PlanetPhysics/Fresnel Formulae

$\int _{0}^{\infty }\!\cos {x^{2}}\,dx\,=\,\int _{0}^{\infty }\!\sin {x^{2}}\,dx\,=\,{\frac {\sqrt {2\pi }}{4}}$

Proof.

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The function \,$z\mapsto e^{-z^{2}}$\, is entire, whence by the fundamental theorem of complex analysis we have

${\begin{matrix}\oint _{\gamma }e^{-z^{2}}\,dz\;=\;0\end{matrix}}$

where $\gamma$ is the perimeter of the circular sector described in the picture.\, We split this contour integral to three portions:

${\begin{matrix}\underbrace {\int _{0}^{R}\!e^{-x^{2}}\,dx} _{I_{1}}+\underbrace {\int _{b}\!e^{-z^{2}}\,dz} _{I_{2}}+\underbrace {\int _{s}\!e^{-z^{2}}\,dz} _{I_{3}}\,=\,0\end{matrix}}$

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 $|I_{2}|=\left|iR\int _{0}^{\frac {\pi }{4}}e^{-R^{2}(\cos 2\varphi +i\sin 2\varphi )}e^{i\varphi }d\varphi \right|\leqq R\!\int _{0}^{\frac {\pi }{4}}\left|e^{-R^{2}(\cos 2\varphi +i\sin 2\varphi )}\right|\cdot \left|e^{i\varphi }\right|\cdot |d\varphi |=R\!\int _{0}^{\frac {\pi }{4}}e^{-R^{2}\cos 2\varphi }d\varphi .$ Comparing the graph of the function \,$\varphi \mapsto \cos 2\varphi$\, with the line through the points \,$(0,\,1)$\, and\, $({\frac {\pi }{4}},\,0)$\, allows us to estimate $\cos 2\varphi$ downwards: $\cos 2\varphi \geqq 1\!-\!{\frac {4\varphi }{\pi }}\quad {\mbox{for}}\quad 0\leqq \varphi \leqq {\frac {\pi }{4}}$ Hence we obtain $|I_{2}|\leqq R\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{e^{R^{2}\cos 2\varphi }}}\leqq R\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{e^{R^{2}(1-{\frac {4\varphi }{\pi }})}}}\leqq {\frac {R}{e^{R^{2}}}}\int _{0}^{\frac {\pi }{4}}e^{{\frac {4R^{2}}{\pi }}\varphi }d\varphi ,$ 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle \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

${\begin{matrix}I_{3}&\quad ={\frac {1\!+\!i}{\sqrt {2}}}\int _{R}^{0}e^{-it^{2}}\,dt=-{\frac {1}{\sqrt {2}}}\int _{0}^{R}(1+i)(\cos {t^{2}}-i\sin {t^{2}})\,dt\\&\quad =-{\frac {1}{\sqrt {2}}}\left(\int _{0}^{R}\sin {t^{2}}\,dt+\int _{0}^{R}\cos {t^{2}}\,dt\right)+{\frac {i}{\sqrt {2}}}\left(\int _{0}^{R}\sin {t^{2}}\,dt-\int _{0}^{R}\cos {t^{2}}\,dt\right).\end{matrix}}$

Thus, letting\, $R\to \infty$,\, the equation (2) implies

${\begin{matrix}{\frac {\sqrt {\pi }}{2}}\!+\!0\!-{\frac {1}{\sqrt {2}}}\left(\int _{0}^{\infty }\!\sin {t^{2}}\,dt+\!\int _{0}^{\infty }\!\cos {t^{2}}\,dt\right)\!+\!{\frac {i}{\sqrt {2}}}\left(\int _{0}^{\infty }\!\sin {t^{2}}\,dt-\!\int _{0}^{\infty }\!\cos {t^{2}}\,dt\right)\;=\;0.\end{matrix}}$

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.