 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-1/13f/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 13F / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Integrate the entire function

e^{i z^{2}}

around the positively oriented sector boundary: the real segment

0 \to R

, the arc

R e^{i θ}

for

0 \leq θ \leq π /4

, and the radial segment back to zero. Cauchy integral theorem gives

\int_{0}^{R} e^{i x^{2}} d x + \int_{arc} e^{i z^{2}} d z - e^{iπ /4} \int_{0}^{R} e^{- r^{2}} d r = 0.

The last term follows from

z = r e^{iπ /4}

and

i z^{2} = - r^{2}

; its minus sign is the return orientation.

On the arc,

∣ e^{i z^{2}} ∣ = e^{- R^{2} s i n (2 θ)}

. The supplied sine bound gives

sin (2 θ) \geq 4 θ / π

throughout this sector, so

\int_{arc} e^{i z^{2}} d z \leq R \int_{0}^{π /4} e^{- 4 R^{2} θ / π} d θ = \frac{π}{4 R} (1 - e^{- R^{2}}) ⟶ 0.

The Gaussian integral then yields

\int_{0}^{\infty} e^{i x^{2}} d x = e^{iπ /4} π /2

. Taking real and imaginary parts proves both Fresnel integrals:

\int_{0}^{\infty} cos x^{2} d x = \int_{0}^{\infty} sin x^{2} d x = \frac{π}{8} .

The contour estimate proves existence of the improper oscillatory integrals as well as their values; absolute convergence is neither asserted nor required.

 Read the full article

 New to topics? Read the docs here!