Double indefinite integrals! How to evaluate this indefinite integral? I tried by polar coordinate or substitution, but I couldn't go through it. Is there any help please?
$$\int e ^{-(x^4 + y^4)} dxdy $$
 A: The definite integral over the entire plane can be computed in terms of the Gamma function:
$$
\begin{align}
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^4-y^4}\,\mathrm{d}x\,\mathrm{d}y
&=\int_{-\infty}^\infty\int_{-\infty}^\infty  e^{-x^4}e^{-y^4}\,\mathrm{d}x\,\mathrm{d}y\\
&=\int_{-\infty}^\infty e^{-x^4}\,\mathrm{d}x\int_{-\infty}^\infty e^{-y^4}\,\mathrm{d}y\\
&=\left(\int_{-\infty}^\infty e^{-x^4}\,\mathrm{d}x\right)^2\\
&=\left(2\int_0^\infty e^{-x^4}\,\mathrm{d}x\right)^2\\
&=4\left(\int_0^\infty e^{-x^4}\,\mathrm{d}x\right)^2\\
&=4\left(\frac14\int_0^\infty e^{-u}u^{-3/4}\,\mathrm{d}u\right)^2\\
&=\frac14\Gamma\left(\frac14\right)^2
\end{align}
$$
The indefinite integral in two dimensions is a function of the domain, not like the anti-derivative in one dimension, where the domain is usually an interval. I am not sure what you are looking for when you ask for the indefinite integral on $\mathbb{R}^2$.
A: $\displaystyle\int_0^\infty e^{-x^n}dx=\Big(\!\tfrac1n\!\Big)!\quad$ However, the indefinite integral does not possess a closed form expression.
See Liouville's theorem and the Risch algorithm for more details.
