# Double Integral Over a Square Region Centered at the Origin Replace $$c$$ with $$d$$ in the figure. I'm trying to evaluate the integral

$$\begin{equation} \mathcal{I}=\int_{-d}^{d}\int_{-d}^{d}\exp\left(-ax^2-bxy-cy^2\right)dxdy \end{equation}$$

If I use polar coordinates transformation, $$x=r\cos x,$$ $$y=r\sin y,$$ we have four double integrals to evaluate over the regions $$\begin{equation} \theta\in [0,\frac{\pi}{4}], \theta\in (\frac{\pi}{4},\frac{\pi}{2}],...,\theta\in (\frac{7\pi}{4},2\pi]. \end{equation}$$ Note that the the region in the second quadrant will yield the same result as the fourth quadrant region. The integral setup for half of the first quadrant is $$\begin{equation} \mathcal{I}_{1}=\int_{0}^{\frac{\pi}{4}}d\theta\int_{0}^{\frac{d}{\cos \theta}}r\exp\left(-ar^2\cos^2\theta-br^2\sin\theta\cos\theta-cr^2\sin^2\theta\right)dr \end{equation}$$

I would appreciate approximate analytical expressions

As it is, we can compute the inner integral (this is the gaussian integral) and face the problem of $$\frac{\sqrt{\pi }}{2 \sqrt{a}}\int_{-d}^{+d}e^{\frac{ \left(b^2-4 a c\right)}{4 a}y^2} \left(\text{erf}\left(\frac{2 a d-b y}{2 \sqrt{a}}\right)+\text{erf}\left(\frac{2 a d+b y}{2 \sqrt{a}}\right)\right)\,dy$$ which does not seem to be possible.
• @Richard.Same problem : the inner integral does not make any problem. It is given by $$\frac{1-\exp \left(-\frac{1}{2} d^2 \sec ^2(\theta ) ((a+c)+(a-c) \cos (2 \theta )+b \sin (2 \theta ))\right)}{(a+c)+(a-c) \cos (2 \theta )+b \sin (2 \theta )}$$ which cannot be integrated wrt to $\theta$ in the general case. Apr 18, 2021 at 4:51