A calculus problem by Kushal Bose

$\displaystyle \begin{aligned} I &=\int_{\mathbb{R}^2} e^{-(x^2+xy+y^2)}\;dx\; dy \\ &= \int_{\mathbb{R}^2} e^{-\left(\large{x}+\dfrac{y}{2}\right)^2-\dfrac{3y^2}{4}}\; dx \; dy \\ &= \dfrac{2}{\sqrt{3}} \int_{\mathbb{R}^2} e^{-(u^2+v^2)}\;du\;dv \quad\text{By the transformation } x+\dfrac{y}{2}=u\quad \dfrac{\sqrt{3}}{2}y=v \\ &= \dfrac{2}{\sqrt{3}}\left(\int_{\mathbb{R}}e^{-x^2}\; dx\right)^2 \\ &= \dfrac{2\pi}{\sqrt{3}}\end{aligned}$

Note: The jacobian of the transformation $x+\dfrac{y}{2}=u,\sqrt{3}y=2v$ is as follows :

$\displaystyle J = \begin{vmatrix} 1 & -\dfrac{1}{\sqrt{3}} \\ 0 & \dfrac{2}{\sqrt{3}} \end{vmatrix} = \dfrac{2}{\sqrt{3}}$

Use polar coordinates, $x = r \cos \theta, y = r \sin \theta, |J| = r$ :

$\large \displaystyle = \int_{-\pi}^{\pi} \int_0^{\infty} r e^{-(r^2 + r^2 \cos \theta \sin \theta) } dr d \theta$

$\large \displaystyle = \int_{-\pi}^{\pi} \left [ \frac{ e^{-r^2 ( 1 + \cos \theta \sin \theta) }}{2 (1 + \cos \theta \sin \theta ) } \right ]_{\infty}^0 d \theta$

$\large \displaystyle = \frac12 \int_{-\pi}^{\pi} \frac{1}{1 + \cos \theta \sin \theta} d \theta$

$\large \displaystyle = \frac12 \cdot 2 \cdot \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1 + \cos \theta \sin \theta} \color{#3D99F6} \cdot \frac{\sec^2 \theta} {\sec^2 \theta} \color{#333333} d \theta$

$\large \displaystyle = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2 \theta}{\sec^2 \theta + \tan \theta} d \theta$

$\color{#20A900} \large \displaystyle u = \tan \theta \implies du = \sec^2 \theta d \theta$

$\large \displaystyle = \int_{-\infty}^{\infty} \frac{1}{u^2 + u + 1} du$

$\large \displaystyle = \int_{-\infty}^{\infty} \frac{1}{(u + \frac12)^2 + \frac34} du$

$\color{#20A900} \large \displaystyle v = u + \frac12 \implies dv = du$

$\large \displaystyle = \int_{-\infty}^{\infty} \frac{1}{v^2 + \frac34} dv$

$\large \displaystyle = \frac43 \int_{-\infty}^{\infty} \frac{1}{ (\frac{2}{\sqrt{3}}v)^2 + 1} dv$

$\large \displaystyle = \frac43 \cdot \frac{\sqrt{3}}{2} \cdot \tan^{-1} (\frac{2}{\sqrt{3}}v) \Bigg | _{-\infty}^{\infty}$

$\large \displaystyle = \frac{2 \sqrt{3}}{3} \cdot \left [ \frac{\pi}{2} - \left ( -\frac{\pi}{2} \right ) \right]$

$\large \displaystyle = \frac{2 \pi \sqrt{3}}{3}$

$\color{#20A900} \large \displaystyle = \boxed {\large \displaystyle \frac{2 \pi}{\sqrt{3}} }$

Then:

$\color{#3D99F6} \large \displaystyle a = 2, b = 2, c = 3, \boxed {\large \displaystyle a + b + c = 7 }$