Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 4 / a / Solution

The function $e^{-z^2}$ is entire, so Cauchy's integral theorem makes its contour integral around the rectangle vanish. The lower horizontal integral tends to $\sqrt{\pi }$ by the Gaussian integral. On a vertical side $z=\pm R+it$, with $t$ between $0$ and $b$, its modulus is $e^{-R^2+t^2}\le e^{-R^2+b^2}$; hence both vertical integrals tend to zero. The upper side has the opposite orientation and contributesConsequently the integral in this expression is $\sqrt{\pi }{e}^{-b^2}$. Its imaginary part vanishes because $e^{-x^2}\sin (2bx)$ is odd, and its real part is twice the positive-half-line integral because the integrand there is even. ThereforeFor negative $b$ the rectangle lies below the real axis and the same cancellation holds; alternatively use the even dependence on $b$.