Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 2 / 13D (Complex Analysis or Complex Methods)

The denominator vanishes when $z=(2j+1)i\pi /2$. For $R>0$, only $z_0=i\pi /2$ lies inside the rectangle. This is a simple pole, since the denominator derivative there is $2$. Its residue isThe positive orientation and the residue theorem consequently give

On the top side put $z=x+i\pi$, with $x$ running from $R$ to $-R$. Since $e^{i(x+i\pi {)}^2/\pi }=-e^{i{x}^2/\pi }{e}^{-2x}$ and $e^{-2(x+i\pi )}=e^{-2x}$, the reversed orientation cancels the minus sign. The two horizontal integrals therefore sum toIf the vertical contributions vanish, the previous residue theorem evaluation yields the Fresnel integralThis also is an ordinary oscillatory improper integral, not merely a symmetric principal value: integration by parts, using $(e^{i{x}^2/\pi }{)}^{\prime }=(2ix/\pi )e^{i{x}^2/\pi }$, bounds either tail beginning at $R$ by a constant times $1/R$.

For $z=R+iy$, $0\le y\le \pi$, the numerator modulus is $e^{-2Ry/\pi }$ and the denominator modulus is at least $1-e^{-2R}$. Hence the magnitude of the right-side contour integral is at mostOn $z=-R+iy$ the corresponding bounds are $e^{2Ry/\pi }$ and $e^{2R}-1$, giving exactly the same bound $\pi /(2R)$. Thus the total vertical contribution is $O(R^{-1})$ and vanishes. This integral estimate handles the endpoints where a pointwise uniform-decay argument would fail.