Does $\int f(u) \, du \int g(v) \, dv = \int \int f(u) g(v) \, du \, dv $? I saw a proof of $\Gamma (\frac{1}{2})=\sqrt{\pi}$ that had the step
$$\left[ \Gamma\left(\frac{1}{2}\right) \right]^2 = \left\{2 \int_0^\infty e^{-u^2} \, du\right\} \left\{ 2 \int_0^\infty e^{-u^2}du\right\}= 4 \int_0^\infty \int_0^\infty e^{-(u^2+v^2)} \, du \, dv$$
Does this always work?
 A: One fact that's being used is this:
$$
5 \int_1^8 f(x) \,dx = \int_1^8 5 f(x)\,dx. \tag 1
$$
The essential thing about the number $5$ that makes this possible is that $5$ does not change as $x$ goes from $1$ to $8$.
$$
\left(2 \int_0^\infty e^{-u^2} \, du\right) \left( 2 \int_0^\infty e^{-u^2} \, du \right)
$$
Another thing you need is this:
$$
\int_a^b f(x)\,dx = \int_a^b f(w)\,dw = \int_a^b f(v) \, dv\quad\text{etc.}
$$
You can change the name of a bound variable.  For example:
$$
\sum_{j=1}^4 j^2 = 1^2 + 2^2 + 3^2 + 4^2 = \sum_{k=1}^4 k^2.
$$
This tells you that
$$
\left(2 \int_0^\infty e^{-u^2} \, du\right) \left( 2 \int_0^\infty e^{-u^2} \, du \right) = \left(2 \int_0^\infty e^{-u^2} \, du\right) \left( 2 \int_0^\infty e^{-v^2} \, dv \right).
$$
Now notice that the first integral -- the one with the $u$ -- does not change as $v$ goes from $0$ to $\infty$, so it is like the number $5$ in line $(1)$ above.  Thus we can write:
$$
2 \int_0^\infty \left(2 \int_0^\infty e^{-u^2} \, du\right) e^{-v^2} \, dv.
$$
Next, observe that $e^{-v^2}$ is like the number $5$ in line $(1)$ above: It does not change as $u$ goes from $0$ to $\infty$.  Thus we can similarly move it to the inside of the integral that is $\Big($in parentheses$\Big)$.
