How to integrate $e^{-x^2 -y^2}$? How do you compute the following integral:$e^{-x^2 -y^2}$? Wolfram already gives the answer as: $\frac{1}{2}  \sqrt{\pi} e^{-y^2}  \text{erf}(x) + C$, but I have no idea how to get there. I tried using double integration, and all I ended up with $\frac{1}{4}xy * e^{(-x^2)-y^2}$.
Wolfram Alpha Calculation
 A: If you want to integrate over both $x$ and $y$, then you can convert it to an integral in polar coordinates  $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}\ dx \ dy=
\int_{0}^{+\infty}\int_0^{2\pi}e^{-r^2}\ r\ d\theta \ dr\\
=2\pi\int_{0}^{+\infty}e^{-r^2}\ r \ dr\\=\pi\int_0^{+\infty}e^{-u}\ du$$
A: You cannot find the indefinite integral with respect to both $x$ and $y$ because it does not make sense. Imagine finding a function $F(x,y)$ as an answer; then what are the properties you expect $F(x,y)$ to have? Since you cannot differentiate simultaneously with respect to $x$ and $y$, integrating with respect to two variables does not make sense.
Of course if you're talking about a definite integral things change, but you have to specify the domain on which you are integrating. 
I'm going to make an educated guess and say that what you want is $$\int_{\mathbb R^2} e^{-x^2 - y^2} \ dx \ dy$$
It should be clear that this is a number and not a function.
To find the value of this integral you can make the usual substitution in polar coordinates to get 
$$\int_0^\infty \int_0^{2\pi} \rho e^{-\rho^2} \ d\theta \ d\rho  = 2\pi \int_0^\infty\rho e^{-\rho^2} \ d\rho$$
And I think you can conclude from here :)
