Fresnel formulae result | Physics Library

The Fresnel formulae presented in the parent entry are

∫ ∫ √ --- ∞ 2 ∞ 2 --2π- cos(x ) dx = sin(x )dx = 4 . 0 0

The proof in the parent entry Fresnel formulae uses the entire function

−z2 z ↦− → e

and integrates it around a circular sector of angle π∕4.

Let

∫ R 2 ∫ 2 ∫ 2 I1 = e−x dx, I2 = e −z dz, I3 = e−z dz, 0 b s

where b is the circular arc and s is the radial segment from Re^iπ∕4 back to the origin. By Cauchy’s integral theorem,

I + I + I = 0. 1 2 3

From the Gaussian integral,

√-- -π-- lRim→∞ I1 = 2 .

For the circular arc, write

z = Reiφ, dz = iReiφ dφ, 0 ≤ φ ≤ π. 4

Then

∫ π∕4 2 |I2| ≤ R e−R cos(2φ)dφ. 0

Since

4φ ( π) cos(2φ ) ≥ 1 − --- 0 ≤ φ ≤ -- , π 4

we obtain

∫ π∕4 2 4φ π ( 2) π |I2| ≤ R e− R (1− π )dφ = --- 1 − e−R < ---. 0 4R 4R

Therefore

lim I = 0. R→ ∞ 2

For the radial segment, use

z = 1 +√-it, dz = 1 +√-idt, R ≥ t ≥ 0. 2 2

Because z² = it²,

2 2 e−z = e−it = cos(t2) − i sin(t2).

Thus

Letting R →∞ in

I1 + I2 + I3 = 0

gives

√ -- ( ∫ ∞ ∫ ∞ ) (∫ ∞ ∫ ∞ ) --π-− √1-- cos(t2) dt + sin (t2)dt + √i-- sin (t2)dt − cos(t2)dt = 0. 2 2 0 0 2 0 0

The imaginary part yields

∫ ∞ ∫ ∞ cos(t2)dt = sin(t2)dt. 0 0

The real part then gives

√ -- ∫ ∞ --π- √ -- 2 2 − 2 0 sin(t) dt = 0,

and hence

∫ ∫ √ -- √ --- ∞ 2 ∞ 2 π 2π sin(x ) dx = cos(x ) dx = -√---= --4--. 0 0 2 2

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