How to integrate $e^{-x^2 -y^2}$?

Question

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

Answer 1

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 :)

Answer 2

If you want to integrate over both $x$ and $y$, then you can convert it to an integral in polar coordinates $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}\ dx \ dy= \int_{0}^{+\infty}\int_0^{2\pi}e^{-r^2}\ r\ d\theta \ dr\\ =2\pi\int_{0}^{+\infty}e^{-r^2}\ r \ dr\\=\pi\int_0^{+\infty}e^{-u}\ du$$