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
]
Best Answer
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.