What property allows me to integrate a gaussian function? Whenever I integrate a gaussian function, I get to a step that makes me a little uncomfortable because I don't fully understand it. The only way I know of to analytically integrate the gaussian function is to multiply two of them together, like so...
$$\int_{- \infty}^{\infty} e^{-x^2}dx\int_{- \infty}^{\infty} e^{-y^2}dy = \int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-x^2-y^2}\,dx\,dy$$ 
My questions is, what allows me to group two different integrands together into one integrand? When is this technique not allowed? Thanks in advance.
 A: You are allowed to do this due to  Fubini's theorem
A: You can use this technique whenever the integrals converge.
A: Two things are special here. Aside from the worry about improper integrals, you can express the double integral 
$$\iint_R f(x,y)\, dA$$
as a product of single integrals when
(1) $R$ is a rectangle, and
(2) $f(x,y)=g(x)h(y)$ for some functions $g$ and $h$. To spare ourselves anomalies, let's assume all functions in question are continuous. 
A: You have
$$
\int_a^b \left( \int_c^d f(x)g(y)\,dx\right)\,dy.
$$
Look at the inside integral:
$$
\int_c^d f(x)g(y)\,dx.
$$
The thing you're integrating is $g(y)$ times something, and you have $x$ going from $c$ to $d$.  As $x$ goes from $c$ to $d$, notice that $g(y)$ DOES NOT CHANGE.  $g(y)$ does not depend on $x$.  And $g(y)$ is a FACTOR of the function you're integrating, i.e. what you're integrating is $g(y)$ times something.  Since this factor does not depend on $x$, it's a CONSTANT as a function of $x$, so it can be pulled out, getting
$$
g(y)\int_c^d f(x)\,dx.
$$
Now you have
$$
\int_a^b g(y)\left(\int_c^d f(x)\,dx\right) \,dy.
$$
Now notice that the expression in parentheses does not depend on $y$, and you're integrating with respect to $y$ as $y$ goes from $a$ to $b$.  In other words, as a function of $y$, that expression is a CONSTANT, so it can likewise be pulled out.  Then you're done.
