Improper Integrals

If $a<a^{\mathrm{\prime }}$, then for every $R>a^{\mathrm{\prime }}$, [L1, L2] ${\int }_a^{R}f={\int }_a^{a^{\mathrm{\prime }}}f+{\int }_{a^{\mathrm{\prime }}}^{R}f\mathrm{.}$ 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 ${\int }_a^{a^{\mathrm{\prime }}}f$. 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 $\epsilon >0$, [L2] makes their respective errors smaller than $\epsilon \mathrm{/}(2(1+\mathrm{\mid }r\mathrm{\mid }))$ and $\epsilon \mathrm{/}(2(1+\mathrm{\mid }s\mathrm{\mid }))$ sufficiently near the end. The triangle inequality then makes the error of the linear combination from $rA+sB$ smaller than $\epsilon$. 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 $R\to \mathrm{\infty }$, then $\mathrm{\mid }F(v)-F(u)\mathrm{\mid }<\epsilon$ 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 $\epsilon >0$, choose a large integer $n$ for which $\mathrm{\mid }F(n)-L\mathrm{\mid }<\epsilon \mathrm{/}2$ and the tail condition is below $\epsilon \mathrm{/}2$. 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 $a+1\mathrm{/}n$ or $b-1\mathrm{/}n$ furnished by [L3]; the identical Cauchy argument applies. Reversing the real line gives the $-\mathrm{\infty }$ form.

At $+\mathrm{\infty }$, $F(R)={\int }_a^{R}f$ 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 $F(n)\to S={\sup }_{n}F(n)$.

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 $-\mathrm{\infty }$ form. Moving the initial finite endpoint is harmless by [L3].

By the Cauchy criterion [L2], remote tail integrals of $\mathrm{\mid }f\mathrm{\mid }$ 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 $I={\int }_a^{\mathrm{\infty }}f$. By [L1], ${\int }_R^{\mathrm{\infty }}f=I-{\int }_a^{R}f$, 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 $\mathrm{\mid }f\mathrm{\mid }\le g$, step 1.1 applied to $\mathrm{\mid }f\mathrm{\mid }$ 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 $L\mathrm{/}2$ gives $L\mathrm{/}2<f\mathrm{/}g<3L\mathrm{/}2$ sufficiently near the singular end. Thus $(L\mathrm{/}2)g\le f\le (3L\mathrm{/}2)g$. Applying [L2] in both directions proves the equivalence.

If $L=0$, eventually $f\mathrm{/}g\le 1$, so $f\le g$. If $f\mathrm{/}g\to +\mathrm{\infty }$, eventually $g\le f$. The one-way conclusions again follow from [L2].

For every integer $k\ge 1$, 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 ${\int }_1^{N}f\le {\int }_1^{R}f\le {\int }_1^{M}f$, 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 $p=m\mathrm{/}q$ 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 $q{t}^{q-m-1}$. The integer-power rule gives primitive $q{t}^{q-m}\mathrm{/}(q-m)$ when $m\mathrm{\ne }q$. Substituting $t=x^{1\mathrm{/}q}$ back gives $x^{1-p}\mathrm{/}(1-p)$. The FTC proves the displayed formula.

On $[2^k,2^{k+1}]$, $1\mathrm{/}x\ge 2^{-(k+1)}$, so its integral is at least $1\mathrm{/}2$. 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 $R^{1-p}\to 0$, so [L1] gives ${\int }_1^R{x}^{-p}dx\to 1\mathrm{/}(p-1)$.

If $p\le 1$, 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 $p\le 0$ compare $x^{-p}\ge 1$ with the constant one.

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

If $p\ge 1$, 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 $\epsilon >0$, take the common smaller truncation scale on which each term is within $\epsilon \mathrm{/}2$ of its limit; the triangle inequality then puts their sum within $\epsilon$ of the sum in [L1].

On the real line, split at zero. As $R\to \mathrm{\infty }$, ${\int }_{-R}^{0}f$ and ${\int }_0^{R}f$ tend separately to their two tail values. The same $\epsilon \mathrm{/}2$ 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 $f(x)=1\mathrm{/}x$ 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 ${\int }_0^1{x}^{-p}dx$ 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 $1\mathrm{/}x$ is not one: take $f(x)=x$, which is continuous and therefore locally integrable. For every $R$ the substitution $x\mapsto -x$ gives ${\int }_{-R}^{0}xdx=-{\int }_0^{R}xdx$, so ${\int }_{-R}^{R}xdx=0$ and the whole-line principal value is $0$; but ${\int }_0^{R}xdx=R^2\mathrm{/}2$ is unbounded in $R$, so the tail ${\int }_0^{\mathrm{\infty }}xdx$ does not converge.

Choose $M$ with $\mathrm{\mid }F\mathrm{\mid }\le M$. In clause 1, apply [L1] on $[u,v]$. Every partial integral ${\int }_u^{x}f=F(x)-F(u)$ has absolute value at most $2M$, so [L1, L2, assume-case first] $\mid {\int }_u^{v}fg\mid \le 2M(g(u)+g(v))\le 4Mg(u)\mathrm{.}$ This tends to zero as $u\to \mathrm{\infty }$, and [L2] proves convergence.

In clause 2 $f$ is continuous, hence integrable on every $[u,v]\subseteq [a,\mathrm{\infty })$, so [L5] gives $F^{\mathrm{\prime }}=f$ there — the hypothesis [L3] requires and which boundedness of $F$ does not supply. With $g$ differentiable and $g^{\mathrm{\prime }}$ 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 $g(R)\to 0$. Also $\mathrm{\mid }F{g}^{\mathrm{\prime }}\mathrm{\mid }\le M\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }$, so [L4] makes $\int F{g}^{\mathrm{\prime }}$ converge. Passing $R\to \mathrm{\infty }$ proves convergence of $\int fg$.

The two clauses are exhausted by steps 1.1–1.2. Reversing orientation proves the $-\mathrm{\infty }$ 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 $\epsilon >0$, the definition of supremum gives $x_0$ with $L-\epsilon <g(x_0)\le L$; monotonicity then gives $L-\epsilon <g(x)\le L$ 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 $h\ge 0$, $h$ is nonincreasing toward the singular end, and $h\to 0$.

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 ${\int }_a^{\mathrm{\infty }}uv$ converges. Then its truncation primitive is bounded by [L3], and $uv$ is continuous. Put $g=1\mathrm{/}v$. Positivity and $v\to \mathrm{\infty }$ give $g\to 0$: for $\epsilon >0$, eventually $v>1\mathrm{/}\epsilon$, hence $0<g<\epsilon$. Also [L1] and the hypothesis give absolute convergence of $\int g^{\mathrm{\prime }}$.

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] ${\int }_{x_j}^{x_{j+1}}\mathrm{\mid }f\mathrm{\mid }g\ge \delta g(x_{j+1})\mathrm{.}$ Also ${\int }_{x_j}^{x_{j+1}}g\le Mg(x_j)$. Since $\int g$ diverges, additivity and [L2] force ${\sum }_{j}g(x_j)$, and hence its shifted tail, to diverge. Thus the block lower bounds for $\int \mathrm{\mid }f\mathrm{\mid }g$ 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 ${\int }_{x_j}^{x_{j+1}}\mathrm{\mid }f\mathrm{\mid }g\ge \delta g(x_j)$, whose partial sums diverge.

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

Assume first that $\varphi$ is nondecreasing, and let $0\le h\le M$. [L1] For a partition $P=\{t_i\}$ of $[c,d]$, transport its points through $\varphi$ and delete repeated image points. If $U_i,u_i$ are the supremum and infimum of ${\varphi }^{\mathrm{\prime }}$ on $[t_{i-1},t_i]$, the mean value theorem gives $u_i\mathrm{\Delta }{t}_i\le \varphi (t_i)-\varphi (t_{i-1})\le U_i\mathrm{\Delta }{t}_i\mathrm{.}$ On a flat interval the image increment is zero and ${\varphi }^{\mathrm{\prime }}=0$ in its interior, so its contribution may be discarded.

Compare the upper sum of $h$ on the transported partition with the upper sum of $(h\circ \varphi ){\varphi }^{\mathrm{\prime }}$ on $P$. [step 1.1, L2, L4] On each nonflat interval the two relevant suprema differ, after multiplication by $\mathrm{\Delta }{t}_i$, by at most $M(U_i-u_i)\mathrm{\Delta }{t}_i$; the identical estimate holds for lower sums. Hence each pair of corresponding sums differs by at most $M{\sum }_i(U_i-u_i)\mathrm{\Delta }{t}_i\mathrm{.}$ Because ${\varphi }^{\mathrm{\prime }}$ is integrable, refinements can make this error arbitrarily small. Taking upper and lower integrals therefore gives $\overline{{\int }_a^b}h=\overline{{\int }_c^d}(h\circ \varphi ){\varphi }^{\mathrm{\prime }},\underset{‾}{{\int }_a^b}h=\underset{‾}{{\int }_c^d}(h\circ \varphi ){\varphi }^{\mathrm{\prime }}\mathrm{.}$ Thus one nonnegative function is integrable exactly when the other is, and their integrals then agree.