9.4: Convergence of Parametrized Integrals and Functions

I. We now consider C-integrals of the form

$$C \int f(t, u) d m(t),$$

where $m$ is Lebesgue or LS measure in $E^{1}.$ Here the variable $u,$ called a parameter, remains fixed in the process of integration; but the end result depends on $u,$ of course.

We assume $f : E^{2} \rightarrow E$ ($E$ complete) even if not stated explicitly. As before, we give our definitions and theorems for the case

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

The other cases $\left(C \int_{-\infty}^{a}, C \int_{a}^{b-}, \text { etc. }\right)$ are analogous; they are treated in Problems 2 and 3. We assume

$$a, b, c, x, t, u, v \in E^{1}$$

throughout, and write "$d t$" for "$d m(t)$" iff $m$ is Lebesgue measure.

If

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

converges for each $u$ in a set $B \subseteq E^{1},$ we can define a map $F : B \rightarrow E$ by

$$F(u)=C \int_{a}^{\infty} f(t, u) d m(t)=\lim _{x \rightarrow \infty} \int_{a}^{x} f(t, u) d m(t).$$

This means that

$$(\forall u \in B)(\forall \varepsilon>0)(\exists b>a)(\forall x \geq b) \quad\left|\int_{a}^{x} f(t, u) d m(t)-F(u)\right|<\varepsilon,$$

so $|F|<\infty$ on $B$.

Here $b$ depends on both $\varepsilon$ and $u$ (convergence is "pointwise"). However, it may occur that one and the same $b$ fits all $u \in B,$ so that $b$ depends on $\varepsilon$ alone. We then say that

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

converges uniformly on $B$ (i.e., for $u \in B$), and write

$$F(u)=C \int_{a}^{\infty} f(t, u) d m(t) \text { (uniformly) on } B.$$

Explicitly, this means that

$$(\forall \varepsilon>0)(\exists b>a)(\forall u \in B)(\forall x \geq b) \quad\left|\int_{a}^{x} f(t, u) d m(t)-F(u)\right|<\varepsilon.$$

Clearly, this implies (1), but not conversely. We now obtain the following.

Theorem $\PageIndex{1}$ (Cauchy criterion)

Suppose

$$\int_{a}^{x} f(t, u) d m(t)$$

exists for $x \geq a$ and $u \in B \subseteq E^{1}.$ (This is automatic if $E \subseteq E^{*};$ see Chapter 8, §5.)

Then

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

converges uniformly on $B$ iff for every $\varepsilon>0,$ there is $b>a$ such that

$$(\forall v, x \in[b, \infty))(\forall u \in B) \quad\left|\int_{v}^{x} f(t, u) d m(t)\right|<\varepsilon,$$

and

$$\left|\int_{a}^{b} f(t, u) d m(t)\right|<\infty.$$

Proof

The necessity of (3) follows as in Theorem 2 of §3. (Verify!)

To prove sufficiency, suppose the desired $b$ exists for every $\varepsilon>0.$ Then for each (fixed) $u \in B$,

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

satisfies Theorem 2 of §3. Hence

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

exists for every $u \in B$ (pointwise). Now, from (3), writing briefly $\int f$ for $\int f(t, u) d m(t),$ we obtain

$$\left|\int_{v}^{x} f\right|=\left|\int_{a}^{x} f-\int_{a}^{v} f\right|<\varepsilon$$

for all $u \in B$ and all $x>v \geq b$.

Making $x \rightarrow \infty$ (with $u$ and $v$ temporarily fixed), we have by (4) that

$$\left|F(u)-\int_{a}^{v} f\right| \leq \varepsilon$$

whenever $v \geq b$.

But by our assumption, $b$ depends on $\varepsilon$ alone (not on $u$). Thus unfixing $u$, we see that (5) establishes the uniform convergence of

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

as required.$\quad \square$

Corollary $\PageIndex{1}$

Under the assumptions of Theorem 1,

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

converges uniformly on $B$ if

$$C \int_{a}^{\infty}|f(t, u)| d m(t)$$

does.

Indeed,

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

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

Let $f : E^{2} \rightarrow E$ and $M : E^{2} \rightarrow E^{*}$ satisfy

$$|f(t, u)| \leq M(t, u)$$

for $u \in B \subseteq E^{1}$ and $t \geq a$.

Then

$$C \int_{a}^{\infty}|f(t, u)| d m(t)$$

converges uniformly on $B$ if

$$C \int_{a}^{\infty} M(t, u) d m(t)$$

does.

Indeed, Theorem 1 applies, with

$$\left|\int_{v}^{x} f\right| \leq \int_{v}^{x} M<\varepsilon.$$

Hence we have the following corollary.

Corollary $\PageIndex{3}$ ("$M$-test")

Let $f : E^{2} \rightarrow E$ and $M : E^{1} \rightarrow E^{*}$ satisfy

$$|f(t, u)| \leq M(t)$$

for $u \in B \subseteq E^{1}$ and $t \geq a.$ Suppose

$$C \int_{a}^{\infty} M(t) d m(t)$$

converges. Then

$$C \int_{a}^{\infty}|f(t, u)| d m(t)$$

converges (uniformly) on $B.$ So does

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

by Corollary 1.

Proof

Set

$$h(t, u)=M(t) \geq|f(t, u)|.$$

Then Corollary 2 applies (with $M$ replaced by $h$ there). Indeed, the convergence of

$$C \int h=C \int M$$

is trivially "uniform" for $u \in B,$ since $M$ does not depend on $u$ at all.$\quad \square$

Note 1. Observe also that, if $h(t, u)$ does not depend on $u,$ then the (pointwise) and (uniform) convergence of $C \int h$ are trivially equivalent.

We also have the following result.

Corollary $\PageIndex{4}$

Suppose

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

converges (pointwise) on $B \subseteq E^{1}.$ Then this convergence is uniform iff

$$\lim _{\nu \rightarrow \infty} C \int_{v}^{\infty} f(t, u) d m(t)=0 \text { (uniformly) on } B,$$

i.e., iff

$$(\forall \varepsilon>0)(\exists b>a)(\forall u \in B)(\forall v \geq b) \quad\left|C \int_{v}^{\infty} f(t, u) d m(t)\right|<\varepsilon.$$

Proof

The proof (based on Theorem 1) is left to the reader, along with that of the following corollary.

Corollary $\PageIndex{5}$

Suppose

$$\int_{a}^{b} f(t, u) d m(t) \neq \pm \infty$$

exists for each $u \in B \subseteq E^{1}$.

Then

$$C \int_{a}^{\infty} f(t, u) d m(t)$$

converges (uniformly) on $B$ iff

$$C \int_{b}^{\infty} f(t, u) d m(t)$$

does.

II. The Abel-Dirichlet tests for uniform convergence of series (Problems 9 and 11 in Chapter 4, §13) have various analogues for C-integrals. We give two of them, using the second law of the mean (Corollary 5 in §1).

First, however, we generalize our definitions, "unstarring" some ideas of Chapter 4, §11. Specifically, given

$$H : E^{2} \rightarrow E \text{ (} E \text { complete),}$$

we say that $H(x, y)$ converges to $F(y),$ uniformly on $B,$ as $x \rightarrow q\left(q \in E^{*}\right),$ and write

$$\lim _{x \rightarrow q} H(x, y)=F(y) \text { (uniformly) on } B$$

iff we have

$$(\forall \varepsilon>0)\left(\exists G_{\neg q}\right)(\forall y \in B)\left(\forall x \in G_{\neg q}\right) \quad|H(x, y)-F(y)|<\varepsilon;$$

hence $|F|<\infty$ on $B$.

If here $q=\infty,$ the deleted globe $G_{\neg q}$ has the form $(b, \infty).$ Thus if

$$H(x, u)=\int_{a}^{x} f(t, u) d t,$$

(6) turns into (2) as a special case. If (6) holds with $"\left(\exists G_{\neg q}\right) "$ and $"(\forall y \in B)"$ interchanged, as in (1), convergence is pointwise only.

As in Chapter 8, §8, we denote by $f(\cdot, y),$ or $f^{y},$ the function of $x$ alone (on $E^{1}$) given by

$$f^{y}(x)=f(x, y).$$

Similarly,

$$f_{x}(y)=f(x, y).$$

Of course, we may replace $f(x, y)$ by $f(t, u)$ or $H(t, u),$ etc.

We use Lebesgue measure in Theorems 2 and 3 below.

Theorem $\PageIndex{2}$

Assume $f, g : E^{2} \rightarrow E^{1}$ satisfy

(i) $C \int_{a}^{\infty} g(t, u) d t$ converges (uniformly) on $B$;

(ii) each $g^{u}(u \in B)$ is $L$-measurable on $A=[a, \infty)$;

(iii) each $f^{u}(u \in B)$ is monotone $(\downarrow$ or $\uparrow)$ on $A;$ and

(iv) $|f|<K \in E^{1}$ (bounded) on $A \times B$.

Then

$$C \int_{a}^{\infty} f(t, u) g(t, u) d t$$

converges uniformly on $B$.

Proof

Given $\varepsilon>0,$ use assumption (i) and Theorem 1 to choose $b>a$ so that

$$\left|L \int_{v}^{x} g(t, u) d t\right|<\frac{\varepsilon}{2 K},$$

written briefly as

$$\left|L \int_{v}^{x} g^{u}\right|<\frac{\varepsilon}{2 K},$$

for all $u \in B$ and $x>v \geq b,$ with $K$ as in (iv).

Hence by (ii), each $g^{u}(u \in B)$ is $L$-integrable on any interval $[v, x] \subset A$, with $x>v \geq b.$ Thus given such $u$ and $[v, x],$ we can use (iii) and Corollary 5 from §1 to find that

$$L \int_{v}^{x} f^{u} g^{u}=f^{u}(v) L \int_{v}^{c} g^{u}+f^{u}(x) L \int_{c}^{x} g^{u}$$

for some $c \in[v, x]$.

Combining with (7) and using (iv), we easily obtain

$$\left|L \int_{v}^{x} f(t, u) g(t, u) d t\right|<\varepsilon$$

whenever $u \in B$ and $x>v \geq b.$ (Verify!)

Our assertion now follows by Theorem 1.$\quad \square$

Theorem $\PageIndex{3}$ (Abel-Dirichlet test)

Let $f, g : E^{2} \rightarrow E^{*}$ satisfy

(a) $\lim _{t \rightarrow \infty} f(t, u)=0$ (uniformly) for $u \in B$;

(b) each $f^{u}(u \in B)$ is nonincreasing $(\downarrow)$ on $A=[0, \infty)$;

(c) each $g^{u}(u \in B)$ is $L$-measurable on $A;$ and

(d) $\left(\exists K \in E^{1}\right)(\forall x \in A)(\forall u \in B)\left|L \int_{a}^{x} g(t, u) d t\right|<K$.

Then

$$C \int_{a}^{\infty} f(t, u) g(t, u) d t$$

converges uniformly on $B$.

Proof Outline

Argue as in Problem 13 of §3, replacing Theorem 2 in §3 by Theorem 1 of the present section.

By Lemma 2 in §1, obtain

$$\left|L \int_{v}^{x} f^{u} g^{u}\right|=\left|f^{u}(v) L \int_{a}^{x} g^{u}\right| \leq K f(v, u)$$

for $u \in B$ and $x>v \geq a$.

Then use assumption (a) to fix $k$ so that

$$|f(t, u)|<\frac{\varepsilon}{2 K}$$

for $t>k$ and $u \in B. \quad \square$

Note 2. Via components, Theorems 2 and 3 extend to the case $g : E^{2} \rightarrow$ $E^{n}\left(C^{n}\right).$

Note 3. While Corollaries 2 and 3 apply to absolute convergence only, Theorems 2 and 3 cover conditional convergence, too (a great advantage!). The theorems also apply if $f$ or $g$ is independent of $u$ (see Note 1). This supersedes Problems 13 and 14 in §3.

Examples

(A) The integral

$$\int_{0}^{\infty} \frac{\sin t u}{t} d t$$

converges uniformly on $B_{\delta}=[\delta, \infty)$ if $\delta>0,$ and pointwise on $B=[0, \infty)$.

Indeed, we can use Theorem 3, with

$$g(t, u)=\sin t u$$

and

$$f(t, u)=\frac{1}{t}, f(0, u)=1,$$

say. Then the limit

$$\lim _{t \rightarrow \infty} \frac{1}{t}=0$$

is trivially uniform for $u \in B_{\delta},$ as $f$ is independent of $u.$ Thus assumption (a) is satisfied. So is (d) because

$$\left|\int_{0}^{x} \sin t u d t\right|=\left|\frac{1}{u} \int_{0}^{x u} \sin \theta d \theta\right| \leq \frac{1}{\delta} \cdot 2.$$

(Explain!) The rest is easy.

Note that Theorem 2 fails here since assumption (i) is not satisfied.

(B) The integral

$$\int_{0}^{\infty} \frac{1}{t} e^{-t u} \sin a t d t$$

converges uniformly on $B=[0, \infty).$ It does so absolutely on $B_{\delta}=[\delta, \infty),$ if $\delta>0.$

Here we shall use Theorem 2 (though Theorem 3 works, too). Set

$$f(t, u)=e^{-t u}$$

and

$$g(t, u)=\frac{\sin a t}{t}, g(0, u)=a.$$

Then

$$\int_{0}^{\infty} g(t, u) d t$$

converges (substitute $x=a t$ in Problem 8 or 15 in §3). Convergence is trivially uniform, by Note 1. Thus assumption (i) holds, and so do the other assumptions. Hence the result.

For absolute convergence on $B_{\delta},$ use Corollary 3 with

$$M(t)=e^{-\delta t},$$

so $M \geq|f g|$.

Note that, quite similarly, one treats C-integrals of the form

$$\int_{a}^{\infty} e^{-t u} g(t) d t, \int_{a}^{\infty} e^{-t^{2} u} g(t) d t, \text { etc.,}$$

provided

$$\int_{a}^{\infty} g(t) d t$$

converges $(a \geq 0)$.

In fact, Theorem 2 states (roughly) that the uniform convergence of $C \int g$ implies that of $C \int f g,$ provided $f$ is monotone (in $t$) and bounded.

III. We conclude with some theorems on uniform convergence of functions $H : E^{2} \rightarrow E$ (see (6)). In Theorem 4, $m$ is again an LS (or Lebesgue) measure in $E^{1};$ the deleted globe $G_{\neg q}^{*}$ is fixed.

Theorem $\PageIndex{4}$

Suppose

$$\lim _{x \rightarrow q} H(x, y)=F(y) \text { (uniformly)}$$

for $y \in B \subseteq E^{1}.$ Then we have the following:

(i) If all $H_{x}\left(x \in G_{\neg q}^{*}\right)$ are continuous or $m$-measurable on $B,$ so also is $F$.

(ii) The same applies to $m$-integrability on $B,$ provided $m B<\infty;$ and then

$$\lim _{x \rightarrow q} \int_{B}\left|H_{x}-F\right|=0;$$

hence

$$\lim _{x \rightarrow q} \int_{B} H_{x}=\int_{B} F=\int_{B}\left(\lim _{x \rightarrow q} H_{x}\right).$$

Formula (8') is known as the rule of passage to the limit under the integral sign.

Proof

(i) Fix a sequence $x_{k} \rightarrow q$ $(x_{k}$ in the deleted globe $G_{\neg q}^{*}),$ and set

$$H_{k}=H_{x_{k}} \quad(k=1,2, \ldots).$$

The uniform convergence

$$H(x, y) \rightarrow F(y)$$

is preserved as $x$ runs over that sequence (see Problem 4). Hence if all $H_{k}$ are continuous or measurable, so is $F$ (Theorem 2 in Chapter 4, §12 and Theorem 4 in Chapter 8, §1. Thus clause (i) is proved.

(ii) Now let all $H_{x}$ be $m$-integrable on $B;$ let

$$m B<\infty.$$

Then the $H_{k}$ are $m$-measurable on $B,$ and so is $F,$ by (i). Also, by (6),

$$(\forall \varepsilon>0)\left(\exists G_{\neg q}\right)\left(\forall x \in G_{\neg q}\right) \quad \int_{B}\left|H_{x}-F\right| \leq \int_{B}(\varepsilon)=\varepsilon m B<\infty,$$

proving (8). Moreover, as

$$\int_{B}\left|H_{x}-F\right|<\infty,$$

$H_{x}-F$ is $m$-integrable on $B,$ and so is

$$F=H_{x}-\left(H_{x}-F\right).$$

Hence

$$\left|\int_{B} H_{x}-\int_{B} F\right|=\left|\int_{B}\left(H_{x}-F\right)\right| \leq \int_{B}\left|H_{x}-F\right| \rightarrow 0,$$

as $x \rightarrow q,$ by (8). Thus (8') is proved, too.$\quad \square$

Quite similarly (keeping $E$ complete and using sequences), we obtain the following result.

Theorem $\PageIndex{5}$

Suppose that

(i) all $H_{x}\left(x \in G_{-q}^{*}\right)$ are continuous and finite on a finite interval $B \subset E^{1}$, and differentiable on $B-Q,$ for a fixed countable set $Q$;

(ii) $\lim _{x \rightarrow q} H\left(x, y_{0}\right) \neq \pm \infty$ exists for some $y_{0} \in B;$ and

(iii) $\lim _{x \rightarrow q} D_{2} H(x, y)=f(y)$ (uniformly) exists on $B-Q$.

Then $f,$ so defined, has a primitive $F$ on $B,$ exact on $B-Q$ (so $F^{\prime}=f$ on $B-Q);$ moreover,

$$F(y)=\lim _{x \rightarrow y} H(x, y) \text { (uniformly) for } y \in B.$$

Outline of Proof

Note that

$$D_{2} H(x, y)=\frac{d}{d y} H_{x}(y).$$

Use Theorem 1 of Chapter 5, §9, with $F_{n}=H_{x_{n}}, x_{n} \rightarrow q. \quad \square$

Note 4. If $x \rightarrow q$ over a path $P$ (clustering at $q$), one must replace $G_{\neg q}$ and $G_{\neg q}^{*}$ by $P \cap G_{\neg q}$ and $P \cap G_{\neg q}^{*}$ in (6) and in Theorems 4 and 5.