The goal is to have an explicit formula for the following integral: $$\int_{\mathbb{R}^2}\exp({x^t N x+ x^tAy}) dx$$ where $x \in \mathbb{R}^2$ is the variable ($x^t$ is the transpose), $y \in \mathbb{C}^2$ is fixed, $N \in Mat(2x2,\mathbb{C})$ is a symmetric matrix with negative definite real part, and $A \in GL(2,\mathbb{R})$.
My first thought was to use a square root of $N$, change variable and complete the square, but the existence/uniqueness of $\sqrt{N}$ is not guaranteed. I would expect the integral to converge because the real part of $N$ is negative definite and the integral of the imaginary part oscillates quickly enough.
Does anyone have any hint/solution? Thanks.
This is equivalent to $$I=\int_{R^{2}}e^{-x^{T}Nx+x^{T}Ay}$$ Where $N$ has a positive definite real part. As $N$ is symmetric one may orthogonally diagonalize it $N=O\Lambda{O}^{T}$. You let $x=Oz$ and $s=O^{T}Ay$ and the integral becomes $$I=\int_{R^{2}}e^{-z^{T}\Lambda{z}+z^{T}s}\det{O}dz=\int_{R^{2}}e^{-z^{T}\Lambda{z}+z^{T}s}dz=\int_{R}e^{-\lambda_{1}z_{1}^{2}+s_{1}z_{1}}dz_{1}\int_{R}e^{-\lambda_{2}z_{2}^{2}+s_{2}z_{2}}dz_{2}$$ $$=\frac{\pi}{\sqrt{\lambda_{1}\lambda_{2}}}e^{\frac{s_{1}^{2}}{4\lambda_{1}}+\frac{s_{2}^{2}}{4\lambda_{2}}}=\frac{\pi}{\sqrt{\det{N}}}e^{\frac{1}{4}s^{T}\Lambda^{-1}s}=\frac{\pi}{\sqrt{\det{N}}}e^{\frac{1}{4}y^{T}A^{T}O\Lambda^{-1}O^{T}Ay}=$$ $$=\frac{\pi}{\sqrt{\det{N}}}e^{\frac{1}{4}y^{T}A^{T}N^{-1}Ay}$$ And here, the fact that $$\int_{\mathbb{R}}e^{-az^{2}+bz}dz=\sqrt{\frac{\pi}{a}}e^{\frac{b^{2}}{4a}}$$ was used.