FresnelFormulae 2009-04-18 05:33:26 2009-04-18 05:33:26 Theorem Fresnel integrals pstricks need to convert to png, rendering for now but need figure too % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{pstricks} \usepackage{pst-plot} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions % % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sR}[0]{\mathbb{R}} \newcommand{\sC}[0]{\mathbb{C}} \newcommand{\sN}[0]{\mathbb{N}} \newcommand{\sZ}[0]{\mathbb{Z}} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\F}{\mathbbmss{F}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand*{\norm}[1]{\lVert #1 \rVert} \newcommand*{\abs}[1]{| #1 |} \newtheorem{thm}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} $$\int_0^\infty\!\cos{x^2}\,dx \,=\, \int_0^\infty\!\sin{x^2}\,dx \,=\, \frac{\sqrt{2\pi}}{4}$$ {\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.