 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-1/13a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The integrand

F (z) = \frac{e ^{i z^{2} / (4 π)}}{e ^{z /2} - e ^{- z /2}}

has one pole inside the rectangle, at

z = 0

1

. The residue theorem therefore gives

\int_{C} F (z) d z = 2 πi

. The vertical-side integrals tend to zero as

R \to \infty

On the horizontal sides, direct substitution

z = x \pm πi

, with the upper side oppositely oriented, shows that their sum is

i e^{- iπ /4} \int_{- R}^{R} e^{i x^{2} / (4 π)} (\frac{1}{1 + e ^{x}} + \frac{e ^{x}}{1 + e ^{x}}) d x .

Taking the limit and comparing with

2 πi

gives

R \to \infty lim \int_{- R}^{R} e^{i x^{2} / (4 π)} d x = 2 π e^{iπ /4} .

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