Fresnel formulae | Physics Library

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

Proof.

Quarter-sector contour used in the proof.

−z2 z ↦− → e

is entire. Hence, by Cauchy’s integral theorem,

∮ − z2 e dz = 0, γ

where γ is the positively oriented boundary of the circular sector shown above. Split the contour integral into three parts:

∫ R ∫ ∫ e−x2 dx+ e−z2 dz + e−z2 dz = 0. 0--- ---- b-- --- s-- --- ◟ ◝I◜1 ◞ ◟ ◝I◜2 ◞ ◟ ◝I◜3 ◞

From the Gaussian integral,

√-- π lRim→∞ I1 = ---. 2

For I₂, parameterize the circular arc by

z = Reiφ, dz = iReiφ dφ, 0 ≤ φ ≤ π. 4

Then

∫ π∕4 |I| ≤ R e−R2cos(2φ)dφ. 2 0

On the interval 0 ≤ φ ≤ π∕4, the function cos(2φ) lies above the chord joining (0, 1) and (π∕4, 0), so

4φ- cos(2φ) ≥ 1 − π .

Therefore

∫ π∕4 − R2 1− 4φ π ( 2) π |I2| ≤ R e ( π )dφ = --- 1 − e−R < ---. 0 4R 4R

Hence

Rli→m∞ I2 = 0.

For I₃, parameterize the diagonal segment by

iπ∕4 1 +√-i 1 +√-i z = e t = 2 t, dz = 2 dt, R ≥ t ≥ 0.

Since z² = it²,

e−z2 = e−it2 = cos(t2) − i sin(t2).

Thus

Letting R →∞ in

I1 + I2 + I3 = 0

gives

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

The imaginary part must vanish, so

∫ ∞ ∫ ∞ 2 2 0 cos(t )dt = 0 sin(t )dt.

Using the real part then gives

√ π- √ --∫ ∞ ----− 2 sin(t2) dt = 0. 2 0

Therefore

∫ ∞ ∫ ∞ √ -- √ --- sin(x2) dx = cos(x2) dx = -√-π-= --2π-. 0 0 2 2 4

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