Fresnel formulae result

The Fresnel formulae presented in the parent entry- that the proof refers to- are as follows:

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

(visible there only in the Tex code mode because of a current quirk/glitch with pstricks vs. Tex).

The remainder of the equations Tex mode correctly specified by pahio in the parent entry entitled Fresnel formulae is precisely quoted here as follows: “The function $$z \mapsto e^{-z^2}$$ is entire, whence by the fundamental theorem of complex analysis we have

$$\oint_\gamma e^{-z^2}\,dz \;=\; 0$$ (1)

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

$$\underbrace{\int_0^R\!e^{-x^2}\,dx}_{I_1}+\underbrace{\int_b\!e^{-z^2}\,dz}_{I_2} +\underbrace{\int_s\!e^{-z^2}\,dz}_{I_3} \,=\,0$$ (2)

By the entry concerning the Gaussian integral, we know that

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

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

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

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:

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

Hence we obtain

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

and moreover

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

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

$$I_3 \quad = \frac{1\!+\!i}{\sqrt{2}}\int_R^0e^{-it^2}\,dt = -\frac{1}{\sqrt{2}}\int_0^R(1+i)(\cos{t^2}-i\sin{t^2})\,dt$$

$$\quad = -\frac{1}{\sqrt{2}}\left(\int_0^R\sin{t^2}\,dt+\int_0^R\c... ...t) +\frac{i}{\sqrt{2}}\left(\int_0^R\sin{t^2}\,dt-\int_0^R\cos{t^2}\,dt\right).$$

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

$$\frac{\sqrt{\pi}}{2}\!+\!0\! -\frac{1}{\sqrt{2}}\left(\int_0^\inf... ...eft(\int_0^\infty\!\sin{t^2}\,dt-\!\int_0^\infty\!\cos{t^2}\,dt\right) \;=\; 0.$$ (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

$$\frac{\sqrt{\pi}}{2}+0-\frac{1}{\sqrt{2}}\!\cdot\!2\!\int_0^\infty\!\sin{t^2}\,dt \,=\, 0.$$

So we get also the result $\int_0^\infty\sin{x^2}\,dx = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{4}.$ "

Contributors to this entry (in most recent order):

• bci1

As of this snapshot date, this entry was owned by bci1.