Improper Integrals

If $a<a'$, then for every $R>a'$, [L1, L2] $\int_a^R f=\int_a^{a'}f+\int_{a'}^R f.$ The first term is fixed and finite. Subtracting it changes the absolute error from a proposed translated limit by exactly the same amount, so [L2] shows that either truncation limit exists exactly when the other does and that their values differ by $\int_a^{a'}f$. The other three one-ended orientations follow by the same identity with endpoints reversed.

Let $s<t$ be two split points on the whole line. Step 1.1 transfers the finite proper integral $\int_s^t f$ from the right tail to the left tail, so the two sums agree. The same calculation around an interior singularity changes only the nonsingular finite portion. Because [L3] continues to require both pieces separately, no cancellation of divergent pieces is introduced.

On every compact truncation, [L1] gives $\int(rf+sg)=r\int f+s\int g$. Let the two truncation integrals tend to $A$ and $B$. Given $\varepsilon>0$, [L2] makes their respective errors smaller than $\varepsilon/(2(1+|r|))$ and $\varepsilon/(2(1+|s|))$ sufficiently near the end. The triangle inequality then makes the error of the linear combination from $rA+sB$ smaller than $\varepsilon$. This proves the formula on every one-ended interval, including $r=0$ or $s=0$.

For a mixed integral, apply step 1.1 to every separately convergent piece and then add the finitely many resulting identities as required by [L3]. No assertion is made when either side would contain an indeterminate difference of divergent quantities.

Put $F(R)=\int_a^R f$. If $F(R)$ has a finite limit as $R\to\infty$, then $|F(v)-F(u)|<\varepsilon$ for all sufficiently large $u,v$. By [L1], this difference is $\int_u^v f$, proving necessity.

Conversely, the tail condition and [L3] make the sequence $F(n)$ Cauchy, hence convergent to some $L$ by [L2]. Given $\varepsilon>0$, choose a large integer $n$ for which $|F(n)-L|<\varepsilon/2$ and the tail condition is below $\varepsilon/2$. For every real $R\ge n$, [L1] gives $F(R)-L=(F(R)-F(n))+(F(n)-L)$, so $F(R)\to L$.

For a finite endpoint use the reciprocal sequence $a+1/n$ or $b-1/n$ furnished by [L3]; the identical Cauchy argument applies. Reversing the real line gives the $-\infty$ form.

At $+\infty$, $F(R)=\int_a^R f$ is nondecreasing by [L1]. If $F(R)$ converges, its range is bounded. Conversely, if its range is bounded above, the integer sequence $F(n)$ is bounded and nondecreasing, so [L2] gives $F(n)\to S=\sup_nF(n)$.

For $n\le R\le n+1$, monotonicity gives $F(n)\le F(R)\le F(n+1)$. Hence $F(R)\to S$. Every real truncation lies below a later integer truncation, so $S$ is also the supremum of the full truncation range.

Reciprocal truncations and the same squeeze prove the finite-endpoint forms; reversing orientation proves the $-\infty$ form. Moving the initial finite endpoint is harmless by [L3].

By the Cauchy criterion [L2], remote tail integrals of $|f|$ are arbitrarily small. The proper inequality [L1] makes the corresponding tail integrals of $f$ no larger in absolute value. A second application of [L2] proves convergence of $\int f$.

Apply [L1] on compact truncations. Along integer truncations at infinity, or reciprocal truncations at a finite endpoint, both sides converge to the corresponding improper values; [L3] passes the inequality to those sequence limits and gives the displayed bound.

For a mixed integral, absolute convergence is required on every piece. Steps 1.1–1.2 apply piecewise, and finite addition completes the claim.

Write $I=\int_a^\infty f$. By [L1], $\int_R^\infty f=I-\int_a^R f$, which tends to zero by the definition of $I$.

The epsilon formulation is exactly [L2]. Reversing orientation or replacing infinite truncations by one-sided finite truncations proves all other stated forms.

Discard the finite portion before the eventual inequality using [L4]. On every remaining compact truncation, [L1] gives $0\le\int f\le\int g$. Convergence of $\int g$ bounds the latter truncations, so [L2] gives convergence of $\int f$.

If $|f|\le g$, step 1.1 applied to $|f|$ proves absolute convergence; [L3] then proves convergence of $f$. The argument depends only on the direction of truncation and therefore proves every endpoint form.

By [L1], tolerance $L/2$ gives $L/2<f/g<3L/2$ sufficiently near the singular end. Thus $(L/2)g\le f\le(3L/2)g$. Applying [L2] in both directions proves the equivalence.

If $L=0$, eventually $f/g\le1$, so $f\le g$. If $f/g\to+\infty$, eventually $g\le f$. The one-way conclusions again follow from [L2].

For every integer $k\ge1$, monotonicity of $f$ and [L1, L2] give $f(k+1)\le\int_k^{k+1}f(x)\,dx\le f(k)$. Adding these inequalities and using interval additivity yields $\sum_{k=2}^{N+1}f(k)\le\int_1^{N+1}f(x)\,dx\le\sum_{k=1}^{N}f(k)$. Thus the series partial sums are bounded exactly when the integer truncation integrals are bounded. By [L2], this is exactly convergence of the corresponding two monotone sequences.

Suppose the integer truncations converge to $I$. Given a sufficiently large integer $N$ and any real $R\ge N$, [L3] supplies an integer $M>R$. Nonnegativity gives $\int_1^Nf\le\int_1^Rf\le\int_1^Mf$, and both integer bounds tend to $I$; hence the full real-parameter limit is $I$. The reverse implication is immediate by restriction to integer truncations. Tail invariance handles finite changes.

Write $p=m/q$ with an integer $m$ and a positive integer $q$. Substituting $x=t^q$ on the positive interval and using [L1]–[L3] reduces the integrand to $qt^{q-m-1}$. The integer-power rule gives primitive $qt^{q-m}/(q-m)$ when $m\ne q$. Substituting $t=x^{1/q}$ back gives $x^{1-p}/(1-p)$. The FTC proves the displayed formula.

On $[2^k,2^{k+1}]$, $1/x\ge2^{-(k+1)}$, so its integral is at least $1/2$. Adding the first $N$ dyadic blocks proves the first lower bound. The intervals $[2^{-(k+1)},2^{-k}]$ give the second in the same way.

If $p>1$, then $R^{1-p}\to0$, so [L1] gives $\int_1^R x^{-p}dx\to1/(p-1)$.

If $p\le1$, the integral at infinity diverges: for $0<p<1$ the formula in [L1] is unbounded by [L2], for $p=1$ use the first dyadic bound, and for $p\le0$ compare $x^{-p}\ge1$ with the constant one.

If $p<1$, then $1-p>0$, so [L1] and $c^{1-p}\to0$ from [L2] give convergence at zero with value $1/(1-p)$.

If $p\ge1$, the zero-endpoint integral diverges: use the unbounded formula and [L2] for $p>1$, and the second dyadic bound for $p=1$.

The alternatives in steps 1.1–1.4 exhaust all rational exponents and establish both thresholds and values.

At an interior point $c$, the two truncated terms in [L3] tend separately to the two finite one-sided values. Given $\varepsilon>0$, take the common smaller truncation scale on which each term is within $\varepsilon/2$ of its limit; the triangle inequality then puts their sum within $\varepsilon$ of the sum in [L1].

On the real line, split at zero. As $R\to\infty$, $\int_{-R}^0f$ and $\int_0^Rf$ tend separately to their two tail values. The same $\varepsilon/2$ estimate shows that their sum tends to the mixed value. Split-point invariance [L2] removes any dependence on zero.

The converses fail, and a witness is available on this page rather than assumed. Take $f(x)=1/x$ on $[-1,1]$ with the interior singularity at $0$. For every $\delta\in(0,1)$ the substitution $x\mapsto-x$ gives $\int_{-1}^{-\delta}x^{-1}\,dx=-\int_{\delta}^{1}x^{-1}\,dx$, so the symmetric truncations cancel exactly and the principal value exists and is $0$. But $\int_0^1x^{-p}\,dx$ converges exactly when $p<1$ by [L4], so at $p=1$ the right-hand one-sided integral diverges, and by the same reflection so does the left-hand one. Hence the principal value can exist while neither one-sided improper integral converges, and the converse of the first claim fails. For the whole line a separate witness is needed, because [L5] admits only a function locally Riemann integrable on all of $\mathbb R$ and $1/x$ is not one: take $f(x)=x$, which is continuous and therefore locally integrable. For every $R$ the substitution $x\mapsto-x$ gives $\int_{-R}^0x\,dx=-\int_0^Rx\,dx$, so $\int_{-R}^Rx\,dx=0$ and the whole-line principal value is $0$; but $\int_0^Rx\,dx=R^2/2$ is unbounded in $R$, so the tail $\int_0^\infty x\,dx$ does not converge.

Choose $M$ with $|F|\le M$. In clause 1, apply [L1] on $[u,v]$. Every partial integral $\int_u^xf=F(x)-F(u)$ has absolute value at most $2M$, so [L1, L2, assume-case first] $\left|\int_u^vfg\right|\le2M(g(u)+g(v))\le4Mg(u).$ This tends to zero as $u\to\infty$, and [L2] proves convergence.

In clause 2 $f$ is continuous, hence integrable on every $[u,v]\subseteq[a,\infty)$, so [L5] gives $F'=f$ there — the hypothesis [L3] requires and which boundedness of $F$ does not supply. With $g$ differentiable and $g'$ integrable on $[u,v]$, [L3] gives the integration-by-parts identity. The boundary term $F(R)g(R)$ tends to zero because $F$ is bounded and $g(R)\to0$. Also $|Fg'|\le M|g'|$, so [L4] makes $\int Fg'$ converge. Passing $R\to\infty$ proves convergence of $\int fg$.

The two clauses are exhausted by steps 1.1–1.2. Reversing orientation proves the $-\infty$ case. At a finite endpoint, use a primitive based at a fixed nonsingular point and take the corresponding one-sided limits; the same estimates are unchanged.

Suppose first that $g$ is nondecreasing and let $L$ be the supremum of its bounded range. Given $\varepsilon>0$, the definition of supremum gives $x_0$ with $L-\varepsilon<g(x_0)\le L$; monotonicity then gives $L-\varepsilon<g(x)\le L$ for every $x\ge x_0$, so $g(x)\to L$. The infimum argument handles a nonincreasing $g$. If $g$ is nonincreasing put $h=g-L$; if it is nondecreasing put $h=L-g$. In either case $h\ge0$, $h$ is nonincreasing toward the singular end, and $h\to0$.

The primitive of $f$ is bounded by [L2], so [L3] makes $\int fh$ converge. Since $fg=Lf\pm fh$, linearity [L4] and convergence of $\int f$ prove convergence of $\int fg$. The oriented endpoint variants are identical.

Suppose for contradiction that $\int_a^\infty uv$ converges. Then its truncation primitive is bounded by [L3], and $uv$ is continuous. Put $g=1/v$. Positivity and $v\to\infty$ give $g\to0$: for $\varepsilon>0$, eventually $v>1/\varepsilon$, hence $0<g<\varepsilon$. Also [L1] and the hypothesis give absolute convergence of $\int g'$.

Apply [L2] with the continuous function $uv$ and multiplier $g$. It yields convergence of $\int(uv)g=\int u$, contradicting the hypothesis. Hence $\int uv$ diverges.

Suppose first that $g$ is nonincreasing. On the $j$th block, $g(x)\ge g(x_{j+1})$, hence [L1, L2, L3] $\int_{x_j}^{x_{j+1}}|f|g\ge\delta g(x_{j+1}).$ Also $\int_{x_j}^{x_{j+1}}g\le M g(x_j)$. Since $\int g$ diverges, additivity and [L2] force $\sum_jg(x_j)$, and hence its shifted tail, to diverge. Thus the block lower bounds for $\int|f|g$ have unbounded partial sums.

If $g$ is nondecreasing, divergence of its integral implies it is positive at some point; thereafter $g(x_j)$ is bounded below by a positive constant. Now $\int_{x_j}^{x_{j+1}}|f|g\ge\delta g(x_j)$, whose partial sums diverge.

In either monotonicity case the nonnegative truncations of $|fg|$ are unbounded, so [L2] proves divergence and the definition rules out absolute convergence.