Line contour integral of complex Gaussian

Question

Say I have the entire function $$f(z)=e^{-\frac{1}{2}z^2}.$$ I would like to consider the integral $$I=\int_\Gamma f(x)dz,$$ where $\Gamma$ is a line with negative slope $<1$ in $\mathbb{C}$ (so if you plot it, it look like $Im(z)=cRe(z)$, $-1<c<0$). Though solvers tell me that $I=\sqrt{2\pi}$ and I can intuitively see why this is true (continuously deform $\Gamma$ into the real line), how would I show this rigorously?

Answer 1

Consider the parts in the left and right half-plane separately - or, by the parity of the integrand, consider only one of the two parts, the other one follows by symmetry.

The slanted line can be parameterised as $t\cdot e^{-i\varphi}$, with $0 < \varphi < \frac{\pi}{4}$. By Cauchy's integral theorem, for every $R > 0$ we have

$$0 = \int_{0}^R \exp\left(-\frac{1}{2}(te^{-i\varphi})^2\right)e^{-i\varphi}\,dt + \int_{-\varphi}^0 \exp\left(-\frac{1}{2}(Re^{i\vartheta})^2\right)iRe^{i\vartheta}\,d\vartheta - \int_0^R e^{-\frac{1}{2}t^2}\,dt.$$

On the circular arc, we can bound the integrand by

$$R\exp\left(-\frac{1}{2}R^2\cos (2\varphi)\right)$$

since $\operatorname{Re} e^{2i\vartheta} = \cos (2\vartheta) \geqslant \cos (2\varphi)$ for $-\varphi \leqslant \vartheta \leqslant 0$, so

$$\left\lvert \int_{-\varphi}^0 \exp\left(-\frac{1}{2}(Re^{i\vartheta})^2\right)iRe^{i\vartheta}\,d\vartheta \right\rvert \leqslant \varphi R\exp\left(-\frac{1}{2}R^2\cos (2\varphi)\right) \xrightarrow{R\to+\infty} 0,$$

and hence

$$\lim_{R\to+\infty} \int_{0}^R \exp\left(-\frac{1}{2}(te^{-i\varphi})^2\right)e^{-i\varphi}\,dt = \lim_{R\to+\infty} \int_0^R e^{-\frac{1}{2}t^2}\,dt = \frac{\sqrt{2\pi}}{2}.$$

Aside from the integral theorem, you need an estimate that shows the integral over the path connecting the two straight line segments tends to $0$ when the segments exhaust the two half-lines.