FresnelFormulaeResult 2009-04-29 09:57:17 2009-04-29 14:25:55 Result Fresnel formulae The immediate failure is the unused legacy dependency: LaTeX Error: File `bbm.sty' not found. The article never actually needs bbm, mathrsfs, the number-set macros, or the theorem environments, so I removed all of that instead of installing another obsolete package. I also cleaned the article content because it was essentially duplicating the parent Fresnel proof and still contained outdated comments about PSTricks rendering. The proof is now written in the same corrected form as object 649, % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{mathrsfs} % 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} The Fresnel formulae presented in the parent entry-- that the proof refers to-- are as follows: $$\int_0^\infty\!\cos{x^2}\,dx \,=\, \int_0^\infty\!\sin{x^2}\,dx \,=\, \frac{\sqrt{2\pi}}{4}$$ (visible there only in the Tex code mode because of a current quirk/glitch with pstricks vs. Tex). The remainder of the equations Tex mode correctly specified by pahio in the parent entry entitled \PMlinkname{Fresnel formulae}{http://planetphysics.org/encyclopedia/FresnelFormulae.html} is precisely quoted here as follows: ``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\, $\int_0^\infty\sin{x^2}\,dx = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{4}.$ "