Evaluating $I(a,n) = \int^{\infty}_{-\infty}{e^{iax - nx^2}}\, dx$ for real $a$ and real $n > 0$ with Jordan's Lemma

Question

Given that $e^{-nx^2}$ does not have any singularities, I believe $I(a,n) = 0$. Is this a correct application of Jordan's Lemma?

$$I(a,n) = \int^{\infty}_{-\infty}{e^{iax - nx^2}}\, dx$$

Answer 1

No, that is not correct. Jordan's lemma doesn't apply, since the $e^{-nz^2}$ part of the integrand is unbounded in both the upper and the lower half plane. In fact, it grows super-exponentially for $\lvert \operatorname{Im} z\rvert \to \infty$; We have $\left\lvert e^{-nz^2}\right\rvert = e^{n\left((\operatorname{Im} z)^2 - (\operatorname{Re} z)^2\right)}$.

We can evaluate the integral in many ways, however. For $a = 0$, the well-known

$$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$$

yields $I(0,n) = \sqrt{\dfrac{\pi}{n}}$ via the substitution $y = \sqrt{n}\cdot x$. Then we can evaluate $I(a,n)$ by noting that Cauchy's integral theorem allows shifting the contour of integration, which gives us

$$\begin{align} I(a,n) &= \int_{-\infty}^\infty e^{iax - nx^2}\,dx\tag{complete the square}\\ &= \int_{-\infty}^\infty e^{-n (x-ia/(2n))^2-a^2/(4n)}\,dx\\ &= e^{-a^2/(4n)}\int_{-\infty}^\infty e^{-n(x-ia/(2n))^2}\,dx \tag{shift contour}\\ &= e^{-a^2/(4n)}\int_{-\infty + ia/(2n)}^{\infty+ia/(2n)} e^{-n(z-ia/(2n))^2}\,dz \tag{$z-ia/(2n) = \operatorname{Re} z$}\\ &= e^{-a^2/(4n)} \int_{-\infty}^\infty e^{-nx^2}\,dx\\ &= e^{-a^2/(4n)} \sqrt{\frac{\pi}{n}}. \end{align}$$