I was looking into a particular method to demonstrate that the following expression is a probability density function for the Gaussian / Normal distribution. (i.e. that the integral = 1) :
$$ f_{X} (x) = \frac{1}{\sqrt{2\pi} \sigma} exp \left( \frac{-(x - \mu)^2}{2\sigma^{2}} \right) $$
Define $$ I = \frac{1}{\sqrt{2\pi} \sigma} \int_{-\infty}^{\infty} exp \left( \frac{-(x - \mu)^2}{2\sigma^{2}} \right) $$
If we take $ y = \frac{x - \mu}{\sigma} \ \ $,$ \ \ \sigma \ dy = dx $, we need to show that:
$$ \sqrt{2\pi} = \int_{-\infty}^{\infty} e^{\frac{-y^{2}}{2}} dy$$
$$ I_{1} (r) = \lim_{r \to \infty} \int_{-r}^{r} e^{\frac{-y^{2}}{2}} dy $$
This is where the "trick" is performed, to approximate a square $ [(-r,-r),(r,-r),(-r,r),(r,r)] \ $ using an inner (radius $r$) and outer circle (radius $\sqrt{2} r$). What I'm having difficult understanding is how we got the expression below for $ I_{1} (r)^{2} $, or in other words, how did we convert the above expression into a double integral. I can't visualize how the expression $ e^\frac{-y^{2}}{2}$, with a Gaussian-looking curve can be"doubled" and then approximated by a square - what is the reasoning behind this?
Just to clarify what I'm saying and where this is headed - here is rest of the derivation.
$$ I_{1} (r)^{2} = \left( \int_{-r}^{r} e^\frac{-y^{2}}{2} dy \right) \left( \int_{-r}^{r} e^\frac{-x^{2}}{2} dx \right) $$
$$ = \int_{-r}^{r} \int_{-r}^{r} e^\frac{-(x^{2}+y^{2})}{2} dx \ dy $$
$$ = \iint\limits_{x^{2} + y^{2} < r^{2} } e^\frac{-(x^{2}+y^{2})}{2} dx \ dy $$
$$ = \int_{0}^{2\pi} \int_{0}^{r} e^\frac{-r^{2}}{2} r \ dr \ \ r \ d\theta $$
$$ = \int_{0}^{2\pi} \big[ -e^\frac{-r^{2}}{2} \big]^{R}_{0} \ d\theta = 2 \pi \big[1- e^\frac{-R^{2}}{2} \big] $$
$$I_{2} (r) \leq I_{1}^{2} (r) \leq I_{2} (\sqrt{2} r) $$
$$ 2 \pi (1-e^\frac{-R^{2}}{2}) \leq I_{1}^{2} (r) \leq 2 \pi (1-e^\frac{-R^{2}}{2}) $$
$$ \lim_{r \to \infty} I_{1}^{2} = 2\pi $$ (via Squeeze Theorem)
$$ I_{1} = \sqrt{2\pi} = \int_{-\infty}^{\infty} e^\frac{-y^{2}}{2} dy $$
Is there a more intuitive way to determine that the expression is a PDF for the Gaussian Distribution?
Things are prety clear here, your notations are the ones that confuse everything.
Let $I_1 (r) = \int \limits _{-r} ^r$ and $I_1 = \lim \limits _{r \to \infty} I_1 (r)$.
Let $D(r)$ be the disk of centerpoint $(0,0)$ and radius $r$. Let $I_2 (r) = \iint \limits _{D(r)}$ (in polar coordinates).
Now $I_1 ^2 = \left( \lim \limits _{r \to \infty} I_1 (r) \right) ^2 = \lim \limits _{r \to \infty} I_1 (r) ^2 = \lim \limits _{r \to \infty} I_2 (r)$ (even though $I_1 (r) ^2 \ne I_2 (r)$), so $I_1 = \sqrt {\lim \limits _{r \to \infty} I_2 (r)}$, and $I_2 (r)$ is really easy to compute.
In simpler words: to compute the integral on $\Bbb R ^2$ you may choose to view $\Bbb R^2$ either as a limit of increasing squares, or as a limit of increasing disks - the result will be the same. You use squares in order to obtain the square of the integral you desire; you use disks next in order to change to polar coordinates where it is easy to perform computations explicitly.