Talk:PlanetPhysics/Fresnel Formulae

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$$\int_0^\infty\!\cos{x^2}\,dx \,=\, \int_0^\infty\!\sin{x^2}\,dx \,=\, \frac{\sqrt{2\pi}}{4}$$

{\em Proof.}

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The \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} \,$\displaystyle z \mapsto e^{-z^2}$\, is entire, whence by the fundamental \htmladdnormallink{theorem}{http://planetphysics.us/encyclopedia/Formula.html} of complex analysis we have \begin{align} \oint_\gamma e^{-z^2}\,dz \;=\; 0 \end{align} where $\gamma$ is the perimeter 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 \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html} 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 \htmladdnormallink{graph}{http://planetphysics.us/encyclopedia/Cod.html} 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.

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