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# 9.3: Improper (Cauchy) Integrals

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Cauchy extended R-integration to unbounded sets and functions as follows.

Given $$f : E^{1} \rightarrow E$$ and assuming that the right-hand side R-integrals and limits exist, define (first for unbounded sets, then for unbounded functions)

(i) $$\int_{a}^{\infty} f=\int_{[a, \infty)} f=\lim _{x \rightarrow \infty} R \int_{a}^{x} f$$;

(ii) $$\int_{-\infty}^{a} f=\int_{(-\infty, a]} f=\lim _{x \rightarrow-\infty} R \int_{x}^{a} f$$.

If both

$\int_{0}^{\infty} f \text { and } \int_{-\infty}^{0} f$

exists, define

$\int_{-\infty}^{\infty} f=\int_{(-\infty, 0)} f+\int_{[0, \infty)} f.$

Now, suppose $$f$$ is unbounded near some $$p \in A=[a, b],$$ i.e., unbounded on $$A \cap G_{\neg p}$$ for every deleted globe $$G_{\neg p}$$ about $$p$$ (such points $$p$$ are called singularities).

Then (again assuming existence of the R-integrals and limits), we define

1. in case of a singularity $$p=a$$, $\int_{a+}^{b} f=\int_{(a, b]} f=\lim _{x \rightarrow a+} R \int_{x}^{b} f;$
2. if $$p=b,$$ then $\int_{a}^{b-} f=\int_{[a, b)} f=\lim _{x \rightarrow b-} R \int_{a}^{x} f;$
3. if $$a<p<b$$ and if $\int_{a}^{p-} f \text { and } \int_{p+}^{b} f$

exist, then

$\int_{a}^{b} f=\int_{a}^{p-} f+\int_{p}^{p} f+\int_{p+}^{b} f.$

The term

$\int_{p}^{p} f=\int_{[p, p]} f$

is necessary if $$R S$$- or $$L S$$-integrals are used.

Finally, if $$A$$ contains several singularities, it must be split into subintervals, each with at most one endpoint singularity; and $$\int_{a}^{b} f$$ is split accordingly. We call all such integrals improper or Cauchy (C) integrals. A C-integral is said to converge iff it exists and is finite.

This theory is greatly enriched if in the above definitions, one replaces $$R$$-integrals by Lebesgue integrals, using Lebesgue or LS measure in $$E^{1}.$$ (This makes sense even when a Lebesgue integral (proper) does exist; see Theorem 1.) Below, $$m$$ shall denote such a measure unless stated otherwise.

C-integrals with respect to $$m$$ will be denoted by

$C \int_{a}^{\infty} f d m, \quad C \int_{[a, b)} f, \quad \text {etc. }$

"Classical" notation:

$C \int f(x) d m(x) \text { or } C \int f(x) d x$

(the latter if $$m$$ is Lebesgue measure). We omit the "C" if confusion with proper integrals $$\int_{a}^{x} f$$ is unlikely.

Note 1. C-integrals are limits of integrals, not integrals proper. Yet they may equal the latter (Theorem 1 below) and then may be used to compute them.

Caution. "Singularities" in $$[a, b]$$ may affect the primitive used in computations (cf. Problem 4 in §1). Then $$[a,b]$$ must be split (see above), and $$C \int_{a}^{b} f$$ splits accordingly. (Additivity applies to C-integrals; see Problem 9, below.)

## Examples

(A) The integral

$L \int_{-1}^{1 / 2} \frac{d x}{x^{2}}$

has a singularity at $$0.$$ By Theorem 1 below, we get

\begin{aligned} L \int_{-1}^{1 / 2} \frac{d x}{x^{2}} &=\int_{-1}^{0-} \frac{d x}{x^{2}}+\int_{0+}^{1 / 2} \frac{d x}{x^{2}} \\ &=\lim _{x \rightarrow 0^{-}}\left(-\frac{1}{x}-1\right)+\lim _{x \rightarrow 0+}\left(-2+\frac{1}{x}\right)=\infty+\infty=\infty. \end{aligned}

(B) We have

$C \int_{1 / 2}^{\infty} \frac{d x}{x^{2}}=\lim _{x \rightarrow \infty}\left(-\frac{1}{x}+2\right)=2.$

Hence

$C \int_{-1}^{\infty} \frac{d x}{x^{2}}=C \int_{-1}^{1 / 2} \frac{d x}{x^{2}}+C \int_{1 / 2}^{\infty} \frac{d x}{x^{2}}=\infty+2=\infty.$

(C) The integral

$L \int_{-1}^{1} \frac{|x|}{x} d x$

has no singularities (consider deleted globes about $$0$$). The primitive $$F(x)=|x|$$ exists (example (b) in Chapter 5, §5); so

$L \int_{-1}^{1} \frac{|x|}{x} d x=\left.|x|\right|_{-1} ^{1}=0.$

In the rest of this section, we state our theorems mainly for

$C \int_{a}^{\infty} f,$

but they apply, with similar proofs, to

$C \int_{-\infty}^{\infty} f, \quad C \int_{a}^{b-} f, \quad \text {etc. }$

The measure $$m$$ is as explained above.

## Theorem $$\PageIndex{1}$$

Let $$A=[a, \infty), f : E^{1} \rightarrow E$$ ($$E$$ complete).

(i) If $$f \geq 0$$ on $$A,$$ then

$C \int_{a}^{\infty} f d m$

exists $$(\leq \infty)$$ and equals

$\int_{A} f d m.$

(ii) The map $$f$$ is $$m$$-integrable on $$A$$ iff

$C \int_{a}^{\infty}|f|<\infty$

and $$f$$ is $$m$$-measurable on $$A;$$ then again,

$C \int_{a}^{\infty} f d m=\int_{A} f d m.$

Proof

(i) Let $$f \geq 0$$ on $$A.$$ By the rules of Chapter 8, §5, $$\int_{A} f$$ is always defined for such $$f;$$ so we may set

$F(x)=\int_{a}^{x} f d m, \quad x \geq a.$

Then by Theorem 1(f) in Chapter 8, §5, $$F \uparrow$$ on $$A;$$ for $$a \leq x \leq y$$ implies

$F(x)=\int_{a}^{x} f \leq \int_{a}^{y} f=F(y).$

Now, by the properties of monotone limits,

$\lim _{x \rightarrow \infty} F(x)=\lim _{x \rightarrow \infty} \int_{a}^{x} f=C \int_{a}^{\infty} f$

exists in $$E^{*};$$ so by Theorem 1 of Chapter 4, §2, it can be found by making $$x$$ run over some sequence $$x_{k} \rightarrow \infty,$$ say, $$x_{k}=k$$.

Thus set

$A_{k}=[a, k], \quad k=1,2, \ldots.$

Then $$\left\{A_{k}\right\} \uparrow$$ and

$\bigcup A_{k}=A=[a, \infty),$

i.e., $$A_{k} \nearrow A$$.

Moreover, by Note 4 in Chapter 8, §5, the set function $$s=\int f$$ is $$\sigma$$-additive and semifinite $$(\geq 0).$$ Thus by Theorem 2 of Chapter 7, §4 (left continuity)

$\int_{A} f d m=\lim _{k \rightarrow \infty} \int_{A_{k}} f=\lim _{k \rightarrow \infty} \int_{a}^{k} f=C \int_{a}^{\infty} f,$

proving (i).

(ii) By clause (i),

$C \int_{a}^{\infty}|f|=\int_{A}|f| d m$

exists, as $$|f| \geq 0.$$ Hence

$C \int_{a}^{\infty}|f|<\infty$

plus measurability amounts to integrability (Theorem 2 of Chapter 8, §6).

Moreover,

$C \int_{a}^{\infty}|f|<\infty$

implies the convergence of $$C \int_{a}^{\infty} f$$ (see Corollary 1 below). Thus as

$\lim _{x \rightarrow \infty} \int_{a}^{x} f$

exists, we proceed exactly as before (here $$s=\int f$$ is finite), proving (ii) also.$$\quad \square$$

Note 2. If $$E \subseteq E^{*},$$ formula (1) results even if $$f$$ is not $$m$$-measurable.

Note 3. While $$f$$ cannot be integrable unless $$|f|$$ is (Corollary 2 of Chapter 8, §6), it can happen that

$C \int f$

converges even if

$C \int|f|=\infty$

(this is called conditional convergence). A case in point is

$C \int_{0}^{\infty} \frac{\sin x}{x} d x;$

see Problem 8.

Thus $$C$$-integrals may be finite where proper integrals are $$\infty$$ or fail to exist (a great advantage!). Yet they are deficient in other respects (see Problem 9(c)).

For our next theorem, we need the previously "starred" Theorem 2 in Chapter 4, (Review it!) As we shall see, C-integrals resemble infinite series.

## Theorem $$\PageIndex{2}$$ (Cauchy criterion)

Let $$A=[a, \infty), f : E^{1} \rightarrow E, E$$ complete.
Suppose

$\int_{a}^{x} f d m$

exists for each $$x \in A.$$ (This is automatic if $$E \subseteq E^{*};$$ see Chapter 8, §5.)

Then

$C \int_{a}^{\infty} f$

converges iff for every $$\varepsilon>0,$$ there is $$b \in A$$ such that

$\left|\int_{v}^{x} f d m\right|<\varepsilon \quad \text {whenever } b \leq v \leq x<\infty,$

and

$\left|\int_{a}^{b} f d m\right|<\infty.$

Proof

By additivity (Chapter 8, §5, Theorem 2; Chapter 8, §7, Theorem 3),

$\int_{a}^{x} f=\int_{a}^{v} f+\int_{v}^{x} f$

if $$a \leq v \leq x<\infty.$$ (In case $$E \subseteq E^{*},$$ this holds even if $$f$$ is not integrable; see Theorem 2, of Chapter 8, §5.)

Now, if

$C \int_{a}^{\infty} f$

converges, let

$r=\lim _{x \rightarrow \infty} \int_{a}^{x} f d m \neq \pm \infty.$

Then for any $$\varepsilon>0,$$ there is some

$b \in[a, \infty)=A$

such that

$\left|\int_{a}^{x} f d m-r\right|<\frac{1}{2} \varepsilon \quad \text { for } x \geq b.$

(Why may we use the standard metric here?)

Taking $$x=b,$$ we get (2'). Also, if $$a \leq b \leq v \leq x,$$ we have

$\left|\int_{a}^{x} f d m-r\right|<\frac{1}{2} \varepsilon$

and

$\left|r-\int_{a}^{\nu} f d m\right|<\frac{1}{2} \varepsilon.$

Hence by the triangle law, (2) follows also. Thus this $$b$$ satisfies (2).

Conversely, suppose such a $$b$$ exists for every given $$\varepsilon>0.$$ Fixing $$b,$$ we thus have (2) and (2'). Now, with $$A=[a, \infty),$$ define $$F : A \rightarrow E$$ by

$F(x)=\int_{a}^{x} f d m,$

so

$C \int_{a}^{\infty} f=\lim _{x \rightarrow \infty} F(x)$

if this limit exists. By (2),

$|F(x)|=\left|\int_{a}^{x} f d m\right| \leq\left|\int_{a}^{b} f d m\right|+\left|\int_{b}^{x} f d m\right|<\left|\int_{a}^{b} f d m\right|+\varepsilon$

if $$x \geq b.$$ Thus $$F$$ is finite on $$[b, \infty),$$ and so we may again use the standard metric

$\rho(F(x), F(v))=|F(x)-F(v)|=\left|\int_{a}^{x} f d m-\int_{a}^{v} f d m\right| \leq\left|\int_{v}^{x} f d m\right|<\varepsilon$

if $$x, v \geq b.$$ The existence of

$C \int_{a}^{\infty} f d m=\lim _{x \rightarrow \infty} F(x) \neq \pm \infty$

now follows by Theorem 2 of Chapter 4, §2. (We shall henceforth presuppose this "starred" theorem.)

Thus all is proved.$$\quad \square$$

## Corollary $$\PageIndex{1}$$

Under the same assumptions as in Theorem 2, the convergence of

$C \int_{a}^{\infty}|f| d m$

implies that of

$C \int_{a}^{\infty} f d m.$

Indeed,

$\left|\int_{v}^{x} f\right| \leq \int_{v}^{x}|f|$

(Theorem 1(g) of Chapter 8, §5, and Problem 10 in Chapter 8, §7).

Note 4. We say that $$C \int f$$ converges absolutely iff $$C \int|f|$$ converges.

## Corollary $$\PageIndex{2}$$ (comparison test)

If $$|f| \leq|g|$$ a.e. on $$A=[a, \infty)$$ for some $$f, g : E^{1} \rightarrow E,$$ then

$C \int_{a}^{\infty}|f| \leq C \int_{a}^{\infty}|g|;$

so the convergence of

$C \int_{a}^{\infty}|g|$

implies that of

$C \int_{a}^{\infty}|f|.$

For as $$|f|,|g| \geq 0,$$ Theorem 1 reduces all to Theorem 1(c) of Chapter 8, §5.

Note 5. As we see, absolutely convergent C-integrals coincide with proper (finite) Lebesgue integrals of nonnegative or $$m$$-measurable maps. For conditional (i.e., nonabsolute) convergence, see Problems 6-9, 13, and 14.

Iterated C-Integrals. Let the product space $$X \times Y$$ of Chapter 8, §8 be

$E^{1} \times E^{1}=E^{2},$

and let $$p=m \times n,$$ where $$m$$ and $$n$$ are Lebesgue measure or LS measures in $$E^{1}$$. Let

$A=[a, b], B=[c, d], \text { and } D=A \times B.$

Then the integral

$\int_{B} \int_{A} f d m d n=\int_{Y} \int_{X} f C_{D} d m d n$

is also written

$\int_{c}^{d} \int_{a}^{b} f d m d n$

or

$\int_{c}^{d} \int_{a}^{b} f(x, y) d m(x) d n(y).$

As usual, we write "$$d x$$" for "$$d m(x)$$" if $$m$$ is Lebesgue measure in $$E^{1};$$ similarly for $$n.$$

We now define

\begin{aligned} C \int_{a}^{\infty} \int_{c}^{\infty} f d n d m &=\lim _{b \rightarrow \infty} \int_{a}^{b}\left(\lim _{d \rightarrow \infty} \int_{c}^{d} f(x, y) d n(y)\right) d m(x) \\ &=C \int_{a}^{\infty} \int_{c}^{\infty} f(x, y) d n(y) d m(x), \end{aligned}

provided the limits and integrals involved exist.

If the integral (3) is finite, we say that it converges. Again, convergence is absolute if it holds also with $$f$$ replaced by $$|f|,$$ and conditional otherwise. Similar definitions apply to

$C \int_{c}^{\infty} \int_{a}^{\infty} f d m d n, C \int_{-\infty}^{b} \int_{c}^{\infty} f d n d m, \text { etc.}$

## Theorem $$\PageIndex{3}$$

Let $$f : E^{2} \rightarrow E^{*}$$ be $$p$$-measurable on $$E^{2}$$ ($$p, m, n$$ as above). Then we have the following.

(i*) The Cauchy integrals

$C \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}|f| d n d m \text { and } C \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}|f| d m d n$

exist $$(\leq \infty),$$ and both equal

$\int_{E^{2}}|f| d p.$

(ii*) If one of these three integrals is finite, then

$C \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f d n d m \text { and } C \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f d m d n$

converge, and both equal

$\int_{E^{2}} f d p.$

(Similarly for $$C \int_{a}^{\infty} \int_{-\infty}^{b} f d n d m,$$ etc.)

Proof

As $$m$$ and $$n$$ are $$\sigma$$-finite (finite on intervals!), $$f$$ surely has $$\sigma$$-finite support.

As $$|f| \geq 0,$$ clause (i*) easily follows from our present Theorem 1(i) and Theorem 3(i) of Chapter 8, §8.

Similarly, clause (ii*) follows from Theorem 3(ii) of the same section.$$\quad \square$$

## Theorem $$\PageIndex{4}$$ (passage to polars)

Let $$p=$$ Lebesgue measure in $$E^{2}.$$ Suppose $$f : E^{2} \rightarrow E^{*}$$ is $$p$$-measurable on $$E^{2}.$$ Set

$F(r, \theta)=f(r \cos \theta, r \sin \theta), \quad r>0.$

Then

(a) $$C \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f d x d y=C \int_{0}^{\infty} r d r \int_{0}^{2 \pi} F d \theta,$$ and

(b) $$C \int_{0}^{\infty} \int_{0}^{\infty} f d x d y=C \int_{0}^{\infty} r d r \int_{0}^{\pi / 2} F d \theta,$$

provided $$f$$ is nonnegative or $$p$$-integrable on $$E^{2}$$ (for (a)) or on $$(0, \infty) \times(0, \infty)$$ (for (b)).

Proof Outline

First let $$f=C_{D},$$ with $$D$$ a "curved rectangle"

$\left\{(r, \theta) | r_{1}<r \leq r_{2}, \theta_{1}<\theta \leq \theta_{2}\right\}$

for some $$r_{1}<r_{2}$$ in $$X=(0, \infty)$$ and $$\theta_{1}<\theta_{2}$$ in $$Y=[0,2 \pi).$$ By elementary geometry (or calculus), the area

$p D=\frac{1}{2}\left(r_{2}^{2}-r_{1}^{2}\right)\left(\theta_{2}-\theta_{1}\right)$

(the difference between two circular sectors).

For $$f=C_{D},$$ formulas (a) and (b) easily follow from

$p D=L \int_{E^{2}} C_{D} d p.$

(Verify!) Now, curved rectangles behave like half-open intervals

$\left(r_{1}, r_{2}\right] \times\left(\theta_{1}, \theta_{2}\right]$

in $$E^{2},$$ since Theorem 1 in Chapter 7, §1, and Lemma 2 of Chapter 7, §2, apply with the same proof. Thus they form a semiring generating the Borel field in $$E^{2}$$.

Hence show (as in Chapter 8, §8 that Theorem 4 holds for $$f=C_{D}(D \in \mathcal{B}$$). Then take $$D \in \mathcal{M}^{*}$$. Next let $$f$$ be elementary and nonnegative, and so on, as in Theorems 2 and 3 in Chapter 8, §8.$$\quad \square$$

## Examples (continued)

(D) Let

$J=L \int_{0}^{\infty} e^{-x^{2}} d x;$

so

\begin{aligned} J^{2} &=\left(C \int_{0}^{\infty} e^{-x^{2}} d x\right)\left(C \int_{0}^{\infty} e^{-y^{2}} d y\right) \\ &=C \int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y. \quad \text {(Why?)} \end{aligned}

Set

$f(x, y)=e^{-\left(x^{2}+y^{2}\right)}$

in Theorem 4(b). Then $$F(r, \theta)=e^{-r^{2}};$$ hence

\begin{aligned} J^{2} &=C \int_{0}^{\infty} r d r\left(\int_{0}^{\frac{\pi}{2}} e^{-r^{2}} d \theta\right) \\ &=C \int_{0}^{\infty} r e^{-r^{2}} d r \cdot \frac{\pi}{2}=-\left.\frac{1}{4} \pi e^{-t}\right|_{0} ^{\infty}=\frac{1}{4} \pi. \end{aligned}

(Here we computed

$\int r e^{-r^{2}} d r$

by substituting $$r^{2}=t$$.) Thus

$C \int_{0}^{\infty} e^{-x^{2}} d x=L \int_{0}^{\infty} e^{-x^{2}} d x=\sqrt{\frac{1}{4} \pi}=\frac{1}{2} \sqrt{\pi}.$