# Double Integral Over a Square Region Centered at the Origin

Replace $$c$$ with $$d$$ in the figure. I'm trying to evaluate the integral

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

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

I would appreciate approximate analytical expressions

I do not know what polar coordinates would give since you did not show your results and attempts.

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.

• Thank you. yeah the error function is a tough nut to crack...worst nightmare for definite integral. Anyway, I made edits for the polar coordinates case...any ideas on how to proceed? Apr 17, 2021 at 19:16
• @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