2. Core. Determine convergence of $\int_1^\infty dx/[x(x+1)]$ and compute its value.
3. Core. Show $\int_1^\infty\cos x/x\,dx$ converges and supply a signed tail bound.
4. Proof. If $f$ is absolutely integrable on $[a,\infty)$, prove $\int_a^R f$ has a limit as $R\to\infty$, and explain why $f(x)\to0$ is not necessary without additional assumptions.

Full solutions

1. It converges exactly when $p<1$, with value $1/(1-p)$ from the primitive and the limit at zero. For $p\ge1$ it diverges.
2. Decompose $1/[x(x+1)]=1/x-1/(x+1)$. The finite integral is $\ln R-\ln(R+1)+\ln2$, tending to $\ln2$. It is nonnegative, hence also absolutely convergent.
3. Parts gives $\int_S^T\cos x/x=[\sin x/x]_S^T+\int_S^T\sin x/x^2$. Its magnitude is at most $1/S+1/T+1/S-1/T=2/S$. This verifies the Cauchy criterion and bounds the signed infinite tail by $2/S$.
4. The triangle inequality bounds every signed tail by the corresponding absolute tail, proving the Cauchy criterion. For a counterexample to necessity, construct continuous triangular spikes of height one centered at positive integers $n\ge2$, each supported on $[n-2^{-n-2},n+2^{-n-2}]$, and zero elsewhere. The supports are disjoint and their total area is $\sum_{n\ge2}2^{-n-2}<\infty$, but the function equals one at every spike center. It is locally integrable and absolutely integrable yet does not tend to zero. Narrow spikes evade any pointwise conclusion from finite total area alone.

Summary and term definitions

An improper integral is a specified limit of ordinary finite integrals. Absolute convergence controls tails without cancellation; conditional convergence uses cancellation that disappears on taking absolute values. A Cauchy principal value couples cutoffs and is not interchangeable with independent one-sided convergence. Record the limiting definition before manipulating divergent expressions.