calculus - Why does combining $int_{-infty}^{infty} e^{-x^2} dxint_{-infty}^{infty} e^{-y^2} dy$ into a $dx,dy$ integral work?

I was looking here: How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$? to find out how to evaluate $\displaystyle\int\limits_{-\infty}^{\infty} e^{-x^2} dx$, and didn't understand why
$$\left(\displaystyle\int\limits_{-\infty}^{\infty} e^{-x^2} dx\right)\left(\int\limits_{-\infty}^{\infty} e^{-y^2} dy\right)=\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx\thinspace dy$$
I know you can use Fubini's Theorem from this: Why does $\left(\int_{-\infty}^{\infty}e^{-t^2} dt \right)^2= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2 + y^2)}dx\,dy$?, but I'm still confused about how exactly you can just multiply two integrals together like that.

A detailed answer would be very nice.

Thanks!

Answer

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2}e^{-y^2} dx dy$$

Because we treat $y$ constant in the first, inner, integral we can pull it out.

$$=\int_{-\infty}^{\infty} e^{-y^2} \int_{-\infty}^{\infty} e^{-x^2} dx dy$$

Now because $\int_{-\infty}^{\infty} e^{-x^2} dx$ is some constant we can pull it out,

$$=\int_{-\infty}^{\infty} e^{-x^2} dx\int_{-\infty}^{\infty} e^{-y^2} dy$$

The result we got is generalizable, given $g(x,y)=f(x)h(y)$ we have,

$$\int_{a}^{b} \int_{c}^{d} g(x,y) dx dy=\int_{a}^{b} h(y) dy \int_{c}^{d} f(x) dx$$

Provided everything we write down exists.