When dealing with coupled variables, one can uncouple them using the trick of representing them as a Gaussian integral:
$$\exp \left\{\frac{b^{2}}{4a}\right\}=\int \frac{\mathrm{d} x}{\sqrt{2 \pi / a}} \exp \left\{-a x^{2} \pm b x\right\}$$
Similarly, when $b$ is complex, we can still linearise using a complex Gaussian:
$$\exp \left(\frac{|b|^{2}}{a}\right)=\int \frac{\mathrm{d} z \mathrm{d} \bar{z}}{2 \mathrm{i} \pi a} \exp \left \{-a \bar{z} z-\bar{b} z-b \bar{z}\right \}$$
However $\text{Re}[a]>0$. I have: $\exp\left(-|b|^2\right)$. Is it impossible to represent it as a gaussian integral?
(Note, I am trying to reproduce the results of Sommers et al. But it seems very unclear how they performed their integrals…cf Sommers, H. J., Crisanti, A., Sompolinsky, H., & Stein, Y. (1988). Spectrum of large random asymmetric matrices. Physical review letters, 60(19), 1895.)
Best Answer
Since we know that $1=\int \frac{du}{\sqrt{\pi}}\exp(-u^{2})$, the negative version of the first case should be done by adding a complex number to the variable, the shift has to be done carefully and more can be read here for example but we will proceed boldly
$ 1 =\int \frac{du}{\sqrt{\pi}} \exp(-u^{2}) =_{(u=\sqrt{a}x\pm i\frac{b}{2\sqrt{a}})}\int \frac{dx}{\sqrt{\pi/a}}\exp(-ax^{2}\pm ibx+\frac{b^2}{4a}) =\exp(\frac{b^2}{4a})\int \frac{dx}{\sqrt{\pi/a}}\exp(-ax^{2}\pm ibx) $
$$\therefore \exp(-\frac{b^2}{4a})=\int\frac{dx}{\sqrt{\pi/a}}\exp(-ax^{2}\pm ibx)$$
Notice that if we would have done the change of variable $u=\sqrt{a}x\pm \frac{b}{2\sqrt{a}}$ we would have got $\exp(\frac{b^2}{4a})=\int\frac{dx}{\sqrt{\pi/a}}\exp(-ax^{2}\pm bx)$ so there is an extra factor of 2 in your formula as I see it, also if we change the constant as $a=\frac{1}{2c}$ we should get $\exp(-\frac{cb^2}{2})=\int\frac{dx}{\sqrt{2c\pi}}\exp(-\frac{x^{2}}{2c}\pm ibx)$ which is know as the Hubbard-Stratonovich transformation
Asumming the formula for the second case is correct then $b=0,a=1$ gives $1=\int \frac{dz d\bar{z}}{2i\pi}\exp(-|z|^2)$ and we can use it as a starting point to get the negative version of this case similarly as we did before
$1 =\int \frac{du d\bar{u}}{2i\pi}\exp(-|u|^2) =_{(u=\sqrt{a}z\pm i\frac{b}{\sqrt{a}})}\int\frac{dz d\bar{z}}{2i\pi a}\exp(-|\sqrt{a}z\pm i\frac{b}{\sqrt{a}}|^2) =\exp(\frac{|b|^2}{4a})\int\frac{dz d\bar{z}}{2i\pi a}\exp(-az\overline{z}\pm i\overline{b}z\pm ib\overline{z}) $
$$\therefore \exp(-\frac{|b|^2}{4a})=\int\frac{dz d\bar{z}}{2i\pi a}\exp(-az\overline{z}\pm i\overline{b}z\pm i b\overline{z})$$