\documentclass{article} \usepackage[T1]{fontenc} \usepackage{textcomp} \renewcommand{\rmdefault}{ptm} \usepackage[scaled=0.92]{helvet} \usepackage[psamsfonts]{amsfonts} \usepackage{amsmath, amsbsy,verbatim} \usepackage[dvips, bookmarks, colorlinks=true, plainpages = false, citecolor = blue, urlcolor = blue, filecolor = blue]{hyperref} \newtheorem{corollary}{Corollary} \newtheorem{definition}{Definition} \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newtheorem{example}{Example} \newcommand{\proof}{\noindent{\sc\bf Proof}\quad } \def\endproof{{\hfill \vbox{\hrule\hbox{% \vrule height1.3ex\hskip0.8ex\vrule}\hrule }}\par} \newcommand{\bbox}{\phantom{1}\hfill{\rule{6pt}{6pt}}} \newcommand{\pd}[2]{\frac{\partial{#1}}{\partial{#2}}} \newcommand{\dst}{\displaystyle} \newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\set}[2]{\left\{#1\, \big|\, #2\right\}} \newcommand{\exer}[1]{\par\noindent{\bf $#1$}.} \newcommand{\boxit}[1]{\bigskip\noindent{\bf #1}\\\vskip-6pt\hskip-\parindent} \newcounter{lcal} \newenvironment{alist}{\begin{list}{\bf (\alph{lcal})} {\topsep 0pt\partopsep 0pt\labelwidth 14pt \labelsep 8pt\leftmargin 22pt\itemsep 0pt \usecounter{lcal}}}{\end{list}} \newcounter{exercise} \newenvironment{exerciselist}{\begin{list}{\bf \arabic{exercise}.} {\topsep 10pt\partopsep 0pt\labelwidth 16pt \labelsep 12pt\leftmargin 28pt \itemsep 8pt\usecounter{exercise}}}{\end{list}} \begin{document} \thispagestyle{empty} \bf \begin{center} {\Huge FUNCTIONS DEFINED BY\\ \medskip IMPROPER INTEGRALS} \vspace{.5in} \huge \bigskip \vspace{.75in} \bf\huge \href{http://ramanujan.math.trinity.edu/wtrench/index.shtml} {William F. Trench} \medskip \\\large Andrew G. Cowles Distinguished Professor Emeritus\\ Department of Mathematics\\ Trinity University \\ San Antonio, Texas, USA\\ \href{mailto:{wtrench@trinity.edu}} {wtrench@trinity.edu} \large \vspace*{.75in} \end{center} \rm \noindent This is a supplement to the author's \href{http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF} {\large Introduction to Real Analysis}. It has been judged to meet the evaluation criteria set by the Editorial Board of the American Institute of Mathematics in connection with the Institute's \href{http://www.aimath.org/textbooks/} {Open Textbook Initiative}. It may be copied, modified, redistributed, translated, and built upon subject to the Creative Commons \href{http://creativecommons.org/licenses/by-nc-sa/3.0/deed.en_G} {Attribution-NonCommercial-ShareAlike 3.0 Unported License}. A complete instructor's solution manual is available by email to \href{mailto:wtrench@trinity.edu} {wtrench@trinity.edu}, subject to verification of the requestor's faculty status. \newpage \rm \section{Foreword} \label{section:foreword} This is a revised version of Section~7.5 of my \emph{Advanced Calculus} (Harper \& Row, 1978). It is a supplement to my textbook \href{http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF} {\emph{Introduction to Real Analysis}}, which is referenced several times here. You should review Section~3.4 (Improper Integrals) of that book before reading this document. \section{Introduction}\label{section:introduction} In Section~7.2 (pp. 462--484) we considered functions of the form $$ F(y)=\int_{a}^{b}f(x,y)\,dx, \quad c \le y \le d. $$ We saw that if $f$ is continuous on $[a,b]\times [c,d]$, then $F$ is continuous on $[c,d]$ (Exercise~7.2.3, p.~481) and that we can reverse the order of integration in $$ \int_{c}^{d}F(y)\,dy=\int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy $$ to evaluate it as $$ \int_{c}^{d}F(y)\,dy=\int_{a}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx $$ (Corollary~7.2.3, p.~466). Here is another important property of $F$. \begin{theorem} \label{theorem:1} If $f$ and $f_{y}$ are continuous on $[a,b]\times [c,d],$ then \begin{equation} \label{eq:1} F(y)=\int_{a}^{b}f(x,y)\,dx, \quad c \le y \le d, \end{equation} is continuously differentiable on $[c,d]$ and $F'(y)$ can be obtained by differentiating \eqref{eq:1} under the integral sign with respect to $y;$ that is, \begin{equation} \label{eq:2} F'(y)=\int_{a}^{b}f_{y}(x,y)\,dx, \quad c \le y \le d. \end{equation} Here $F'(a)$ and $f_{y}(x,a)$ are derivatives from the right and $F'(b)$ and $f_{y}(x,b)$ are derivatives from the left$.$ \end{theorem} \proof If $y$ and $y+\Delta y$ are in $[c,d]$ and $\Delta y\ne0$, then \begin{equation} \label{eq:3} \frac{F(y+\Delta y)-F(y)}{\Delta y}= \int_{a}^{b}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\,dx. \end{equation} From the mean value theorem (Theorem~2.3.11, p.~83), if $x\in[a,b]$ and $y$, $y+\Delta y\in[c,d]$, there is a $y(x)$ between $y$ and $y+\Delta y$ such that $$ f(x,y+\Delta y)-f(x,y)=f_{y}(x,y)\Delta y= f_{y}(x,y(x))\Delta y+(f_{y}(x,y(x)-f_{y}(x,y))\Delta y. $$ From this and \eqref{eq:3}, \begin{equation} \label{eq:4} \left|\frac{F(y+\Delta y)-F(y)}{\Delta y}-\int_{a}^{b}f_{y}(x,y)\,dx\right| \le \int_{a}^{b} |f_{y}(x,y(x))-f_{y}(x,y)|\,dx. \end{equation} Now suppose $\epsilon>0$. Since $f_{y}$ is uniformly continuous on the compact set $[a,b]\times [c,d]$ (Corollary~5.2.14, p.~314) and $y(x)$ is between $y$ and $y+\Delta y$, there is a $\delta>0$ such that if $|\Delta|<\delta$ then $$ |f_{y}(x,y)-f_{y}(x,y(x))|<\epsilon,\quad (x,y)\in[a,b]\times [c,d]. $$ This and \eqref{eq:4} imply that $$ \left|\frac{F(y+\Delta y-F(y))}{\Delta y}-\int_{a}^{b}f_{y}(x,y)\,dx\right|<\epsilon(b-a) $$ if $y$ and $y+\Delta y$ are in $[c,d]$ and $0<|\Delta y|<\delta$. This implies \eqref{eq:2}. Since the integral in \eqref{eq:2} is continuous on $[c,d]$ (Exercise~7.2.3, p.~481, with $f$ replaced by $f_{y}$), $F'$ is continuous on $[c,d]$. \endproof \begin{example} \label{example:1} \rm Since $$ f(x,y)=\cos xy\text{\quad and\quad} f_{y}(x,y)=-x\sin xy $$ are continuous for all $(x,y)$, Theorem~\ref{theorem:1} implies that if \begin{equation} \label{eq:5} F(y)=\int_{0}^{\pi} \cos xy\,dx,\quad -\infty|y|$.) This provides a convenient way to evaluate the integral in \eqref{eq:6}: integrating the right side of \eqref{eq:5} with respect to $x$ yields $$ F(y)=\frac{\sin xy}{y}\bigg|_{x=0}^{\pi}=\frac{\sin\pi y}{y}, \quad y\ne0. $$ Differentiating this and using \eqref{eq:6} yields $$ \int_{0}^{\pi}x\sin xy\,dx =\frac{\sin \pi y}{y^{2}}- \frac{\pi\cos \pi y}{y}, \quad y\ne0. $$ To verify this, use integration by parts. \bbox \end{example} We will study the continuity, differentiability, and integrability of $$ F(y)=\int_{a}^{b}f(x,y)\,dx,\quad y\in S, $$ where $S$ is an interval or a union of intervals, and $F$ is a convergent improper integral for each $y\in S$. If the domain of $f$ is $[a,b)\times S$ where $-\infty0$, there is an $r=r_{0}(y)$ (which also depends on $\epsilon$) such that \begin{equation} \label{eq:8} \left|F(y)-\int_{a}^{r}f(x,y)\,dx\right|= \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad r_{0}(y)\le y0;\\ \end{cases} $$ therefore, $F$ is discontinuous at $y=0$. \end{example} \begin{example} \label{example:3} \rm The function $$ f(x,y)=y^{3}e^{-y^{2}x} $$ is continuous on $[0,\infty)\times (-\infty,\infty)$. Let $$ F(y)=\int_{0}^{\infty}f(x,y)\,dx= \int_{0}^{\infty}y^{3}e^{-y^{2}x}\,dx =y,\quad -\infty0,$ there is an $r_{0}\in[a,b)$ such that \begin{equation} \label{eq:9} |G(r)-G(r_{1})|<\epsilon,\quad r_{0}\le r,r_{1}0$ there is an $r_{0}\in [a,b)$ such that $$ |G(r)-L|<\frac{\epsilon}{2} \text{\quad and\quad} |G(r_{1})-L|<\frac{\epsilon}{2},\quad r_{0}\le r,r_{1}0,$ there is an $r_{0}\in(a,b]$ such that $$ |G(r)-G(r_{1})|<\epsilon,\quad a0,$ there is an $r_{0} \in [a,b)$ such that $$ \left|\int_{a}^{b}f(x,y)\,dx-\int_{a}^{r}f(x,y)\,dx\right| < \epsilon,\quad y\in S, \quad r_{0}\le r0,$ there is an $r_{0} \in [a,b)$ such that \begin{equation} \label{eq:12} \left|\int_{r}^{r_{1}}f(x,y)\,dx\right|< \epsilon, \quad y\in S,\quad r_{0}\le r,r_{1}0$. From Definition~\ref{definition:1}, there is an $r_{0}\in [a,b)$ such that \begin{equation} \label{eq:13} \left|\int_{r}^{b}f(x,y)\,dx\right| <\frac{\epsilon}{2} \text{\, and\,} \left|\int_{r_{1}}^{b}f(x,y)\,dx\right| <\frac{\epsilon}{2} ,\quad y\in S, \quad r_{0}\le r,r_{1}0$ then, for each $y\in S$, there is an $r_{0}(y) \in [a,b)$ such that \begin{equation} \label{eq:15} \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad y\in S,\quad r_{0}(y)\le r< b. \end{equation} For each $y\in S$, choose $r_{1}(y)\ge \max[{r_{0}(y),r_{0}}]$. (Recall \eqref{eq:14}). Then $$ \int_{r}^{b}f(x,y)\,dx = \int_{r}^{r_{1}(y)}f(x,y)\,dx+ \int_{r_{1}(y)}^{b}f(x,y)\,dx, \quad $$ so \eqref{eq:12}, \eqref{eq:15}, and the triangle inequality imply that $$ \left|\int_{r}^{b} f(x,y)\,dx\right|< 2\epsilon, \quad y\in S, \quad r_{0}\le r0$; however, it does not converge uniformly on any neighborhood of $y=0$, since, for any $r>0$, $e^{-r|y|}>\frac{1}{2}$ if $|y|$ is sufficiently small. \end{example} \begin{definition} \label{definition:2} If the improper integral $$ \int_{a}^{b}f(x,y)\,dx=\lim_{r\to a+}\int_{r}^{b}f(x,y)\,dx $$ converges on $S,$ it is said to converge uniformly $($or be uniformly convergent$)$ on $S$ if$,$ for each $\epsilon>0,$ there is an $r_{0} \in (a,b]$ such that $$ \left|\int_{a}^{b}f(x,y)\,dx-\int_{r}^{b}f(x,y)\,dx\right| <\epsilon, \quad y\in S,\quad a0,$ there is an $r_{0}\in (a,b]$ such that $$ \left|\int_{r_{1}}^{r}f(x,y)\,dx\right|< \epsilon,\quad y\in S,\quad a 0$, so we consider the improper integrals $$ F_{1}(y)=\int_{0}^{1}x^{-1/2}e^{-xy}\,dx \text{\quad and\quad} F_{2}(y)=\int_{1}^{\infty}x^{-1/2}e^{-xy}\,dx $$ separately. Moreover, we could just as well define \begin{equation}\label{eq:16} F_{1}(y)=\int_{0}^{c}x^{-1/2}e^{-xy}\,dx \text{\quad and\quad} F_{2}(y)=\int_{c}^{\infty}x^{-1/2}e^{-xy}\,dx, \end{equation} where $c$ is any positive number. Definition~\ref{definition:2} applies to $F_{1}$. If $00$. It does not converge uniformly on $(0,\rho)$, since the change of variable $u=xy$ yields $$ \int_{r}^{r_{1}}x^{-1/2}e^{-xy}\,dx=y^{-1/2} \int_{ry}^{r_{1}y}u^{-1/2}e^{-u}\,du, $$ which, for any fixed $r>0$, can be made arbitrarily large by taking $y$ sufficiently small and $r=1/y$. Therefore we conclude that $F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0.$ Note that that the constant $c$ in \eqref{eq:16} plays no role in this argument. \end{example} \begin{example} \label{example:6} \rm Suppose we take \begin{equation} \label{eq:17} \int_{0}^{\infty}\frac{\sin u}{u}\,du =\frac{\pi}{2} \end{equation} as given (Exercise~\ref{exer:31}{\bf(b)}). Substituting $u=xy$ with $y>0$ yields \begin{equation} \label{eq:18} \int_{0}^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2},\quad y>0. \end{equation} What about uniform convergence? Since $(\sin xy)/x$ is continuous at $x=0$, Definition~\ref{definition:1} and Theorem~\ref{theorem:4} apply here. If $00$, then $$ \int_{r}^{r_{1}}\frac{\sin xy}{x}\,dx=-\frac{1}{y} \left(\frac{\cos xy}{x}\biggr|_{r}^{r_{1}}+ \int_{r}^{r_{1}}\frac{\cos xy}{x^{2}}\,dx\right), \text{\, so\quad} \left|\int_{r}^{r_{1}}\frac{\sin xy}{x}\,dx\right|<\frac{3}{ry}. $$ Therefore \eqref{eq:18} converges uniformly on $[\rho,\infty)$ if $\rho>0$. On the other hand, from \eqref{eq:17}, there is a $\delta>0$ such that $$ \int_{u_{0}}^{\infty}\frac{\sin u}{u}\,du>\frac{\pi}{4}, \quad 0 \le u_{0}<\delta. $$ This and \eqref{eq:18} imply that $$ \int_{r}^{\infty}\frac{\sin xy}{x}\,dx=\int_{yr}^{\infty}\frac{\sin u}{u}\,du >\frac{\pi}{4} $$ for any $r>0$ if $0 0$. \end{example} \section{ Absolutely Uniformly Convergent Improper Integrals}\label{section:absolutely} \begin{definition}{\bf$($Absolute Uniform Convergence I$)$} \label{definition:4} The improper integral $$ \int_{a}^{b}f(x,y)\,dx=\lim_{r\to b-}\int_{a}^{r}f(x,y)\,dx $$ is said to converge absolutely uniformly on $S$ if the improper integral $$ \int_{a}^{b}|f(x,y)|\,dx=\lim_{r\to b-}\int_{a}^{r}|f(x,y)|\,dx $$ converges uniformly on $S$; that is, if, for each $\epsilon>0$, there is an $r_{0}\in [a,b)$ such that $$ \left|\int_{a}^{b}|f(x,y)|\,dx-\int_{a}^{r}|f(x,y)|\,dx\right| <\epsilon, \quad y\in S,\quad r_{0}0$, there is an $r_{0}\in [a,b)$ such that $$ \int_{r}^{r_{1}}|f(x,y)|\,dx<\epsilon,\quad y\in S,\quad r_{0}\le r0$ there is an $r_{0}\in [a,b)$ such that $$ L-\epsilon < \int_{a}^{r}M(x)\,dx \le L,\quad r_{0}p_{0}$ and $r\ge a_{0}$, then \begin{eqnarray*} \int_{r}^{\infty}e^{-px} |g(x,y)|\,dx &=& \int_{r}^{\infty} e^{-(p-p_{0})x}e^{-p_{0}x}|g(x,y)|\,dx\\ &\le& K\int_{r}^{\infty} e^{-(p-p_{0})x}\,dx= \frac{K e^{-(p-p_{0})r}}{p-p_{0}}, \end{eqnarray*} so $\int_{0}^{\infty}e^{-px} g(x,y)\,dx $ converges absolutely on $S$. For example, since $$ |x^{\alpha}\sin xy|0$, Theorem~\ref{theorem:4} implies that $\int_{0}^{\infty}e^{-px}x^{\alpha}\sin xy\,dx$ and $\int_{0}^{\infty}e^{-px}x^{\alpha}\cos xy\,dx$ converge absolutely uniformly on $(-\infty,\infty)$ if $p>0$ and $\alpha~\ge~0$. As a matter of fact, $\int_{0}^{\infty}e^{-px}x^{\alpha}\sin xy\,dx$ converges absolutely on $(-\infty,\infty)$ if $p>0$ and $\alpha>-1$. (Why?) \end{example} \begin{definition}{\bf$($Absolute Uniform Convergence II$)$} \label{definition:5} The improper integral $$ \int_{a}^{b}f(x,y)\,dx=\lim_{r\to a+}\int_{r}^{b}f(x,y)\,dx $$ is said to converge absolutely uniformly on $S$ if the improper integral $$ \int_{a}^{b}|f(x,y)|\,dx=\lim_{r\to a+}\int_{r}^{b}|f(x,y)|\,dx $$ converges uniformly on $S$; that is, if, for each $\epsilon>0$, there is an $r_{0}\in (a,b]$ such that $$ \left|\int_{a}^{b}|f(x,y)|\,dx-\int_{r}^{b}|f(x,y)|\,dx\right| <\epsilon, \quad y\in S, \quad a\beta-1$. To see this, note that if $0-1$ and $$ G(y)=\int_{0}^{1}x^{\alpha}\sin xy \,dx $$ converges absolutely uniformly on $(-\infty,\infty)$ if $\alpha>-2$. \end{example} By recalling Theorem~4.4.15 (p.~246), you can see why we associate Theorems~\ref{theorem:6} and \ref{theorem:7} with Weierstrass. \section{Dirichlet's Tests} \label{section:dirichlet} Weierstrass's test is useful and important, but it has a basic shortcoming: it applies only to absolutely uniformly convergent improper integrals. The next theorem applies in some cases where $\int_{a}^{b}f(x,y)\,dx$ converges uniformly on $S$, but $\int_{a}^{b}|f(x,y)|\,dx$ does not. \begin{theorem} \label{theorem:8} $(${\bf \href{http://www-history.mcs.st-and.ac.uk/Biographies/Dirichlet.html} {Dirichlet}'s Test for Uniform Convergence I}$)$ If $g,$ $g_{x},$ and $h$ are continuous on $[a,b)\times S,$ then $$ \int_{a}^{b}g(x,y)h(x,y)\,dx $$ converges uniformly on $S$ if the following conditions are satisfied$:$ \begin{alist} \item % a $\dst{\lim_{x\to b-}\left\{\sup_{y\in S}|g(x,y)|\right\}=0};$ \item % b There is a constant $M$ such that $$ \sup_{y\in S}\left|\int_{a}^{x}h(u,y)\,du\right|< M, \quad a\le x0$. From assumption {\bf (a)}, there is an $r_{0} \in [a,b)$ such that $|g(x,y)|<\epsilon$ on $S$ if $r_{0}\le x 0. $$ The obvious inequality $$ \left|\frac{\cos xy}{x+y}\right|\le \frac{1}{x+y} $$ is useless here, since $$ \int_{0}^{\infty}\frac{dx}{x+y}=\infty. $$ However, integration by parts yields \begin{eqnarray*} \int_{r}^{r_{1}}\frac{\cos xy}{x+y}\,dx &=& \frac{\sin xy}{y(x+y)}\biggr|_{r}^{r_{1}}+ \int_{r}^{r_{1}}\frac{\sin xy}{y(x+y)^{2}}\,dx\\ &=&\frac{\sin r_{1}y}{y(r_{1}+y)}-\frac{\sin ry}{y(r+y)} +\int_{r}^{r_{1}}\frac{\sin xy}{y(x+y)^{2}}\,dx. \end{eqnarray*} Therefore, if $0< r0$. Now Theorem~\ref{theorem:4} implies that $I(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. \end{example} We leave the proof of the following theorem to you (Exercise~\ref{exer:10}). \begin{theorem} \label{theorem:9} $(${\bf Dirichlet's Test for Uniform Convergence II}$)$ If $g,$ $g_{x},$ and $h$ are continuous on $(a,b]\times S,$ then $$ \int_{a}^{b}g(x,y)h(x,y)\,dx $$ converges uniformly on $S$ if the following conditions are satisfied$:$ \begin{alist} \item % a $\dst{\lim_{x\to a+}\left\{\sup_{y\in S}|g(x,y)|\right\}=0};$ \item % b There is a constant $M$ such that $$ \sup_{y\in S}\left|\int_{x}^{b}h(u,y)\,du\right| \le M, \quad a< x\le b; $$ \item % c $\int_{a}^{b}|g_{x}(x,y)|\,dx$ converges uniformly on $S$. \end{alist} \end{theorem} By recalling Theorems~3.4.10 (p.~163), 4.3.20 (p.~217), and 4.4.16 (p.~248), you can see why we associate Theorems~\ref{theorem:8} and \ref{theorem:9} with Dirichlet. \section{Consequences of uniform convergence}\label{section:consequences} \begin{theorem} \label{theorem:10} If $f=f(x,y)$ is continuous on either $[a,b)\times [c,d]$ or $(a,b]\times [c,d]$ and \begin{equation} \label{eq:23} F(y)=\int_{a}^{b}f(x,y)\,dx \end{equation} converges uniformly on $[c,d],$ then $F$ is continuous on $[c,d].$ Moreover$,$ \begin{equation} \label{eq:24} \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy =\int_{a}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx. \end{equation} \end{theorem} \proof We will assume that $f$ is continuous on $(a,b]\times [c,d]$. You can consider the other case (Exercise~\ref{exer:14}). We will first show that $F$ in \eqref{eq:23} is continuous on $[c,d]$. Since $F$ converges uniformly on $[c,d]$, Definition~\ref{definition:1} (specifically, \eqref{eq:11}) implies that if $\epsilon>0$, there is an $r \in [a,b)$ such that $$ \left|\int_{r}^{b}f(x,y)\,dx\right|< \epsilon, \quad c \le y \le d. $$ Therefore, if $c\le y, y_{0}\le d]$, then \begin{eqnarray*} |F(y)-F(y_{0})|&=& \left|\int_{a}^{b}f(x,y)\,dx-\int_{a}^{b}f(x,y_{0})\,dx\right|\\ &\le&\left|\int_{a}^{r}[f(x,y)-f(x,y_{0})]\,dx\right|+ \left|\int_{r}^{b}f(x,y)\,dx\right|\\ &&+\left|\int_{r}^{b}f(x,y_{0})\,dx\right|, \end{eqnarray*} so \begin{equation}\label{eq:25} |F(y)-F(y_{0})| \le \int_{a}^{r}|f(x,y)-f(x,y_{0})|\,dx +2\epsilon. \end{equation} Since $f$ is uniformly continuous on the compact set $[a,r]\times [c,d]$ (Corollary~5.2.14, p.~314), there is a $\delta>0$ such that $$ |f(x,y)-f(x,y_{0})|<\epsilon $$ if $(x,y)$ and $(x,y_{0})$ are in $[a,r]\times [c,d]$ and $|y-y_{0}|<\delta$. This and \eqref{eq:25} imply that $$ |F(y)-F(y_{0})|<(r-a)\epsilon +2\epsilon<(b-a+2)\epsilon $$ if $y$ and $y_{0}$ are in $[c,d]$ and $|y-y_{0}|<\delta$. Therefore $F$ is continuous on $[c,d]$, so the integral on left side of \eqref{eq:24} exists. Denote \begin{equation} \label{eq:26} I= \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy. \end{equation} We will show that the improper integral on the right side of \eqref{eq:24} converges to $I$. To this end, denote $$ I(r)= \int_{a}^{r}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx. $$ Since we can reverse the order of integration of the continuous function $f$ over the rectangle $[a,r]\times [c,d]$ (Corollary~7.2.2, p.~466), $$ I(r)=\int_{c}^{d}\left(\int_{a}^{r}f(x,y)\,dx\right)\,dy. $$ From this and \eqref{eq:26}, $$ I-I(r)=\int_{c}^{d}\left(\int_{r}^{b}f(x,y)\,dx\right)\,dy. $$ Now suppose $\epsilon>0$. Since $\int_{a}^{b}f(x,y)\,dx$ converges uniformly on $[c,d]$, there is an $r_{0}\in (a,b]$ such that $$ \left|\int_{r}^{b}f(x,y)\,dx\right|<\epsilon, \quad r_{0}0, $$ and the convergence is uniform on $[\rho,\infty)$ if $\rho>0$. Therefore Theorem~\ref{theorem:10} implies that if $00, $$ it follows that $$ \int_{0}^{\infty}\frac{e^{-xy_{1}}-e^{-xy_{2}}}{x}\,dx= \log\frac{y_{2}}{y_{1}}, \quad y_{2} \ge y_{1}>0. $$ \end{example} \begin{example} \label{example:11} \rm From Example~\ref{example:6}, $$ \int_{0}^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2}, \quad y>0, $$ and the convergence is uniform on $[\rho,\infty)$ if $\rho>0$. Therefore, Theorem~\ref{theorem:10} implies that if $00$. Since we have assumed that $\lim_{r\to b-}F_{r}(y_{0})=F(y_{0})$ exists, there is an $r_{0}$ in $(a,b)$ such that $$ |F_{r}(y_{0})-F(y_{0})|<\epsilon,\quad r_{0}0. $$ Since $$ \int_{0}^{r}e^{-yx^{2}}\,dx=\frac{1}{\sqrt{y}} \int_{0}^{r\sqrt{y}} e^{-t^{2}}\,dt, $$ it follows that $$ I(y)=\frac{1}{\sqrt{y}}\int_{0}^{\infty}e^{-t^{2}}\,dt, $$ and the convergence is uniform on $[\rho,\infty)$ if $\rho>0$ (Exercise~\ref{exer:8}{\bf(i)}). To evaluate the last integral, denote $J(\rho)=\int_{0}^{\rho}e^{-t^{2}}\,dt$; then $$ J^{2}(\rho)=\left(\int_{0}^{\rho}e^{-u^{2}}\,du\right) \left(\int_{0}^{\rho}e^{-v^{2}}\,dv\right) =\int_{0}^{\rho}\int_{0}^{\rho}e^{-(u^{2}+v^{2})}\,du\,dv. $$ Transforming to polar coordinates $r=r\cos\theta$, $v=r\sin\theta$ yields $$ J^{2}(\rho)=\int_{0}^{\pi/2}\int_{0}^{\rho} re^{-r^{2}}\,dr\,d\theta =\frac{\pi(1-e^{-\rho^{2}})}{4}, \text{\quad so\quad} J(\rho)=\frac{\sqrt{\pi(1-e^{-\rho^{2}})}}{2}. $$ Therefore $$ \int_{0}^{\infty}e^{-t^{2}}\,dt=\lim_{\rho\to\infty}J(\rho)= \frac{\sqrt{\pi}}{2}\text{\quad and\quad} \int_{0}^{\infty}e^{-yx^{2}}\,dx= \frac{1}{2}\sqrt{\frac{\pi}{y}}, \quad y>0. $$ Differentiating this $n$ times with respect to $y$ yields $$ \int_{0}^{\infty}x^{2n}e^{-yx^{2}}\,dx= \frac{1\cdot3\cdots(2n-1)\sqrt{\pi}}{2^{n}y^{n+1/2}}\quad y>0,\quad n=1,2,3, \dots, $$ where Theorem~\ref{theorem:11} justifies the differentiation for every $n$, since all these integrals converge uniformly on $[\rho,\infty)$ if $\rho>0$ (Exercise~\ref{exer:8}(i)). \end{example} Some advice for applying this theorem: Be sure to check first that $F(y_{0})=\int_{a}^{b}f(x,y_{0})\,dx$ converges for at least one value of $y$. If so, differentiate $\int_{a}^{b}f(x,y)\,dx$ formally to obtain $\int_{a}^{b}f_{y}(x,y)\,dx$. Then $F'(y)=\int_{a}^{b}f_{y}(x,y)\,dx$ if $y$ is in some interval on which this improper integral converges uniformly. \place % \section{Applications to Laplace transforms} \label{section:laplace} \medskip The \href{http://www-history.mcs.st-and.ac.uk/Biographies/Laplace.html} {\emph{Laplace}} \emph{transform} of a function $f$ locally integrable on $[0,\infty)$ is $$ F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx $$ for all $s$ such that integral converges. Laplace transforms are widely applied in mathematics, particularly in solving differential equations. We leave it to you to prove the following theorem (Exercise~\ref{exer:26}). \begin{theorem} \label{theorem:12} Suppose $f$ is locally integrable on $[0,\infty)$ and $|f(x)|\le M e^{s_{0}x}$ for sufficiently large $x$. Then the Laplace transform of $F$ converges uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}$. \end{theorem} \begin{theorem} \label{theorem:13} If $f$ is continuous on $[0,\infty)$ and $H(x)=\int_{0}^{\infty}e^{-s_{0}u}f(u)\,du$ is bounded on $[0,\infty),$ then the Laplace transform of $f$ converges uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}.$ \end{theorem} \proof If $0\le r\le r_{1}$, $$ \int_{r}^{r_{1}}e^{-sx}f(x)\,dx =\int_{r}^{r_{1}}e^{-(s-s_{0})x}e^{-s_{0}x}f(x)\,dt =\int_{r}^{r_{1}}e^{-(s-s_{0})t}H'(x)\,dt. $$ Integration by parts yields $$ \int_{r}^{r_{1}}e^{-sx}f(x)\,dt=e^{-(s-s_{0})x}H(x)\biggr|_{r}^{r_{1}} +(s-s_{0})\int_{r}^{r_{1}}e^{-(s-s_{0})x} H(x)\,dx. $$ Therefore, if $|H(x)|\le M$, then \begin{eqnarray*} \left|\int_{r}^{r_{1}}e^{-sx}f(x)\,dx\right|&\le& M\left|e^{-(s-s_{0})r_{1}} +e^{-(s-s_{0})r} +(s-s_{0})\int_{r}^{r_{1}}e^{-(s-s_{0})x}\,dx\right|\\ &\le &3Me^{-(s-s_{0})r}\le 3Me^{-(s_{1}-s_{0})r},\quad s\ge s_{1}. \end{eqnarray*} Now Theorem~\ref{theorem:4} implies that $F(s)$ converges uniformly on $[s_{1},\infty)$. The following theorem draws a considerably stonger conclusion from the same assumptions. \begin{theorem} \label{theorem:14} If $f$ is continuous on $[0,\infty)$ and $$ H(x)=\int_{0}^{x}e^{-s_{0}u}f(u)\,du $$ is bounded on $[0,\infty),$ then the Laplace transform of $f$ is infinitely differentiable on $(s_{0},\infty),$ with \begin{equation} \label{eq:30} F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty} e^{-sx} x^{n}f(x)\,dx; \end{equation} that is, the $n$-th derivative of the Laplace transform of $f(x)$ is the Laplace transform of $(-1)^{n}x^{n}f(x)$. \end{theorem} \proof First we will show that the integrals $$ I_{n}(s)=\int_{0}^{\infty}e^{-sx}x^{n}f(x)\,dx,\quad n=0,1,2, \dots $$ all converge uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}$. If $0s_{0}$ (check!), $$ \left|\int_{r}^{r_{1}}e^{-sx}x^{n}f(x)\,dx\right|<3Me^{-(s-s_{0})r}r^{n},\quad ns_{0}$. Now Theorem~\ref{theorem:11} implies that $F_{n+1}=-F_{n}'$, and an easy induction proof yields \eqref{eq:30} (Exercise~\ref{exer:25}). \endproof \begin{example} \label{example:13} \rm Here we apply Theorem~\ref{theorem:12} with $f(x)=\cos ax$ ($a\ne0$) and $s_{0}=0$. Since $$ \int_{0}^{x}\cos au\,du=\frac{\sin ax}{a} $$ is bounded on $(0,\infty)$, Theorem~\ref{theorem:12} implies that $$ F(s)=\int_{0}^{\infty}e^{-sx}\cos ax\,dx $$ converges and \begin{equation} \label{eq:31} F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty}e^{-sx}x^{n}\cos ax\,dx, \quad s>0. \end{equation} (Note that this is also true if $a=0$.) Elementary integration yields $$ F(s)=\frac{s}{s^{2}+a^{2}}. $$ Hence, from \eqref{eq:31}, $$ \int_{0}^{\infty}e^{-sx}x^{n}\cos ax=(-1)^{n}\frac{d^n}{ds^n} \frac{s}{s^{2}+a^{2}}, \quad n=0,1, \dots. $$ \end{example} \newpage \section{Exercises} \begin{exerciselist} \item\label{exer:1} Suppose $g$ and $h$ are differentiable on $[a,b]$, with $$ a \le g(y) \le b \text{\quad and\quad} a \le h(y) \le b, \quad c \le y \le d. $$ Let $f$ and $f_{y}$ be continuous on $[a,b]\times [c,d]$. Derive \emph{Liebniz's rule}: \begin{eqnarray*} \frac{d}{dy}\int_{g(y)}^{h(y)}f(x,y)\,dx &=&f(h(y),y)h'(y)-f(g(y),y)g'(y)\\&&+\int_{g(y)}^{h(y)}f_{y}(x,y)\,dx. \end{eqnarray*} (Hint: Define $H(y,u,v)=\int_{u}^{v}f(x,y)\,dx$ and use the chain rule.) \item\label{exer:2} Adapt the proof of Theorem~\ref{theorem:2} to prove Theorem~\ref{theorem:3}. \item\label{exer:3} Adapt the proof of Theorem~\ref{theorem:4} to prove Theorem~\ref{theorem:5}. \item\label{exer:4} Show that Definition~\ref{definition:3} is independent of $c$; that is, if $\int_{a}^{c}f(x,y)\,dx$ and $\int_{c}^{b}f(x,y)\,dx$ both converge uniformly on $S$ for some $c\in (a,b)$, then they both converge uniformly on $S$ and every $c\in (a,b)$. \item\label{exer:5} \begin{alist} \item % a Show that if $f$ is bounded on $[a,b]\times [c,d]$ and $\int_{a}^{b}f(x,y)\,dx$ exists as a proper integral for each $y\in [c,d]$, then it converges uniformly on $[c,d]$ according to all of Definition~\ref{definition:1}--\ref{definition:3}. \item % b Give an example to show that the boundedness of $f$ is essential in {\bf(a)}. \end{alist} \item\label{exer:6} Working directly from Definition~\ref{definition:1}, discuss uniform convergence of the following integrals: \begin{tabular}{ll} {\bf(a)} $\dst{\int_{0}^{\infty}\frac{dx}{1+y^{2}x^{2}}\,dx}$ & {\bf(b)} $\dst{\int_{0}^{\infty}e^{-xy}x^{2}\,dx}$ \\ \\ {\bf(c)} $\dst{\int_{0}^{\infty}x^{2n}e^{-yx^{2}}\,dx}$ & {\bf(d)} $\dst{\int_{0}^{\infty}\sin xy^{2}\,dx}$ \\\\ {\bf(e)} $\dst{\int_{0}^{\infty}(3y^{2}-2xy)e^{-y^{2}x}\,dx}$ & {\bf(f)} $\dst{\int_{0}^{\infty}(2xy-y^{2}x^{2})e^{-xy}\,dx}$ \end{tabular} \item\label{exer:7} Adapt the proof of Theorem~\ref{theorem:6} to prove Theorem~\ref{theorem:7}. \item\label{exer:8} Use Weierstrass's test to show that the integral converges uniformly on $S:$ \begin{alist} \item % a $\dst{\int_{0}^{\infty}e^{-xy}\sin x\,dx}$,\quad $S=[\rho,\infty)$,\quad $\rho>0$ \item % b $\dst{\int_{0}^{\infty}\dst{\frac{\sin x}{x^{y}}}\,dx}$,\quad $S=[c,d]$, \quad $10$,\quad $S=(-\infty,\infty)$ \item % d $\dst{\int_{0}^{1}\frac{e^{xy}}{(1-x)^{y}}}\,dx$, \quad $S=(-\infty,b)$,\quad $b<1$ \item % e $\dst{\int_{-\infty}^{\infty}\frac{\cos xy}{1+x^{2}y^{2}}}\,dx$,\quad $S=(-\infty,-\rho]\cup[\rho,\infty)$,\quad $\rho>0$. \item % f $\dst{\int_{1}^{\infty}e^{-x/y}\,dx}$,\quad $S=[\rho,\infty)$,\quad $\rho>0$ \item % g $\dst{\int_{-\infty}^{\infty}e^{xy}e^{-x^{2}}\,dx}$,\quad $S=[-\rho,\rho]$,\quad $\rho>0$ \item % h $\dst{\int_{0}^{\infty}\frac{\cos xy-\cos ax}{x^{2}}\,dx}$,\quad $S=(-\infty,\infty)$ \item % i $\dst{\int_{0}^{\infty}x^{2n}e^{-yx^{2}}\,dx}$,\quad $S=[\rho,\infty)$,\quad $\rho>0$, \quad $n=0$, $1$, $2$,\dots \end{alist} \item\label{exer:9} \begin{alist} \item % a Show that $$ \Gamma(y)=\int_{0}^{\infty} x^{y-1}e^{-x}\,dx $$ converges if $y>0$, and uniformly on $[c,d]$ if $00, \quad n=1,2,3, \dots. $$ How can this be used to define $\Gamma(y)$ in a natural way for all $y\ne0$, $-1$, $-2$, \dots? (This function is called the \emph{gamma function}.) \item % c Show that $\Gamma(n+1)=n!$ if $n$ is a positive integer. \item % d Show that $$ \int_{0}^{\infty}e^{-st}t^{\alpha}\,dt =s^{-\alpha-1}\Gamma(\alpha+1), \quad \alpha>-1, \quad s>0. $$ \end{alist} \item\label{exer:10} Show that Theorem~\ref{theorem:8} remains valid with assumption {\bf(c)} replaced by the assumption that $|g_{x}(x,y)|$ is monotonic with respect to $x$ for all $y\in S$. \item\label{exer:11} Adapt the proof of Theorem~\ref{theorem:8} to prove Theorem~\ref{theorem:9}. \item\label{exer:12} Use Dirichlet's test to show that the following integrals converge uniformly on $S=[\rho,\infty)$ if $\rho>0$: \begin{tabular}{ll} {\bf(a)} $\dst{\int_{1}^{\infty}\frac{\sin xy}{x^{y}}\,dx}$& {\bf(b)} $\dst{\int_{2}^{\infty}\frac{\sin xy}{\log x}\,dx}$\\\\ {\bf(c)} $\dst{\int_{0}^{\infty}\frac{\cos xy}{x+y^{2}}\,dx}$& {\bf(d)} $\dst{\int_{1}^{\infty}\frac{\sin xy}{1+xy}\,dx}$ \end{tabular} \item\label{exer:13} Suppose $g,$ $g_{x}$ and $h$ are continuous on $[a,b)\times S,$ and denote $H(x,y)=\int_{a}^{x}h(u,y)\,du,$ $a\le x0$ \item % b $F(y)=\dst{\int_{0}^{\infty}x^{y}\,dx}$, $y>-1$;\quad $I=\dst{\int_{0}^{\infty}\frac{x^{a}-x^{b}}{\log x}\,dx}$, \quad $a$, $b>-1$ \item % c $F(y)=\dst{\int_{0}^{\infty}e^{-xy}\cos x\,dx}$,\quad $y>0$ $I=\dst{\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\cos x\,dx}$,\quad $a$, $b>0$ \item % d $F(y)=\dst{\int_{0}^{\infty}e^{-xy}\sin x\,dx}$, \quad $y>0$ $I=\dst{\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\sin x\,dx}$, \quad $a$, $b>0$ \item % e $F(y)=\dst{\int_{0}^{\infty}e^{-x}\sin xy\,dx}$;\, $I=\dst{\int_{0}^{\infty}e^{-x}\dst\frac{1-\cos ax}{x}}\,dx$ \item % f $F(y)=\dst{\int_{0}^{\infty}e^{-x}\cos xy\,dx}$;\, $I=\dst{\int_{0}^{\infty}e^{-x}\dst\frac{\sin ax}{x}}\,dx$ \end{alist} \item\label{exer:19} Use Theorem~\ref{theorem:11} to evaluate: \begin{alist} \item % a $\dst{\int_{0}^{1}(\log x)^{n}x^{y}\,dx}$, \quad $y>-1$,\quad $n=0$, $1$, $2$,\dots . \item % b $\dst{\int_{0}^{\infty}\dst{\frac{dx}{(x^{2}+y)^{n+1}}}\,dx}$,\quad $y>0$,\quad $n=0$, $1$, $2$, \dots. \item % c $\dst{\int_{0}^{\infty}x^{2n+1}e^{-yx^{2}}\,dx}$, \quad $y>0$, \quad $n=0$, $1$, $2$,\dots. \item % e $\dst{\int_{0}^{\infty}xy^{x}\,dx}$, \quad $00. $$ \item\label{exer:30} Suppose $f$ is continuous on $[0,\infty)$ and $$ F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx $$ converges for $s = s_{0}$. Show that $\lim_{s\to\infty}F(s)=0$. (Hint: Integrate by parts.) \item\label{exer:31} \begin{alist} \item % a Starting from the result of Exercise~\ref{exer:18}{\bf(d)}, let $b\to\infty$ and invoke Exercise~\ref{exer:30} to evaluate $$ \int_{0}^{\infty}e^{-ax} \frac{\sin x}{x}\,dx, \quad a>0. $$ \item % b Use {\bf(a)} and Exercise~\ref{exer:28} to show that $$ \int_{0}^{\infty} \frac{\sin x}{x}\,dx =\frac{\pi}{2}. $$ \end{alist} \item\label{exer:32} \begin{alist} \item % a Suppose $f$ is continuously differentiable on $[0,\infty)$ and $$ |f(x)| \le Me^{s_{0}x}, \quad 0\le x\le \infty. $$ Show that $$ G(s)=\int_{0}^{\infty} e^{-sx}f'(x)\,dx $$ converges uniformly on $[s_{1},\infty)$ if $s_{1}>s_{0}$. (Hint: Integrate by parts.) \item % b Show from part {\bf(a)} that $$ G(s)=\int_{0}^{\infty} e^{-sx}xe^{x^{2}}\sin e^{x^{2}}\,dx $$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. (Notice that this does not follow from Theorem~\ref{theorem:6} or \ref{theorem:8}.) \end{alist} \item\label{exer:33} Suppose $f$ is continuous on $[0,\infty)$, $$ \lim_{x\to0+}\frac{f(x)}{x} $$ exists, and $$ F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx $$ converges for $s=s_{0}$. Show that $$ \int_{s_{0}}^{\infty}F(u)\,du=\int_{0}^{\infty}e^{-s_{0}x}\frac{f(x)}{x}\,dx. $$ \end{exerciselist} \newpage \bigskip \section{Answers to selected exercises}\label{section:answers} \bigskip \noindent {\bf\ref{exer:5}. (b)} If $f(x,y)=1/y$ for $y\ne0$ and $f(x,0)=1$, then $\int_{a}^{b}f(x,y)\,dx$ does not converge uniformly on $[0,d]$ for any $d>0$. \bigskip \noindent {\bf\ref{exer:6}.} {\bf(a)}, {\bf(d)}, and {\bf(e)} converge uniformly on $(-\infty,\rho]\cup[\rho,\infty)$ if $\rho>0$;\, {\bf(b)}, {\bf(c)}, and {\bf(f)} converge uniformly on $[\rho,\infty)$ if $\rho>0$. \bigskip \noindent {\bf\ref{exer:17}.} Let $C(y)=\dst{\int_{1}^{\infty}\frac{\cos xy}{x}\,dx}$ and $S(y)=\dst{\int_{1}^{\infty}\frac{\sin xy}{x}\,dx}$. Then $C(0)=\infty$ and $S(0)=0$, while $S(y)=\pi/2$ if $y\ne0$. \bigskip \noindent {\bf\ref{exer:18}.} {\bf(a)} $F(y)=\dst{\frac{\pi}{2|y|}}$;\quad $I=\dst{\frac{\pi}{2}\log\frac{a}{b}}$ \quad {\bf(b)} $F(y)=\dst{\frac{1}{y+1}}$;\quad $I=\dst{\log\frac{a+1}{b+1}}$ \bigskip {\bf(c)} $F(y)=\dst{\frac{y}{y^{2}+1}}$;\quad $I=\dst{\frac{1}{2}\,\frac{b^{2}+1}{a^{2}+1}}$ {\bf(d)} $F(y)=\dst{\frac{1}{y^{2}+1}}$;\quad $I=\tan^{-1}b-\tan^{-1}a$ {\bf(e)} $F(y)=\dst{\frac{y}{y^{2}+1}}$;\quad $I=\dst{\frac{1}{2}}\log(1+a^{2})$ {\bf(f)} $F(y)=\dst{\frac{1}{y^{2}+1}}$;\quad $I=\tan^{-1}a$ \bigskip \noindent {\bf\ref{exer:19}.} {\bf(a)} $(-1)^{n}n!(y+1)^{-n-1}$ \quad {\bf(b)} $\pi2^{-2n-1}\dst{\binom{2n}{n}}y^{-n-1/2}$ {\bf(c)} $\dst{\frac{n!}{2y^{n+1}}}$ $(\log y)^{-2}$ {\bf(d)} $\dst{\frac{1}{(\log x)^{2}}}$ \noindent {\bf\ref{exer:22}.} $\dst{\int_{-\infty}^{\infty}|x^{n}f(x)|\,dx<\infty}$ \bigskip \noindent {\bf\ref{exer:24}.} No; the integral defining $F$ diverges for all $y$. \bigskip \noindent {\bf\ref{exer:31}.} {\bf(a)}\, $\dst{\frac{\pi}{2}}-\tan^{-1}a$ \end{document} \newpage \setlength{\parindent}{0pt} \centerline{\large Beginning of manual} {\bf 1.} If $H(y,u,v)=\dst{\int_{u}^{v}f(x,y)\,dx}$ then $$ H_{u}(y,u,v)=-f(u,y), \quad H_{v}(y,u,v)=f(v,y), $$ and, by Theorem~1, $H_{y}(u,v,y) =\dst{\int_{u}^{v}f_{y}(x,y)\,dx}$. If $$ F(y)=H(y,g(y),h(y))=\int_{g(y)}^{h(y)}f(x,y)\,dx, $$ then \begin{eqnarray*} F'(y)&=&H_{v}(y, g(y),h(y))h'(y)+H_{u}(y,g(y),h(y))g'(y)+ H_{y}(y,g(y),h(y))\\ &=& f(h(y),y)h'(y)-f(g(y),y)g'(y) +\int_{g(y)}^{h(y)} f_{y}(x,y)\,dx. \end{eqnarray*} \medskip {\bf 2.} {\bf Theorem 3 (Cauchy Criterion for Convergence of an Improper Integral II)} \it Suppose $g$ is integrable on every finite closed subinterval of $(a,b]$ and denote $$ G(r)=\int_{r}^{b}g(x)\,dx,\quad a< r\le b. $$ Then the improper integral $\int_{a}^{b}g(x)\,dx$ converges if and only if$,$ for each $\epsilon >0,$ there is an $r_{0}\in(a,b]$ such that \begin{equation}\tag{A} |G(r)-G(r_{1})|\le\epsilon,\quad a0$ there is an $r_{0}\in (a,b]$ such that $$ |G(r)-L|<\frac{\epsilon}{2} \text{\quad and\quad} |G(r_{1})-L|<\frac{\epsilon}{2}, \quad a< r,r_{1}\le r_{0}. $$ Therefore, \begin{eqnarray*} |G(r)-G(r_{1})|&=&|(G(r)-L)-(G(r_{1})-L)|\\ &\le& |G(r)-L|+|G(r_{1})-L|\le \epsilon,\quad a< r,r_{1}\le r_{0}. \end{eqnarray*} For sufficiency, (A) implies that $$ |G(r)|= |G(r_{1})+(G(r)-G(r_{1}))|\le |G(r_{1})|+|G(r)-G(r_{1})|\le |G(r_{1})|+\epsilon, $$ $a< r_{1}\le r_{0}$. Since $G$ is also bounded on the compact set $[r_{0},b]$ (Theorem~5.2.11, p.~313), $G$ is bounded on $(a,b]$. Therefore the monotonic functions $$ \overline{G}(r)=\sup\set{G(r_{1})}{a0,$ there is an $r_{0}\in (a,b]$ such that \begin{equation}\tag{A} \left|\int_{r_{1}}^{r}f(x,y)\,dx\right|< \epsilon, \quad y\in S, \quad a 0$. From Definition~2, there is an $r_{0}\in (a,b]$ such that \begin{equation} \tag{B} \left|\int_{a}^{r}f(x,y)\,dx\right| <\frac{\epsilon}{2} \text{\quad and\quad} \left|\int_{a}^{r_{1}}f(x,y)\,dx\right| <\frac{\epsilon}{2},\, y\in S, \, a< r,r_{1}\le r_{0}. \end{equation} Since $$ \int_{r_{1}}^{r}f(x,y)\,dx= \int_{r_{1}}^{b}f(x,y)\,dx- \int_{r}^{b}f(x,y)\,dx $$ (B) and the triangle inequality imply (A). For the converse, denote $$ F(y)=\int_{r}^{b}f(x,y)\,dx. $$ Since (A) implies that $$ |F(r,y)-F(r_{1},y)|\le \epsilon, \quad y\in S, \quad a< r, r_{1}\le r_{0}, $$ Theorem~2 with $G(r)=F(r,y)$ ($y$ fixed but arbitrary in $S$) implies that $\int_{a}^{b} f(x,y)\,dx$ converges pointwise for $y\in S$. Therefore, if $\epsilon>0$ then, for each $y\in S$, there is an $r_{0}(y) \in (a,b]$ such that \begin{equation} \tag{C} \left|\int_{a}^{r}f(x,y)\,dx\right|\le \epsilon, \quad y\in S,\quad a0$ there are points $r_{0}$ and $s_{0}$ in $(a,b)$ such that $$ \left|\int_{r}^{r_{1}}f(x,y)\,dx\right|\le \epsilon,\quad y\in S,\quad r_{0}\le r,r_{1}0$ and $\epsilon >0$, choose $r$ so that $\dst{\int_{r}^{\infty}\frac{du}{1+u^{2}}}< \rho\epsilon$. Then $\dst{\frac{1}{|y|}\int_{r}^{\infty}\frac{du}{1+u^{2}}}<\epsilon$ if $|y|\ge \rho$, so $I(y)$ converges uniformly on $(-\infty, \rho]\bigcup [\rho,\infty)$ if $\rho>0$. \medskip {\bf(b)} $I(y)=\infty$ if $y\le0$. If $y>0$ let $u=xy$; then $I(y)=\dst{\frac{1}{y^{3}}\int_{0}^{\infty}e^{-u}u^{2}\,du}$. If $\rho>0$ and $\epsilon >0$, choose $r$ so that $\dst{\int_{r}^{\infty}e^{-u}u^{2}\,du}<\rho^{3}\epsilon$. Then $\dst{\frac{1}{y^{3}}\int_{r}^{\infty}e^{-u}u^{2}\,du}<\epsilon$ if $y\ge \rho$, so $I(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. \medskip {\bf(c)} $I(y)=\infty$ if $y\le0$. If $y>0$ let $u=xy^{1/2}$; then $I(y)=\dst{y^{-n-1/2}\int_{0}^{\infty}u^{2n}e^{-u}\,du}$. If $\rho>0$ and $\epsilon >0$, we can choose $r$ so that $\dst{\int_{r}^{\infty}u^{2n}e^{-u}\,du}<\epsilon \rho^{n+1/2}$. Then $y^{-n-1/2}\dst{\int_{r}^{\infty}u^{2n}e^{-u}\,du}<\epsilon$ if $y\ge \rho$, so $I(y)$ converges uniformly on $S=[\rho,\infty)$ if $\rho>0$. \medskip {\bf(d)} Since $I(-y)=-I(y)$, it suffices to assume that $y>0$. If $u=yx^{2}$ then $I(y)=\dst{\frac{1}{2\sqrt{y}}\int_{0}^{\infty}\frac{\sin u\,du}{\sqrt{u}}}$. From Example 3.4.14 (p.~162), this integral converges conditionally. If $\rho>0$ and $\epsilon >0$, we can choose $r$ so that $\dst{\left|\int_{r}^{\infty}\frac{\sin u\,du}{\sqrt{u}}\right|}<2\epsilon\sqrt{\rho}$, so $I(y)$ converges uniformly on $(-\infty,-\rho]\bigcup[\rho,\infty)$ if $\rho>0.$ \medskip {\bf(e)} If $u=y^{2}x$ then $\dst{I(y)=3\int_{0}^{\infty}e^{-u}\,du -\frac{2}{y^{3}}\int_{0}^{\infty} ue^{-u}\,du}$. If $\rho>0$, we can choose $r$ so that $\dst{3\int_{r}^{\infty}e^{-u}\,du<\frac{\epsilon}{2}}$ and $\dst{\int_{r}^{\rho}ue^{-u}\,du<\frac{\rho^{3}\epsilon}{2}}$. Then $$ \left|3\int_{r}^{\infty}e^{-u}\,du -\frac{3}{y^{3}}\int_{r}^{\infty} ue^{-u}\,du\right|<\epsilon \text{\quad if\quad} |y|\ge \rho, $$ so $I(y)$ converges uniformly on $(-\infty, \rho]\bigcup [\rho,\infty)$ if $\rho>0$. \medskip {\bf(f)} $I(y)=-\infty$ if $y\le0$. If $y>0$, let $u=xy$; then $I(y)=\dst{\frac{1}{y}\int_{0}^{\infty}(2u-u^{2})e^{-u}\,du}$. If $\rho>0$ and $\epsilon >0$, we can choose $r$ so that $\dst{\int_{r}^{\infty}|2u-u^{2}|e^{-u}\,du}<\epsilon \rho$, so $I(y)$ converges uniformly on $S=[\rho,\infty)$ if $\rho>0$. \medskip {\bf 7.} {\bf Theorem~7 $($Weierstrass's Test for Absolute Uniform Convergence II$)$} Suppose $f=f(x,y)$ is locally integrable $(a,b]$ and, for some $b_{0}\in (a,b],$ \begin{equation}\tag{A} |f(x,y)| \le M(x), \: y\in S, \: x\in (a,b_{0}], \end{equation} where $$ \int_{a}^{b_{0}}M(x)\,dx<\infty. $$ Then $\int_{a}^{b}f(x,y)\,dx$ converges absolutely uniformly on $S.$ \proof Denote $\int_{a}^{b_{0}}M(x)\,dx=L<\infty$. By definition, for each $\epsilon>0$ there is an $r_{0}\in (a,b_{0}]$ such that $$ L-\epsilon \le \int_{r}^{b_{0}}M(x)\,dx \le L,\quad a1$, $$ \frac{|\sin x|}{x^{y}}\le x^{-c} \text{\quad and\quad} \int_{1}^{\infty}x^{-c} \,dx=\frac{1}{c-1}\text{\quad if \quad} $$ $I_{2}(y)$ converges absolutely uniformly on $S$. \medskip {\bf (c)} If $x\ge 1$ then $\dst{e^{-px}\left|\frac{\sin xy}{x}\right|\le e^{-px}}$ for all $y$ and $\dst{\int_{1}^{\infty}e^{-px}\,dx<\infty}$, since $p>0$. \medskip {\bf(d)} $\dst{\frac{e^{xy}}{(1-x)^{y}}}\le \dst{\frac{e^{b}}{(1-x)^{b}}}$, if $0\le x<1$ and $ y\le b$, and $\dst{\int_{0}^{1}(1-x)^{-b}\,dx}<\infty$ if $b<1$. \medskip {\bf(e)} If $|y|\ge \rho>0$ then $\dst{\left|\frac{\cos xy}{1+x^{2}y^{2}}\right|\le \frac{1}{1+\rho^{2}x^{2}}}$ for all $x$, and $\dst{\int_{0}^{\infty}\frac{dx}{1+\rho^{2}x^{2}}}<\infty$. \medskip {\bf(f)} If $y\ge \rho>0$ then $e^{-x/y}\le e^{-x/\rho}$ and $\dst{\int_{0}^{\infty}e^{-x/\rho}\,dx<\infty}$. \medskip {\bf(g)} If $|y|\le \rho$ then $e^{xy}e^{-x^{2}}\le e^{x\rho}e^{-x^{2}}$ and $\dst{\int_{-\infty}^{\infty}e^{x\rho}e^{-x^{2}}\,dx}<\infty$. \medskip {\bf(h)} If $|x|\ge 1$ then $|\cos xy-\cos ax|\le 2$ and $\dst{\int_{1}^{\infty}\frac{2\,dx}{x^{2}}}<\infty$. \medskip \noindent {\bf(i)} If $y\ge \rho>0$ then $e^{-yx^{2}}\le e^{-\rho x^{2}}$ and $\dst{\int_{0}^{\infty} x^{2n}e^{-\rho x^{2}}\,dx<\infty}$. \medskip {\bf 9. (a)} If $00$, $\int_{0}^{1}x^{y-1}e^{-x}\,dx$ converges uniformly on $[c,\infty)$ if $c>0$, by Theorem~7. If $x>1$ then $|x^{y-1}e^{-x}|\le x^{d-1}e^{-x}$ if $y\le d$. Therefore, since $\dst{\int_{1}^{\infty}x^{d-1}e^{-x}}\,dx<\infty$, $\dst{\int_{1}^{\infty}x^{y-1}e^{-x}}\,dx$ converges uniformly on $(-\infty,d]$ for every $d$, by Theorem~6. Hence, $\dst{\int_{0}^{\infty}x^{y-1}e^{-x}\,dx}$ converges uniformly on $[c,d]$ if $c>0$. \medskip {\bf (b)} If $y>0$ then \begin{equation} \Gamma(y)=\int_{0}^{\infty}x^{y-1}e^{-x}\,dx =\frac{x^{y}e^{-x}}{y}\biggr|_{0}^{\infty}+ \frac{1}{y}\int_{0}^{\infty}x^{y}e^{-x}\,dx =\frac{\Gamma(y+1)}{y}. \tag{A} \end{equation} Therefore \begin{equation} \Gamma(y)=\frac{\Gamma(y+n)}{y(y+1)\cdots (y+n-1)} \tag{B} \end{equation} is true when $n=1$. Now suppose it is true for given positive integer $n$, and replace $y$ by $y+n$ in (A): $$ \Gamma(y+n)=\frac{\Gamma(y+n+1)}{y+n}. $$ Substituting this into (B) yields $$ \Gamma(y)=\frac{\Gamma(y+n+1)}{y(y+1)\cdots (y+n)}, $$ which completes the induction. If $-n 0$ there is an $r_{0}\in [a,b)$ such that $$ |g(s,y)|\le \epsilon,\quad y\in S,\quad r_{0}\le s0$. From assumption {\bf (a)}, there is an $r_{0} \in [a,b)$ such that $|g(x,y)|<\epsilon$ on $S$ if $a< x \le r_{0} \le b$. From assumption {\bf(c)} and Theorem~5, there is an $s_{0}\in (a,b]$ such that $$ \int_{r_{1}}^{r}|g_{x}(x,y)|\,dx<\epsilon,\quad y\in S,\quad a0. $$ Now Theorem~4 implies that $F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. \medskip {\bf (b)} Denote $F(y)=\dst{\int_{2}^{\infty}\frac{\sin xy}{\log x}\,dx}$ and, with $2\le r< r_{1}$, $$ F(r,r_{1},y)=\int_{r}^{r_{1}}\frac{\sin xy}{\log x}\,dx =-\frac{\cos xy}{y\log x}\biggr|_{r}^{r_{1}}- \frac{1}{y}\int_{r}^{r_{1}}\frac{\cos xy}{x(\log x)^{2}}\,dx. $$ Therefore $$ |F(r,r_{1},y)|\le \frac{1}{y}\left|\frac{2}{\log r}+\int_{r}^{r_{1}} \frac{dx}{x(\log x)^{2}}\right|\le \frac{3}{y\log r}. $$ Now Theorem~4 implies that $F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. \medskip {\bf (c)} Denote $F(y)=\dst{\int_{0}^{\infty}\frac{\cos xy}{x+y^{2}}}$, and, with $00$. \medskip {\bf (d)} Denote $F(y)=\dst{\int_{0}^{\infty}\frac{\sin xy}{1+xy}}$, and, with $00$. \medskip {\bf 13.} Integration by parts yields \begin{eqnarray*} \int_{r}^{r_{1}}g(x,y)h(x,y)\,dx&=&\int_{r}^{r_{1}}g(x,y)H_{x}(x,y)\,dx\\ &=&g(r_{1},y)H(r_{1},y)-g(r,y)H(r,y)\\ &&-\int_{r}^{r_{1}}g_{x}(x,y)H(x,y)\,dx, \end{eqnarray*} so $$ \left|\int_{r}^{r_{1}}g(x,y)h(x,y)\,dx\right|\le 2\sup_{x\ge r}\left\{\left\{\sup_{y\in S}|g(x,y)H(x,y)|\right\}\right\}+ \left|\int_{r}^{r_{1}}g_{x}(x,y)H(x,y)\,dx\right|. $$ Now suppose $\epsilon\ge 0$. From our first assmption, there is an $s_{0}\in [a,b)$ such that $$ \sup_{x\ge r}\left\{\left\{\sup_{y\in S}|g(x,y)H(x,y)|\right\}\right\}<\epsilon, \quad s_{0}\le r0$, there is an $r \in (a,b]$ such that $$ \left|\int_{a}^{r}f(x,y)\,dx\right|\le \epsilon, \quad c \le y \le d. $$ Therefore, if $y$ and $y_{0}$ are in $[c,d]$, then \begin{eqnarray*} |F(y)-F(y_{0})|&=& \left|\int_{a}^{b}f(x,y)\,dx-\int_{a}^{b}f(x,y_{0})\,dx\right|\\ &\le&\left|\int_{r}^{b}[f(x,y)-f(x,y_{0})]\,dx\right|+ \left|\int_{a}^{r}f(x,y)\,dx\right|\\ &&+\left|\int_{a}^{r}f(x,y_{0})\,dx\right|\\ \end{eqnarray*} so \begin{equation}\tag{C} |F(y)-F(y_{0})| \le \int_{r}^{b}|f(x,y)-f(x,y_{0})|\,dx +2\epsilon. \end{equation} Since $f$ is uniformly continuous on the compact set $[r,b]\times [c,d]$ (Corollary~5.2.14, p.~314), there is a $\delta>0$ such that $$ |f(x,y)-f(x,y_{0})|<\epsilon $$ if $(x,y)$ and $(x,y_{0})$ are in $[r,b]\times [c,d]$ and $|y-y_{0}|<\delta$. This and (C) imply that $$ |F(y)-F(y_{0})|<(r-a)\epsilon +2\epsilon<(b-a+2)\epsilon $$ if $y$ and $y_{0}$ are in $[c,d]$ and $|y-y_{0}|<\delta$. Therefore $F$ is continuous on $[c,d]$, so the integral on left side of (B) exists. Denote \begin{equation}\tag{D} I= \int_{c}^{d}\left(\int_{a}^{b}f(x,y)\,dx\right)\,dy. \end{equation} We will show that the improper integral on the right side of (B) converges to $I$. To this end, denote $$ I(r)= \int_{r}^{b}\left(\int_{c}^{d}f(x,y)\,dy\right)\,dx. $$ Since we can reverse the order of integration of the continuous function $f$ over the rectangle $[r,b]\times [c,d]$ (Corollary~7.2.2, p.~466), $$ I(r)=\int_{c}^{d}\left(\int_{r}^{b}f(x,y)\,dx\right)\,dy. $$ From this and (D), $$ I-I(r)=\int_{c}^{d}\left(\int_{a}^{r}f(x,y)\,dx\right)\,dy. $$ Now suppose $\epsilon>0$. Since $\int_{a}^{b}f(x,y)\,dx$ converges uniformly on $[c,d]$, there is an $r_{0}\in (a,b]$ such that $$ \left|\int_{a}^{r}f(x,y)\,dx\right|<\epsilon, \quad a0$. Since we have assumed that $\lim_{r\to a+}F_{r}(y_{0})=F(y_{0})$ exists, there is an $r_{0}$ in $(a,b)$ such that $$ |F_{r}(y_{0})-F(y_{0})|<\epsilon,\quad a0$ make the change of variable $u=xy$ to see that $$ C(y)=\lim_{r\to\infty}\int_{y}^{ry}\frac{\cos u}{u}\,du= \int_{y}^{\infty}\frac{\cos u}{u}\,du $$ and $$ S(y)=\lim_{r\to\infty}\int_{y}^{ry}\frac{\sin u}{u}\,du. S(y)=\int_{y}^{\infty}\frac{\sin u}{u}\,du. $$ Therefore $\lim_{y\to 0+}C(y)=\infty$, so $C$ is not continuous at $y=0$. Since $S(0)=0$ and $\lim_{y\to 0+}S(y)= \dst{\int_{0}^{\infty}\frac{\sin u}{u}\,du}\ne 0$, $S$ is not continuous at $y=0$. \medskip {\bf 18. (a)} The integral diverges if $y=0$. If $y\ne0$ substitute $u=|y|x$ to obtain \begin{equation} F(y)=\int_{0}^{\infty}\frac{dx}{1+x^{2}y^{2}}= \frac{1}{|y|}\int_{0}^{\infty}\frac{du}{1+u^{2}} =\frac{1}{|y|}\tan^{-1}u\biggr|_{0}^{\infty}=\frac{\pi}{2|y|}, \tag{A} \end{equation} so $F(y)$ converges for all $y\ne0$. To test for uniform convergence, suppose $|y|>0$ and $00$ there is an $\alpha>0$ such that $\dst{\frac{1}{\rho}\int_{\alpha}^{\infty}\frac{du}{1+u^{2}}}<\epsilon$. Therefore $\dst{\int_{r}^{r_{1}}\frac{dx}{1+x^{2}y^{2}}}<\epsilon$ if $\alpha/\rho0$. \medskip To evaluate $$ I=\dst{\int_{0}^{\infty}\frac{\tan^{-1}ax-\tan^{-1}bx}{x}\,dx}, $$ we note that $$ \frac{\tan^{-1}ax-\tan^{-1}bx}{x}=\int_{b}^{a}\frac{dy}{1+x^{2}y^{2}}. $$ Therefore $$ I=\int_{0}^{\infty}\,dx \int_{b}^{a}\frac{dy}{1+x^{2}y^{2}} =\int_{b}^{a}\,dy\int_{0}^{\infty}\frac{dx}{1+x^{2}y^{2}} =\frac{\pi}{2}\int_{b}^{a}\frac{dy}{y}=\frac{\pi}{2}\log\frac{a}{b}, $$ where the second equality is valid because of the uniform convergence of $F(y)$ on the closed interval with endpoints $a$ and $b$, and the third equality follows from (A). \medskip {\bf (b)} $F(y)$ is a proper integral if $y\ge 0$ and it diverges if $y\le -1$. If $-1-1, $$ it follows that $$ \left|\int_{0}^{r}x^{y}\,dx\right|\le \frac{r^{\rho+1}}{\rho +1} \text{\quad if\quad} 0-1$. \medskip Now Theorem~11 implies that $$ I=\dst{\int_{0}^{1}\frac{x^{a}-x^{b}}{\log x}\,dx} = \int_{0}^{1}\,dx \int_{b}^{a}x^{y}\,dy =\int_{b}^{a}\,dy\int_{0}^{1}x^{y}\,dx =\int_{b}^{a}\frac{dy}{y+1}=\log\frac{a+1}{b+1}. $$ \medskip {\bf (c)} $\dst{F(y)=\int_{0}^{\infty} e^{-yx}\cos x \,dx=\frac{y}{y^{2}+1}}$. Since $$ \left|\int_{r}^{\infty}e^{-yx} \cos x\,dx\right|\le \int_{r}^{\infty}e^{-xy}\,dx=\frac{e^{-yr}}{y}, $$ Theorem~4 (or Theorem~6) implies that $F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. Therefore, Theorem implies that if $a$, $b>0$ then \begin{eqnarray*} I&=&\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\cos x\,dx =\int_{0}^{\infty}\cos x\,dx\int_{a}^{b}e^{-yx}\,dy \\ &=&\int_{a}^{b} \,dy\int_{0}^{\infty}e^{-yx}\cos x\,dx =\int_{a}^{b}\frac{y}{y^{2}+1}\,dy=\frac{1}{2}\log\frac{b^{2}+1}{a^{2}+1}. \end{eqnarray*} \medskip {\bf (d)} $\dst{F(y)=\int_{0}^{\infty} e^{-yx}\sin x \,dx=\frac{1}{y^{2}+1}}$. Since $$ \left|\int_{r}^{\infty}e^{-yx} \sin x\,dx\right|\le \int_{r}^{\infty}e^{-yx}\,dx=\frac{e^{-yr}}{y}, $$ Theorem~4 (or Theorem~6) implies that $F(y)$ converges uniformly on every $[\rho,\infty)$ if $\rho>0$. Therefore, if $a$, $b>0$ then Theorem~11 implies that \begin{eqnarray*} I&=&\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\sin x\,dx =\int_{0}^{\infty}\sin x\,dx\int_{a}^{b}e^{-yx}\,dy \\ &=&\int_{a}^{b} \,dy\int_{0}^{\infty}e^{-yx}\sin x\,dx =\int_{a}^{b}\frac{1}{y^{2}+1}\,dy=\tan^{-1}b-\tan^{-1}a. \end{eqnarray*} \medskip {\bf(e)} $\dst{F(y)=\int_{0}^{\infty} e^{-x}\sin xy \,dx=\frac{y}{y^{2}+1}}$. Since $$ \left|\int_{r}^{\infty}e^{-x} \sin xy\,dx\right|\le \int_{r}^{\infty}e^{-x}\,dx=e^{-r}, $$ Theorem~4 (or Theorem~6) implies that $F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. Therefore Theorem~11 implies that \begin{eqnarray*} I&=&\int_{0}^{\infty}e^{-x}\frac{1-\cos ax}{x}\,dx =\int_{0}^{\infty}e^{-x}\,dx\int_{0}^{a}\sin xy\,dy\\ &=&\int_{0}^{a}\,dy\int_{0}^{\infty}e^{-x}\sin xy\,dx =\int_{0}^{a} \frac{y}{y^{2}+1}\,dy=\frac{1}{2}\log(1+a^{2}). \end{eqnarray*} \medskip {\bf(f)} $\dst{F(y)=\int_{0}^{\infty} e^{-x}\cos xy \,dx=\frac{1}{y^{2}+1}}$. Since $$ \left|\int_{r}^{\infty}e^{-x} \cos xy\,dx\right|\le \int_{r}^{\infty}e^{-x}\,dx=e^{-r}, $$ Theorem~4 (or Theorem~6) implies that $F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. Therefore Theorem~11 implies that \begin{eqnarray*} I&=&\int_{0}^{\infty}e^{-x}\sin ax\,dx =\int_{0}^{\infty}e^{-x}\,dx\int_{0}^{a}\cos xy\,dy\\ &=&\int_{0}^{a}\,dy\int_{0}^{\infty}e^{-x}\cos xy\,dx =\int_{0}^{a} \frac{1}{y^{2}+1}\,dy=\tan^{-1}a. \end{eqnarray*} \medskip {\bf 19. (a)} We start with \begin{equation} F(y)=\int_{0}^{1} x^{y}\,dx =\frac{1}{y+1}\quad y>-1. \tag{A} \end{equation} Formally differentiating this yields \begin{equation} F^{(n)}(y)=\int_{0}^{1}(\log x)^{n} x^{y}\,dx =\frac{(-1)^{n}n!}{(y+1)^{n+1}},\quad y>-1. \tag{B} \end{equation} To justify this we will show by induction that the improper integrals $$ I_{n}(y)=\int_{0}^{1}(\log x)^{n} x^{y}\,dx,\quad n=0,1,2, \dots $$ converge uniformly on $[\rho,\infty)$ if $\rho>-1$. We begin with $n=0$:. $$ \int_{0}^{r}x^{y}\,dx = \frac{x^{y+1}}{y+1}\biggr|_{r_{1}}^{r}=\frac{r^{y+1}}{y+1}\le \frac{r^{y+1}}{\rho+1},\quad -1<\rho\le y. $$ so $I_{0}(y)=F(y)$ converges uniformly on $[\rho,\infty)$ if $\rho>-1$. Now suppose that $I_{n}(y)$ converges uniformly on $[\rho,\infty)$. Integrating by parts yields \begin{eqnarray*} \int_{r_{1}}^{r}(\log x)^{n+1}x^{y}\,dx&=& \frac{r^{y+1}(\log r)^{n+1}-r_{1}^{y+1}(\log r_{1})^{n+1}} {y+1}\\ &&-\frac{n+1}{y+1} \int_{r_{1}}^{r}(\log x)^{n}x^{y}\,dx, \quad -1-1$. To this end, note from (C) that \begin{equation}\tag{D} \left|\int_{0}^{r}(\log x)^{n+1} x^{y}\,dx\right| \le \left|\frac{r^{\rho+1}(\log r)^{n+1}}{\rho+1}\right| +\frac{n+1}{\rho+1}\left|\int_{0}^{r}(\log x)^{n}x^{y}\,dx\right| \end{equation} if $y\ge \rho$, Now suppose $\epsilon>0$. Since $\dst{\lim_{r\to0+}r^{\rho+1}(\log r)^{n+1}=0}$, there is an $r_{1}\in (0,1)$ such that $$ \left|\frac{r^{\rho+1}(\log r)^{n+1}}{\rho+1}\right| \le \frac{\epsilon}{2} \text{\quad if \quad} 00. \tag{A} \end{equation} Formally differentiating this yields yields \begin{eqnarray*} \int_{0}^{\infty}\frac{dx}{(x^{2}+y)^{n+1}} &=&\frac{\pi}{2n+1}1\cdot 3\cdots(2n-1)y^{-n-1/2} =\frac{\pi}{2^{2n+1}}\frac{(2n)!}{n!}y^{-n-1/2}\\ &=&\frac{\pi}{2^{2n+1}}\binom{2n}{n}y^{-n-1/2},\quad y>0. \end{eqnarray*} Theorem~11 implies that the formal differentiation is legitimate, since, if $y\ge 0$ and $r>0$, then $$ \int_{r}^{\infty}\frac{dx}{(x^{2}+y)^{n+1}}\le \int_{r}^{\infty}x^{-2n-2}dx=\frac{r^{-2n-1}}{(2n-1)}; $$ hence, the improper integrals $\dst{\int_{0}^{\infty}\frac{dx}{(x^{2}+y)^{n+1}}}$, $n=0$, $1$, $2$, \dots converge uniformly on $[0,\infty)$. \medskip {\bf (c)} Denote $I_{n}(y)=\dst{\int_{0}^{\infty}x^{2n+1}e^{-yx^{2}}\,dx}$. Then $$ I_{0}(y)=\int_{0}^{\infty}xe^{-yx^{2}}= \frac{1}{2}\int_{0}^{\infty}2xe^{-yx^{2}}\,dx =-\frac{1}{2y}e^{-yx^{2}}\biggr|_{0}^{\infty}=\frac{1}{2y}. $$ Since $$ \int_{r}^{\infty}x^{2n+1}e^{-yx^{2}} \,dx\le \int_{r}^{\infty}x^{2n+1}e^{-\rho x^{2}} \,dx\text{\quad if\quad} 0<\rho\le r, $$ if $n\ge 0$, we can differentiate $I_{n}$ formally with respect to $y\in (0,\infty)$ to obtain $$ I_{n}(y)=(-1)^{n}I_{0}^{(n)}=\frac{n!}{2y^{n+1}}. $$ {\bf (d)} Denote \begin{eqnarray*} I(y)&=&\int_{0}^{\infty}y^{x}\,dx =\int_{0}^{\infty}e^{x\log y}\,dx =\frac{1}{\log y}\int_{0}^{\infty}(\log y) y^{x}\,dx \\ &=&\frac{y^{x}}{\log y}\biggr|_{0}^{\infty}=-\frac{1}{\log y}\quad 0r>1$ and $\rho_{2}<1$ then $$ \int_{r}^{\infty}xy^{x-1}\,dx<\int_{r}^{\infty}x\rho_{2}^{x-1}\,dx =\frac{1}{\rho_{2}}\int_{r}^{\infty}x\rho_{2}^{x}\,dx, $$ Since $$ \lim_{x\to\infty}\frac{1}{\rho_{2}}\int_{r}^{\infty}x\rho_{2}^{x}\,dx =0 $$ Theorem~7 implies that $J_{2}(y)$ converges uniformly on $[0,\rho_{2}]$. Therefore, if $0<\rho_{1}<\rho_{2}<1$ then $\dst{\int_{0}^{\infty}xy^{x-1}}$ converges uniformly on $[\rho_{1},\rho_{2}]$. Now Theorem~11 implies that \begin{equation}\tag{A} I'(y)=\int_{0}^{\infty}xy^{x-1}\,dx,\quad 00$. Note that if $\lim_{x\to\infty} x^{n}f(x)=0$, then $\lim_{x\to\infty} x^{k}f(x)=0$, $k=0$, $1$, $2$,\dots $n$. If $0\le k\le n$, then \begin{equation} \tag{A} \int_{r}^{r_{1}}x^{k}f(x)\cos xy\,dx= \frac{1}{y}\left[x^{k}f(x)\sin xy\biggr|_{r}^{r_{1}}- \int_{r}^{r_{1}}(x^{k}f(x))'\sin xy\,dx\right]. \end{equation} Henceforth $k$ is fixed. Our assumptions imply that if $\rho>0$ and $\epsilon>0$ then there is an $r_{0}\in [a,\infty)$ such that $$ \int_{r_{0}}^{\infty}|(x^{k}f(x))'|\,dx<\rho\epsilon \text{\quad and \quad} |x^{k}f(x)|<\rho\epsilon,\quad x\ge r_{0}. $$ Therefore (A) implies that $$ \left|\int_{r}^{r_{1}}x^{k}f(x)\cos xy\,dx\right|<3\epsilon,\quad r_{0}\le rs_{0}, $$ which is true by the definition of $F$ for $n=0$. If $P_{n}$ is true, then Theorems~11 and 13 imply that \begin{eqnarray*} F^{(n+1)}(s)&=&(-1)^{n}\frac{d}{ds} \int_{0}^{\infty}e^{-sx}x^{n}f(x)\,dx=(-1)^{n} \int_{0}^{\infty}\frac{d}{ds}\left(e^{-sx}x^{n}f(x)\right)\,dx\\ &=&(-1)^{n+1}\int_{0}^{\infty}e^{-sx}x^{n+1}f(x)\,dx, \end{eqnarray*} so $P_{n}$ implies $P_{n+1}$, which completes the induction proof. \medskip {\bf 26.} Let $G(x)=\dst{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}$. If $s>s_{0}$ then \begin{equation}\tag{A} F(s)=\int_{0}^{\infty}e^{-sx}f(x)\,dx =\int_{0}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx =(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}G(x)\,dx \end{equation} (integration by parts). Since $\dst{(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}\,dx=1}$, (A) implies that \begin{equation}\tag{B} F(s)-F(s_{0})=(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}(G(x)-F(s_{0}))\,dx. \end{equation} Now suppose $\epsilon>0$. Since $F(s_{0})=\dst{\int_{0}^{\infty}e^{-s_{0}t} f(t)\,dt}=\lim_{t\to\infty}G(x)$, there is an $r$ such that $|G(x)-F(s_{0})|<\epsilon$ if $x\ge r$; hence, from (B), then \begin{eqnarray*} |F(s)-F(s_{0})|&\le& (s-s_{0})\int_{0}^{r}e^{-(s-s_{0})x} |G(x)-F(s_{0})|+\epsilon(s-s_{0})\int_{r}^{\infty} e^{-(s-s_{0})x}\,dx\\ &<& (s-s_{0})\int_{0}^{r}e^{-(s-s_{0})x} |G(x)-F(s_{0})|+\epsilon. \end{eqnarray*} Since $r$ is fixed, we can let $s\to s_{0}^{+}$ to conclude that $\limsup_{s\to s_{0}+}|F(s)-F(s_{0})|\le \epsilon$, which implies that $\lim_{s\to S_{0}+}F(s)=F(s_{0})$. \medskip {\bf 26.} If $s\ge s_{1}>s_{0}$ then $$ |e^{-sx}f(x)|= |e^{-(s-s_{0})x}e^{s_{0}x}f(x)|\le M e^{-(s-s_{0})x} \le M e^{-(s_{1}-s_{0})x}. $$ Since $$ \int_{0}^{\infty}Me^{-(s_{1}-s_{0})x}\,dx=\frac{M}{s_{1}-s_{0}}<\infty, $$ Theorem~6 implies the stated conclusion. \medskip {\bf 27.} In Theorem~13 we assumed only that $\int_{0}^{x}e^{-s_{0}u}f(u)\,du$ is bounded; here we are assuming that $\int_{0}^{\infty}e^{-s_{0}u}f(u)\,du$ is convergent. \medskip Let $$ G(x)=\int_{x}^{\infty}e^{-s_{0}t}f(t)\,dt \text{\quad and\quad} H(x)=\sup\set{|G(t)|}{t\ge x}. $$ Then \begin{equation}\tag{A} |G(x)|\le H(x)\text{\quad and \quad} \lim_{x\to\infty}G(x)=\lim_{x\to\infty}H(x)=0, \end{equation} since $\int_{0}^{\infty}e^{-s_{0}x}f(x)\,dx$ converges. Since $f$ is continuous on $[0,\infty)$, $G'(x)=-e^{-s_{0}x}f(x)$. Integration by parts yields \begin{eqnarray*} \int_{r}^{\infty}e^{-sx}f(x)\,dx&=& \int_{r}^{\infty}e^{-(s-s_{0})x}(e^{-s_{0}x}f(x))\,dx =-\int_{0}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx\\ &=&-e^{-(s-s_{0})x}G(x)\biggr|_{r}^{\infty} +(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}G(x)\,dx\\ &=&e^{-(s-s_{0})r}G(r)+(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}G(x)\,dx,\quad s\ge s_{0}. \end{eqnarray*} Therefore \begin{eqnarray*} \left|\int_{r}^{\infty}e^{-sx}f(x)\,dx\right|&\le& |G(r)|e^{-(s-s_{0})r}+H(r)(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}\,dx\\ &=&(G(r)+H(r))e^{-(s-s_{0})r}\le 2H(r)e^{-(s-s_{0})}, \quad s\ge s_{0}, \end{eqnarray*} so (A) implies that $F(s)$ converges uniformly on $[s_{0},\infty)$. \medskip {\bf 28.} From Theorem~13, $F(s)=\dst{\int_{0}^{\infty}e^{-sx}f(x)\, dx}$ converges for all $s>s_{0}$. Denote $G(x)=\dst{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}$, $x\ge 0$. Then \begin{eqnarray*} F(s)&=&\int_{0}^{\infty}e^{-(s-s_{0})x}e^{-s_{0}x}f(x)\,dx= \int_{0}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx \\ &=&(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}G(x)\,dx \end{eqnarray*} (integration by parts). Since $\dst{(s-s_{0})\int_{0}^{\infty}e^{-(s-s_{0})x}\,dx}=1$, $$ F(s)-F(s_{0})=\int_{0}^{\infty}e^{-(s-s_{0})x}(G(x)-F(s_{0}))\,dx $$ If $\epsilon>0$ there is an $R$ such that $|G(x)-F(s_{0})|<\epsilon$ if $x\ge R$. Therefore, if $s>s_{0}$ then \begin{eqnarray*} |F(s)-F(s_{0})|&<& (s-s_{0})\int_{0}^{R}e^{-(s-s_{0})x}|G(x)-F(s_{0})|\,dx+\epsilon\\ &<&(s-s_{0})\int_{0}^{R}|G(x)-F(s_{0})|\,dx+\epsilon. \end{eqnarray*} Hence $\limsup_{s\to s_{0}+}|F(s)-F(s_{0})|\le \epsilon$. Since $\epsilon$ is arbitrary, this implies that \\ $\lim_{s\to s_{0}+}|F(s)-F(s_{0})|=0$. \medskip {\bf 29.} The assumptions of Exercise~28 imply that $\int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx$ converges for every $r>0$. Since $$ \int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx=\int_{0}^{\infty}e^{-s(r+x)}f(x+r)\,dx =e^{-sr}\int_{0}^{\infty}e^{-sx}f(x+r)\,dx, $$ we can apply the result of Exercise~30 with $f(x)$ replaced by $f(x+r)$, to conclude that \begin{eqnarray*} \lim_{s\to s_{0}+}\int_{r}^{\infty}e^{-sx}f(x)\,dx&=& e^{-s_{0}r}\int_{0}^{\infty}e^{-s_{0}x}f(x+r)\,dx\\ &=&\int_{0}^{\infty}e^{-s_{0}(x+r)}f(x+r)\,dx\\ &=&\int_{r}^{\infty}e^{-s_{0}x}f(x)\,dx. \end{eqnarray*} \medskip {\bf 30.} If $G(x)=\dst{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}$, then $|G(x)|\le M$ on $[0,\infty)$ for some $M$. If $\epsilon>0$, there is an $r>0$ such that \begin{equation}\tag{A} \int_{0}^{r}e^{-s_{0}x}|f(x)|\,dx <\epsilon. \end{equation} If $s>s_{0}$, then \begin{eqnarray*} \int_{r}^{\infty}e^{-sx}f(x)\,dx&=&\int_{r}^{\infty}e^{-(s-s_{0})x}G'(x)\,dx \\ &=&e^{-(s-s_{0})x}G(x)\biggr|_{r}^{\infty} +(s-s_{0})\int_{r}^{\infty}G(x)e^{-(s-s_{0})x}\,dx\\ &=&-e^{-(s-s_{0})r}G(r) +(s-s_{0})\int_{r}^{\infty}G(x)e^{-(s-s_{0})x}\,dx. \end{eqnarray*} Therefore, since $|G(x)|\le M$, \begin{eqnarray*} \left|\int_{r}^{\infty}e^{-sx}f(x)\,dx\right| &\le&Me^{-(s-s_{0})r}+M(s-s_{0})\int_{r}^{\infty}e^{-(s-s_{0})x}\,dx\\ &=&M\left(e^{-(s-s_{0})r}-e^{-(s-s_{0})x}\biggr|_{r}^{\infty}\right) =2Me^{-(s-s_{0})r}. \end{eqnarray*} This and (A) imply that $$ \left|\int_{0}^{\infty}e^{-sx}f(x)\,dx\right|\le \epsilon+2Me^{-(s-s_{0})r}. $$ Therefore, $$ \limsup_{s\to\infty} \left|\int_{0}^{\infty}e^{-sx}f(x)\,dx\right|\le \epsilon. $$ Since $\epsilon$ is arbitrary, this implies that $$ \limsup_{s\to\infty}\int_{0}^{\infty}e^{-sx}f(x)\,dx=0, $$ \medskip {\bf 31. (a)} From Exercise~18{\bf(d)}, $\dst{\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}\sin x\,dx} =\tan^{-1}b-\tan^{-1}a.$ From Exercise~30, letting $b\to\infty$ yields $$ \int_{0}^{\infty}e^{-ax}\frac{\sin x}{x}\,dx= \frac{\pi}{2}-\tan^{-1}a, \text{\quad so \bf{(b)}\quad} \int_{0}^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}. $$ \medskip {\bf 32. (a)} Integrating by parts yields \begin{equation}\tag{A} \int_{0}^{r}e^{-sx}f'(x)\,dt =e^{-sr}f(r)-f(0) +\int_{0}^{r}se^{-sx}f(x)\,dx. \end{equation} Suppose $s\ge s_{1}>s_{0}$. Since $|f(x)|\le Me^{s_{0}x}$, $e^{-sr}|f(r)|\le Me^{-(s_{1}-s_{0})r}$. Therefore $e^{-sr}f(r)=0$ converges uniformly to zero on $[s_{1},\infty)$. Since \begin{eqnarray*} \left|\int_{r}^{\infty}se^{-sx}f(x)\,dx \right|&\le& M|s|\int_{r}^{\infty}e^{-(s-s_{0})x}\,dx\le \frac{M|s|e^{-(s_{1}-s_{0})r}}{s-s_{0}}\\ &\le&\frac{M(s-s_{0}+|s_{0}|)e^{-(s_{1}-s_{0})r}}{s-s_{0}} \le M\left(1+\frac{|s_{0}|}{s_{1}-s_{0}}\right)e^{-(s_{1}-s_{0})r}, \end{eqnarray*} it follows that $\dst{\int_{r}^{\infty}se^{-sx}f(x)\,dx}$ converges to zero uniformly on $[s_{1},\infty)$. Since this implies that $\dst{\int_{0}^{r}se^{-sx}f(x)\,dx}$ converges uniformly on $[s_{1},\infty)$, (A) implies that $G(s)$ converges uniformly on $[s_{1},\infty)$. \medskip {\bf(b)} In this case let $f'(x)=xe^{x^{2}}\sin e^{x^{2}}$, so $f(x)=-\dst{\frac{1}{2}}\cos e^{x^{2}}$. Since $|\cos e^{x^{2}}|\le 1$ for all $x$, the hypotheses stated in (a) hold with $s_{0}=0$. Therefore $G(s)$ converges uniformly on $[\rho,\infty)$ if $\rho>0$. \medskip {\bf 33.} We will first show that $\dst{\int_{0}^{\infty}e^{-s_{0}x}\frac{f(x)}{x}\,dx}$ converges. Denote $G(x)=\dst{\int_{0}^{x}e^{-s_{0}t}f(t)\,dt}$. Since $F(s_{0})$ is convergent; say $|G(x)|\le M$, $0\le x<\infty$. If $0