# 9.4: Convergence of Parametrized Integrals and Functions

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

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.

9.4: Convergence of Parametrized Integrals and Functions is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.