# PlanetPhysics/Fresnel Formulae

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

Proof.

\begin{pspicture}(-1,-1)(6.2,4.2) \psaxes[Dx=10,Dy=10]{->}(0,0)(-1,-1)(6,4) \rput(-0.2,4.1){${\displaystyle y}$} \rput(6.1,-0.2){${\displaystyle x}$} \rput(-0.17,-0.22){0} \rput(5,-0.3){${\displaystyle R}$} \rput(0.7,0.3){${\displaystyle {\frac {\pi }{4}}}$} \rput(1.4,1.7){${\displaystyle s}$} \rput(4.7,2.2){${\displaystyle b}$} \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}

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

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

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

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

By the entry concerning the Gaussian integral, we know that ${\displaystyle \lim _{R\to \infty }I_{1}={\frac {\sqrt {\pi }}{2}}.}$

For handling ${\displaystyle I_{2}}$, we use the substitution ${\displaystyle z\,:=\,Re^{i\varphi }=R(\cos \varphi +i\sin \varphi ),\quad dz\,=\,iRe^{i\varphi }\,d\varphi \quad (0\leqq \varphi \leqq {\frac {\pi }{4}}).}$ Using also de Moivre's formula we can write ${\displaystyle |I_{2}|=\left|iR\int _{0}^{\frac {\pi }{4}}e^{-R^{2}(\cos 2\varphi +i\sin 2\varphi )}e^{i\varphi }d\varphi \right|\leqq R\!\int _{0}^{\frac {\pi }{4}}\left|e^{-R^{2}(\cos 2\varphi +i\sin 2\varphi )}\right|\cdot \left|e^{i\varphi }\right|\cdot |d\varphi |=R\!\int _{0}^{\frac {\pi }{4}}e^{-R^{2}\cos 2\varphi }d\varphi .}$ Comparing the graph of the function \,${\displaystyle \varphi \mapsto \cos 2\varphi }$\, with the line through the points \,${\displaystyle (0,\,1)}$\, and\, ${\displaystyle ({\frac {\pi }{4}},\,0)}$\, allows us to estimate ${\displaystyle \cos 2\varphi }$ downwards: ${\displaystyle \cos 2\varphi \geqq 1\!-\!{\frac {4\varphi }{\pi }}\quad {\mbox{for}}\quad 0\leqq \varphi \leqq {\frac {\pi }{4}}}$ Hence we obtain ${\displaystyle |I_{2}|\leqq R\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{e^{R^{2}\cos 2\varphi }}}\leqq R\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{e^{R^{2}(1-{\frac {4\varphi }{\pi }})}}}\leqq {\frac {R}{e^{R^{2}}}}\int _{0}^{\frac {\pi }{4}}e^{{\frac {4R^{2}}{\pi }}\varphi }d\varphi ,}$ and moreover ${\displaystyle |I_{2}|\leqq {\frac {\pi }{4Re^{R^{2}}}}(e^{R^{2}}-1)<{\frac {\pi e^{R^{2}}}{4Re^{R^{2}}}}={\frac {\pi }{4R}}\;\to 0\quad {\mbox{as}}\quad R\to \infty .}$ Therefore $\displaystyle \lim_{R\to\infty}I_2 = 0.\\$

Then make to ${\displaystyle I_{3}}$ the substitution ${\displaystyle z\;:=\;{\frac {1\!+\!i}{\sqrt {2}}}t,\quad dz\,=\,{\frac {1\!+\!i}{\sqrt {2}}}dt\quad (R\geqq t\geqq 0).}$ It yields

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

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

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

Because the imaginary part vanishes, we infer that\, ${\displaystyle \int _{0}^{\infty }\cos {x^{2}}\,dx=\int _{0}^{\infty }\sin {x^{2}}\,dx}$,\, whence (3) reads ${\displaystyle {\frac {\sqrt {\pi }}{2}}+0-{\frac {1}{\sqrt {2}}}\!\cdot \!2\!\int _{0}^{\infty }\!\sin {t^{2}}\,dt\,=\,0.}$ So we get also the result\, ${\displaystyle \int _{0}^{\infty }\sin {x^{2}}\,dx={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {\pi }}{2}}={\frac {\sqrt {2\pi }}{4}}}$,\, Q.E.D.