 Solution

Cauchy's integral theorem states that a holomorphic function on a simply connected domain has zero contour integral around every closed piecewise smooth curve in that domain. One proof begins with the Cauchy theorem for a triangle. Subdivide a triangle into four equal triangles. Their oriented boundary integrals sum to the original integral because internal edges cancel, so choose a nested subtriangle carrying at least one quarter of the preceding integral's modulus. At stage $n$, its perimeter and diameter are $P/2^n,d/2^n$, and the nested triangles converge to a point $z_0$. Complex differentiability gives $f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$, with $ϵ(z)\to 0$. The affine terms have primitives and integrate to zero. Therefore $|I_n|\le \sup |ϵ|Pd/4^n$, while $|I_0|\le 4^n|I_n|$. Letting $n\to \infty$ proves $I_0=0$. Triangulation cancels internal edges for polygonal contours; local primitives and deformation through the domain extend the result to the stated piecewise smooth curves. Simple connectedness ensures the necessary contractions do not pass through a missing point.

Apply the theorem to $e^{-a{z}^2}$ on the rectangle between heights zero and $\rho$, with vertical sides at $\mathrm{Re}z=\pm R$. On either vertical side,The modulus of each side integral is at most $|\rho |e^{-a{R}^2+a{\rho }^2}\to 0$. Both horizontal integrals converge absolutely. The two horizontal orientations are opposite, so the limit gives $I(\rho ,a)=I(0,a)$, also for negative $\rho$. This is Gaussian contour translation.

Now complete the square:The substitution $z=x-i\nu /(2a)$ moves the path to height $-\nu /(2a)$. Its Gaussian integral equals the real-axis integral by the preceding result. HenceThis also gives the Fourier transform of a Gaussian in the convention with no prefactor in the forward transform.