About the "mixed" form of Gauss and Fresnel integrals

Question

How to integrate the "mixed" form of Gauss and Fresnel integrals as following?

$$\int_{-\infty}^{+\infty} {e^{-x^2-ia(x+b)^2} dx} $$

where $a \in R, b \in R$.

[EDIT]

As Claude Leibovici pointed out, by completing the square in the exponent as following:

$$-x^2-ia(x+b)^2 = -(1+ia)x^2 -i(2ab)x -i(ab^2) \\ = -(1+ia) (x + \frac{iab}{1+ia})^2 + c$$

where $c = -i(ab^2) - \frac{a^2b^2}{1+ia}$, the above integral is reduced to:

$$e^c \int_{-\infty - id}^{+\infty - id} {e^{-(1+ia) y^2} dy}$$

where $d = \frac{ab^2}{1 + a^2}$. This integral is equal to:

$$I_0 \equiv e^c \int_{-\infty}^{+\infty} {e^{-(1+ia) y^2} dy}$$. This is justified by the integration over the following contour to be equal to zero:

$$y \in (-\infty, +\infty) \cup (+\infty, +\infty -id) \cup (+\infty -id, -\infty -id) \cup (-\infty -id, -\infty)$$

The complex Gauss integral $I_0$ is still in the form of "mixed" real Gauss and Fresnel integrals. The question is how to derive and justify that the result can be written as $e^c\sqrt{\frac{\pi}{1+ia}}$?

Answer 1

The standard way to prove the following result is to use the two-variable (x & y) integration, transforming to the polar coordinate.

$$I_C \equiv \int_{-\infty}^{+\infty}{e^{-\alpha x^2}dx} = \sqrt{\frac{\pi}{\alpha}}, \alpha \in C, re(\alpha) > 0$$

The answer to this question is to find a simple way to prove the above result, using the standard Gaussian integral as following.

$$I_G \equiv \int_{=\infty}^{+\infty}{e^{-x^2}dx} = 2 \int_{0}^{+\infty}{e^{-x^2}dx} = \sqrt{\pi}$$

To derive the result for $\large I_C$, first we transform the real variable $\large x$ into the complex variable $\large y \equiv \sqrt{\alpha}x \equiv re^{i\theta}$:

$$I_C =\frac{2}{\sqrt{\alpha}} \int_{y \in \Omega_{\theta}^+} {e^{-y^2}dy} $$

where $\Omega_{\theta}^+ = \{re^{i\theta} | r \in (0, +\infty), \theta = atan(\frac{Im(\sqrt{\alpha})}{Re(\sqrt{\alpha})})\}$.

Now we do the Cauchy contour integration of $\large e^{-y^2}$ over

$$y \in (0, +\infty) \cup (+\infty, +\infty e^{i\theta}) \cup \Omega_{\theta}^+$$

which is equal to zero. The integral over $(+\infty, +\infty e^{i\theta})$ vanishes. This leaves that

$$\int_{0}^{+\infty}{e^{-y^2}dy} - \int_{y \in \Omega_{\theta}^+}{e^{-y^2}dy} = \frac{1}{2}I_G - \frac{\sqrt{\alpha}}{2}I_C = 0$$

Therefore we have proved

$$I_C = \frac{1}{\sqrt{\alpha}} I_G = \sqrt{\frac{\pi}{\alpha}}$$

Answer 2

In general it holds that $$\int_{-\infty}^{\infty} e^{-u^2x^2}dx=\frac{\sqrt{\pi}}{u}$$ Whenever $|\Im u|\leqslant |\Re u|$ (this can be graphically represented as the argument of $u$ being in a sector of at most $\pi/4$ radians.)

You can then integrate through a closed path of the form $[-r,r]$, $[-ru,ru]$ (slanted line) and two circular arcs from $r$ to $ru$ and from $-r$ to $-ru$. The integral over the arcs vanish as $r\to \infty$ because of the condition on the argument of $u$, and the integral over the slanted line is $u$ times the integral we want, while the integral over the $x$-axis is $\sqrt {\pi}$, giving the result.