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[parent] Fresnel formulae (Theorem)

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

Proof.


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

$\displaystyle \oint_\gamma e^{-z^2}\,dz \;=\; 0$ (1)
where $\gamma$ is the perimeter of the circular sector described in the picture.  We split this contour integral to three portions:
$\displaystyle \underbrace{\int_0^R\!e^{-x^2}\,dx}_{I_1}+\underbrace{\int_b\!e^{-z^2}\,dz}_{I_2} +\underbrace{\int_s\!e^{-z^2}\,dz}_{I_3} \,=\,0$ (2)
By the entry concerning the Gaussian integral, we know that

$\displaystyle \lim_{R\to\infty}I_1 = \frac{\sqrt{\pi}}{2}.$

For handling $I_2$, 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

$\displaystyle \vert I_2\vert = \left\vert iR\int_0^{\frac{\pi}{4}}e^{-R^2(\cos2... ...dot\vert d\varphi\vert = R\!\int_0^{\frac{\pi}{4}}e^{-R^2\cos2\varphi}d\varphi.$
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:

$\displaystyle \cos2\varphi \geqq 1\!-\!\frac{4\varphi}{\pi}$   for$\displaystyle \quad 0 \leqq \varphi \leqq \frac{\pi}{4}$
Hence we obtain

$\displaystyle \vert I_2\vert \leqq R\int_0^{\frac{\pi}{4}}\frac{d\varphi}{e^{R^... ... \frac{R}{e^{R^2}} \int_0^{\frac{\pi}{4}} e^{\frac{4R^2}{\pi}\varphi} d\varphi,$
and moreover

$\displaystyle \vert I_2\vert \leqq \frac{\pi}{4Re^{R^2}}(e^{R^2}-1) < \frac{\pi e^{R^2}}{4Re^{R^2}} = \frac{\pi}{4R} \; \to 0$   as$\displaystyle \quad R \to \infty.$
Therefore

$\displaystyle \lim_{R\to\infty}I_2 = 0.\\ $

Then make to $I_3$ 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
$\displaystyle I_3$ $\displaystyle \quad = \frac{1\!+\!i}{\sqrt{2}}\int_R^0e^{-it^2}\,dt = -\frac{1}{\sqrt{2}}\int_0^R(1+i)(\cos{t^2}-i\sin{t^2})\,dt$    
  $\displaystyle \quad = -\frac{1}{\sqrt{2}}\left(\int_0^R\sin{t^2}\,dt+\int_0^R\c... ...t) +\frac{i}{\sqrt{2}}\left(\int_0^R\sin{t^2}\,dt-\int_0^R\cos{t^2}\,dt\right).$    
Thus, letting  $R \to \infty$,  the equation (2) implies
$\displaystyle \frac{\sqrt{\pi}}{2}\!+\!0\! -\frac{1}{\sqrt{2}}\left(\int_0^\inf... ...eft(\int_0^\infty\!\sin{t^2}\,dt-\!\int_0^\infty\!\cos{t^2}\,dt\right) \;=\; 0.$ (3)
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

$\displaystyle \frac{\sqrt{\pi}}{2}+0-\frac{1}{\sqrt{2}}\!\cdot\!2\!\int_0^\infty\!\sin{t^2}\,dt \,=\, 0.$
So we get also the result  $\int_0^\infty\sin{x^2}\,dx = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{4}$,  Q.E.D.



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