We might know the trick to calculate a gaussian integral
$G_2 = \int_{-\infty}^{\infty} e^{-x^2}\,\mathrm{d}x = \sqrt{\pi}$
by instead computing $G^2_2$ and taking advantage of the 2D-plane:
$G_2^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2-y^2}\, \mathrm{d}x \,\mathrm{d}x$ $= \iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,\mathrm{d}(x,y) $
We then transform to polar coordinates to end up with:
$\int_{0}^{\tau}\int_{0}^{\infty}re^{-r^2}\, \mathrm{d}r \,\mathrm{d}\theta$
which is an integral that can be solved. After having solved this integral, the original integral $G_2$ can be obtained by taking the square root
$G_2^2 = \int_{0}^{\tau}\int_{0}^{\infty}r e^{-r^2}\text{ d}r \text{ d}\theta = \frac{1}{2}\int_{0}^{\tau}\int_{0}^{\infty}e^{-s}\text{ d}r \text{ d}\theta = -\frac{1}{2}\tau(-e^\infty+e^0) = \pi \quad \therefore\quad G_2 = \sqrt{\pi}$
where $\tau = 2\pi$
It is a cool trick, but I currently have the integral $G_3 = \int_{-\infty}^{\infty} e^{-x^3} \text{ d}x$, and I was wondering if a similar trick could be used. The trick above makes use of polar coordinates with the pythagorean norm $r^2= x^2+y^2$, but could my integral be solved using a different norm, say $r^3= x^3+y^3$.
In general, could the gaussian integral $G_n$ be solved using $r^n=a^n+b^n$ or using other exotic norms?
(edit: if it can't be used to solve integrals, why would this technique fail with other norms?)