PlanetPhysics/Fresnel Formulae

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos x^{2}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin x^{2}dx=\frac{\sqrt{2\pi }}{4}$

Proof.

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The [[../Bijective/|function]] \,$z\mapsto e^{-z^2}$\, is entire, whence by the fundamental [[../Formula/|theorem]] of complex analysis we have

$\begin{array}{c}{\oint }_{\gamma }{e}^{-z^2}dz=0\end{array}$

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

$\begin{array}{c}\underset{I_1}{\underbrace{\underset{0}{\overset{R}{\int }}{e}^{-x^2}dx}}+\underset{I_2}{\underbrace{\underset{b}{\int }{e}^{-z^2}dz}}+\underset{I_3}{\underbrace{\underset{s}{\int }{e}^{-z^2}dz}}=0\end{array}$

By the entry concerning the Gaussian integral, we know that $\underset{R\to \mathrm{\infty }}{\lim }{I}_1=\frac{\sqrt{\pi }}{2}.$

For handling $I_2$, we use the substitution $z:=R{e}^{i\phi }=R(\cos \phi +i\sin \phi ),dz=iR{e}^{i\phi }d\phi (0\leqq \phi \leqq \frac{\pi }{4}).$ Using also de Moivre's [[../Formula/|formula]] we can write $|I_2|=\left|iR\underset{0}{\overset{\frac{\pi }{4}}{\int }}{e}^{-R^2(\cos 2\phi +i\sin 2\phi )}{e}^{i\phi }d\phi \right|\leqq R\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left|e^{-R^2(\cos 2\phi +i\sin 2\phi )}\right|\cdot \left|e^{i\phi }\right|\cdot |d\phi |=R\underset{0}{\overset{\frac{\pi }{4}}{\int }}{e}^{-R^2\cos 2\phi }d\phi .$ Comparing the [[../Cod/|graph]] of the function \,$\phi \mapsto \cos 2\phi$\, with the line through the points \,$(0,1)$\, and\, $(\frac{\pi }{4},0)$\, allows us to estimate $\cos 2\phi$ downwards: $\cos 2\phi \geqq 1-\frac{4\phi }{\pi }\text{for}0\leqq \phi \leqq \frac{\pi }{4}$ Hence we obtain $|I_2|\leqq R\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{d\phi }{e^{R^2\cos 2\phi }}\leqq R\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{d\phi }{e^{R^2(1-\frac{4\phi }{\pi })}}\leqq \frac{R}{e^{R^2}}\underset{0}{\overset{\frac{\pi }{4}}{\int }}{e}^{\frac{4R^2}{\pi }\phi }d\phi ,$ and moreover $|I_2|\leqq \frac{\pi }{4R{e}^{R^2}}(e^{R^2}-1)<\frac{\pi e^{R^2}}{4R{e}^{R^2}}=\frac{\pi }{4R}\to 0\text{as}R\to \mathrm{\infty }.$ Therefore Failed to parse (syntax error): {\displaystyle \lim_{R\to\infty}I_2 = 0.\\}

Then make to $I_3$ the substitution $z:=\frac{1+i}{\sqrt{2}}t,dz=\frac{1+i}{\sqrt{2}}dt(R\geqq t\geqq 0).$ It yields

$\begin{array}{cc}{I}_3& =\frac{1+i}{\sqrt{2}}\underset{R}{\overset{0}{\int }}{e}^{-i{t}^2}dt=-\frac{1}{\sqrt{2}}\underset{0}{\overset{R}{\int }}(1+i)(\cos t^2-i\sin t^2)dt\\ & =-\frac{1}{\sqrt{2}}(\underset{0}{\overset{R}{\int }}\sin t^{2}dt+\underset{0}{\overset{R}{\int }}\cos t^{2}dt)+\frac{i}{\sqrt{2}}(\underset{0}{\overset{R}{\int }}\sin t^{2}dt-\underset{0}{\overset{R}{\int }}\cos t^{2}dt).\end{array}$

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

$\begin{array}{c}\frac{\sqrt{\pi }}{2}+0-\frac{1}{\sqrt{2}}(\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin t^{2}dt+\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos t^{2}dt)+\frac{i}{\sqrt{2}}(\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin t^{2}dt-\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos t^{2}dt)=0.\end{array}$

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