= Solution
The integrand
$$
F(z)=\frac{e^{iz^2/(4\pi)}}{e^{z/2}-e^{-z/2}}
$$
has one pole inside the rectangle, at $z=0$, with residue $1$. The <residue theorem> therefore gives $\int_CF(z)\,dz=2\pi i$. The vertical-side integrals tend to zero as $R\to\infty$.
On the horizontal sides, direct substitution $z=x\pm\pi i$, with the upper side oppositely oriented, shows that their sum is
$$
i e^{-i\pi/4}\int_{-R}^{R}e^{ix^2/(4\pi)}
\left(\frac1{1+e^x}+\frac{e^x}{1+e^x}\right)dx.
$$
Taking the limit and comparing with $2\pi i$ gives
$$
\boxed{\lim_{R\to\infty}\int_{-R}^{R}e^{ix^2/(4\pi)}\,dx
=2\pi e^{i\pi/4}.}
$$
Back to article page
