9.3: Improper (Cauchy) Integrals
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
- in case of a singularity \(p=a\), \[\int_{a+}^{b} f=\int_{(a, b]} f=\lim _{x \rightarrow a+} R \int_{x}^{b} f;\]
- if \(p=b,\) then \[\int_{a}^{b-} f=\int_{[a, b)} f=\lim _{x \rightarrow b-} R \int_{a}^{x} f;\]
- 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.)
(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.
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.
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.\(\quad \square\)
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.
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
\[C \int_{c}^{\infty} \int_{a}^{\infty} f d m d n, C \int_{-\infty}^{b} \int_{c}^{\infty} f d n d m, \text { etc.}\]
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.\(\quad \square\)
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.\(\quad \square\)
(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
\[C \int_{0}^{\infty} e^{-x^{2}} d x=L \int_{0}^{\infty} e^{-x^{2}} d x=\sqrt{\frac{1}{4} \pi}=\frac{1}{2} \sqrt{\pi}.\]