FRIntegrals 2009-04-17 20:22:47 2009-04-17 20:22:47 Topic % this is the default PlanetPhysics preamble. as your \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} \usepackage{pstricks} \usepackage{pst-plot} % define commands here \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} % define commands here \usepackage{amsmath, amssymb, amsfonts, amsthm, amscd, latexsym} \usepackage{xypic} \usepackage[mathscr]{eucal} \theoremstyle{plain} \newtheorem{lemma}{Lemma}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}{Corollary}[section] \theoremstyle{definition} \newtheorem{definition}{Definition}[section] \newtheorem{example}{Example}[section] %\theoremstyle{remark} \newtheorem{remark}{Remark}[section] \newtheorem*{notation}{Notation} \newtheorem*{claim}{Claim} \renewcommand{\thefootnote}{\ensuremath{\fnsymbol{footnote%%@ }}} \numberwithin{equation}{section} \newcommand{\Ad}{{\rm Ad}} \newcommand{\Aut}{{\rm Aut}} \newcommand{\Cl}{{\rm Cl}} \newcommand{\Co}{{\rm Co}} \newcommand{\DES}{{\rm DES}} \newcommand{\Diff}{{\rm Diff}} \newcommand{\Dom}{{\rm Dom}} \newcommand{\Hol}{{\rm Hol}} \newcommand{\Mon}{{\rm Mon}} \newcommand{\Hom}{{\rm Hom}} \newcommand{\Ker}{{\rm Ker}} \newcommand{\Ind}{{\rm Ind}} \newcommand{\IM}{{\rm Im}} \newcommand{\Is}{{\rm Is}} \newcommand{\ID}{{\rm id}} \newcommand{\GL}{{\rm GL}} \newcommand{\Iso}{{\rm Iso}} \newcommand{\Sem}{{\rm Sem}} \newcommand{\St}{{\rm St}} \newcommand{\Sym}{{\rm Sym}} \newcommand{\SU}{{\rm SU}} \newcommand{\Tor}{{\rm Tor}} \newcommand{\U}{{\rm U}} \newcommand{\A}{\mathcal A} \newcommand{\Ce}{\mathcal C} \newcommand{\D}{\mathcal D} \newcommand{\E}{\mathcal E} \newcommand{\F}{\mathcal F} \newcommand{\G}{\mathcal G} \newcommand{\Q}{\mathcal Q} \newcommand{\R}{\mathcal R} \newcommand{\cS}{\mathcal S} \newcommand{\cU}{\mathcal U} \newcommand{\W}{\mathcal W} \newcommand{\bA}{\mathbb{A}} \newcommand{\bB}{\mathbb{B}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bD}{\mathbb{D}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bG}{\mathbb{G}} \newcommand{\bK}{\mathbb{K}} \newcommand{\bM}{\mathbb{M}} \newcommand{\bN}{\mathbb{N}} \newcommand{\bO}{\mathbb{O}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bV}{\mathbb{V}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bfE}{\mathbf{E}} \newcommand{\bfX}{\mathbf{X}} \newcommand{\bfY}{\mathbf{Y}} \newcommand{\bfZ}{\mathbf{Z}} \renewcommand{\O}{\Omega} \renewcommand{\o}{\omega} \newcommand{\vp}{\varphi} \newcommand{\vep}{\varepsilon} \newcommand{\diag}{{\rm diag}} \newcommand{\grp}{{\mathbb G}} \newcommand{\dgrp}{{\mathbb D}} \newcommand{\desp}{{\mathbb D^{\rm{es}}}} \newcommand{\Geod}{{\rm Geod}} \newcommand{\geod}{{\rm geod}} \newcommand{\hgr}{{\mathbb H}} \newcommand{\mgr}{{\mathbb M}} \newcommand{\ob}{{\rm Ob}} \newcommand{\obg}{{\rm Ob(\mathbb G)}} \newcommand{\obgp}{{\rm Ob(\mathbb G')}} \newcommand{\obh}{{\rm Ob(\mathbb H)}} \newcommand{\Osmooth}{{\Omega^{\infty}(X,*)}} \newcommand{\ghomotop}{{\rho_2^{\square}}} \newcommand{\gcalp}{{\mathbb G(\mathcal P)}} \newcommand{\rf}{{R_{\mathcal F}}} \newcommand{\glob}{{\rm glob}} \newcommand{\loc}{{\rm loc}} \newcommand{\TOP}{{\rm TOP}} \newcommand{\wti}{\widetilde} \newcommand{\what}{\widehat} \renewcommand{\a}{\alpha} \newcommand{\be}{\beta} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta} \newcommand{\del}{\partial} \newcommand{\ka}{\kappa} \newcommand{\si}{\sigma} \newcommand{\ta}{\tau} \newcommand{\lra}{{\longrightarrow}} \newcommand{\ra}{{\rightarrow}} \newcommand{\rat}{{\rightarrowtail}} \newcommand{\oset}[1]{\overset {#1}{\ra}} \newcommand{\osetl}[1]{\overset {#1}{\lra}} \newcommand{\hr}{{\hookrightarrow}} $$\int_0^\infty\!\cos{x^2}\,dx \,=\, \int_0^\infty\!\sin{x^2}\,dx \,=\, \frac{\sqrt{2\pi}}{4}$$ {\em Proof.} \begin{center} \begin{pspicture}(-1,-1)(6.2,4.2) \psaxes[Dx=10,Dy=10]{->}(0,0)(-1,-1)(6,4) \rput(-0.2,4.1){$y$} \rput(6.1,-0.2){$x$} \rput(-0.17,-0.22){0} \rput(5,-0.3){$R$} \rput(0.7,0.3){$\frac{\pi}{4}$} \rput(1.4,1.7){$s$} \rput(4.7,2.2){$b$} \psline[linecolor=blue,linewidth=0.05]{->}(0,0)(5,0) \psline[linecolor=blue,linewidth=0.04]{->}(3.55,3.55)(0,0) \psarc[linecolor=blue,linewidth=0.04]{->}(0,0){5}{-1}{45} \psarc(0,0){0.5}{0}{45} \end{pspicture} \end{center} The function \,$\displaystyle z \mapsto e^{-z^2}$\, is entire, whence by the fundamental theorem of complex analysis we have \begin{align} \oint_\gamma e^{-z^2}\,dz \;=\; 0 \end{align} where $\gamma$ is the \PMlinkname{perimeter}{BasicPolygon} of the circular sector described in the picture.\, We split this contour integral to three portions: \begin{align} \underbrace{\int_0^R\!e^{-x^2}\,dx}_{I_1}+\underbrace{\int_b\!e^{-z^2}\,dz}_{I_2} +\underbrace{\int_s\!e^{-z^2}\,dz}_{I_3} \,=\,0 \end{align} By the entry concerning the Gaussian integral, we know that $$\lim_{R\to\infty}I_1 = \frac{\sqrt{\pi}}{2}.$$ For handling $I_2$, we use the substitution $$z \,:=\, Re^{i\varphi} = R(\cos\varphi+i\sin\varphi), \quad dz \,=\,iRe^{i\varphi}\,d\varphi \quad (0 \leqq \varphi \leqq \frac{\pi}{4}).$$ Using also de Moivre's formula we can write $$|I_2| = \left|iR\int_0^{\frac{\pi}{4}}e^{-R^2(\cos2\varphi+i\sin2\varphi)}e^{i\varphi}d\varphi\right| \leqq R\!\int_0^{\frac{\pi}{4}}\left|e^{-R^2(\cos2\varphi+i\sin2\varphi)}\right|\cdot\left|e^{i\varphi}\right|\cdot|d\varphi| = R\!\int_0^{\frac{\pi}{4}}e^{-R^2\cos2\varphi}d\varphi.$$ Comparing the graph of the function \,$\varphi \mapsto \cos2\varphi$\, with the line through the points \,$(0,\,1)$\, and\, $(\frac{\pi}{4},\,0)$\, allows us to estimate $\cos2\varphi$ downwards: $$\cos2\varphi \geqq 1\!-\!\frac{4\varphi}{\pi} \quad\mbox{for}\quad 0 \leqq \varphi \leqq \frac{\pi}{4}$$ Hence we obtain $$|I_2| \leqq R\int_0^{\frac{\pi}{4}}\frac{d\varphi}{e^{R^2\cos2\varphi}} \leqq R\int_0^{\frac{\pi}{4}}\frac{d\varphi}{e^{R^2(1-\frac{4\varphi}{\pi})}} \leqq \frac{R}{e^{R^2}} \int_0^{\frac{\pi}{4}} e^{\frac{4R^2}{\pi}\varphi} d\varphi,$$ and moreover $$|I_2| \leqq \frac{\pi}{4Re^{R^2}}(e^{R^2}-1) < \frac{\pi e^{R^2}}{4Re^{R^2}} = \frac{\pi}{4R} \; \to 0 \quad \mbox{as} \quad R \to \infty.$$ Therefore $$\lim_{R\to\infty}I_2 = 0.\\$$ Then make to $I_3$ the substitution $$z \;:=\; \frac{1\!+\!i}{\sqrt{2}}t, \quad dz \,=\, \frac{1\!+\!i}{\sqrt{2}}dt \quad(R \geqq t \geqq 0).$$ It yields \begin{align*} I_3 &\quad = \frac{1\!+\!i}{\sqrt{2}}\int_R^0e^{-it^2}\,dt = -\frac{1}{\sqrt{2}}\int_0^R(1+i)(\cos{t^2}-i\sin{t^2})\,dt \\ &\quad = -\frac{1}{\sqrt{2}}\left(\int_0^R\sin{t^2}\,dt+\int_0^R\cos{t^2}\,dt\right) +\frac{i}{\sqrt{2}}\left(\int_0^R\sin{t^2}\,dt-\int_0^R\cos{t^2}\,dt\right). \end{align*} Thus, letting\, $R \to \infty$,\, the equation (2) implies \begin{align} \frac{\sqrt{\pi}}{2}\!+\!0\! -\frac{1}{\sqrt{2}}\left(\int_0^\infty\!\sin{t^2}\,dt+\!\int_0^\infty\!\cos{t^2}\,dt\right)\! +\!\frac{i}{\sqrt{2}}\left(\int_0^\infty\!\sin{t^2}\,dt-\!\int_0^\infty\!\cos{t^2}\,dt\right) \;=\; 0. \end{align} Because the imaginary part vanishes, we infer that\, $\int_0^\infty\cos{x^2}\,dx = \int_0^\infty\sin{x^2}\,dx$,\, whence (3) reads $$\frac{\sqrt{\pi}}{2}+0-\frac{1}{\sqrt{2}}\!\cdot\!2\!\int_0^\infty\!\sin{t^2}\,dt \,=\, 0.$$ So we get also the result\, $\int_0^\infty\sin{x^2}\,dx = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{4}$,\, Q.E.D.