Gaussian integral with complex coefficients

Question

We are all familiar with the standard generalized Gaussian integral $$\int_{\mathbb R} dx \ \exp\left(-ax^2+bx\right)=\sqrt\frac{\pi}{a}\exp\left(\frac{b^2}{4a}\right) $$ where $a,b\in \mathbb R$. I wonder if there is a general formula for the complex case $$\int_{\mathbb R} dx \ \exp\left(-\alpha x^2+\beta x\right)=(?)$$ for $\alpha,\beta\in \mathbb C$. I have seen a less general version where $\alpha$ is complex with non-zero real part and $\beta$ purely imaginary, but I haven't been able to find a formula for the integral above.

Answer 1

I will consider the case in which $\operatorname{Re}(\alpha)=0$ as the case with real part bigger than zero is equal to the real case, while $\operatorname{Re}(\alpha) <0$ implies that the integral diverges (see e.g this ).

We can rewrite the integrand as

$$ \exp\left(-i a\left(x-\frac{i\beta}{2a}\right)^2-\frac{i\beta^2}{4a}\right) $$ So the starting integral becames, with a substitution $$ I=e^{-\frac{i\beta^2}{4a}}\int_\mathbb{R} e^{-ia y^2} dy=2e^{-\frac{\beta^2}{4\alpha^2}}\int_0^\infty e^{-ia y^2} dy $$ Suppose $a<0$ and consider the integral $$ 0=\int_C e^{-iaz^2} dz=\int_0^R e^{-ia y^2} dy+\int_S e^{-iaz^2} dz+\int_D e^{-iaz^2} dz $$ Where $C$ is the path in the picture The integral is zero because the function is holomorphic,

Over $D$: $z=r e^{i \frac{\pi}{4}}$ and so $dz=e^{i \frac{\pi}{4}} dr$

So $$ \int_D e^{-iaz^2} dz=e^{i\frac{\pi}{4}}\int_R^0 e^{ar^2}dr $$

Over $S$: $$ \lim_{R \to \infty}\left\| \int_S e^{-iaz^2} dz\right\|\le\lim_{R \to \infty} \int_0^{\frac{\pi}{4}} \|\exp(-ia(r e^{i\theta})^2) i R e^{i \theta}\|d \theta= \lim_{R \to \infty} R \int_0^{\frac{\pi}{2}} \exp(a R^2 \sin(\theta')) d \theta' $$ As $a<0$ is it possible to show that such limit converges to zero (see e.g this )

So in the end $$ \int_0^\infty e^{-ia y^2}=e^{i\frac{\pi}{4}} \int_0^\infty e^{ar^2} dr=e^{i\frac{\pi}{4}}\sqrt{\frac{\pi}{4|a|}} $$

If $a>0$ you can repeat the process by taking $C$ the sector in the quadrant $\operatorname{Re}(z)<0$, $\operatorname{Im}(z)<0$. You will obtain $$ \int_0^\infty e^{-ia y^2}=e^{-i\frac{\pi}{4}}\sqrt{\frac{\pi}{4a}} $$ So in definite the integral, for $\alpha=ia$, is $$ I= \exp\left(-\frac{i\beta^2}{4 a}-i \operatorname{sgn}(a)\frac{\pi}{4}\right) \sqrt{\frac{\pi}{|a|}} $$

EDIT this is taken by some handwritten notes i have. I found this that is very similar to the source of these notes ( that i was not able to found and prbably was in italian)

PS : this wikipedia article that i have quoted overlooks the fact that the square root of a complex number has $2$ values (it talk about it in the linked article about Fresnel integrals), this is the reason for which all the machinery I wrote is necessary.