The Normal Distribution

The probability density of a normal random variable with mean $\mu$ and variance $\sigma^2$ is $$\begin{align} p(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \end{align}$$

One way to show that the probability density integrates to one is to consider the following integral and switch to polar co-oridinates.
$$\begin{align} &\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\int_{-\infty}^{\infty}e^{-\frac{y^2}{2}} dx dy \\ &\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2 + y^2}{2}} dx dy \end{align}$$
switching to polar co-ordinates $x = r \cos \theta$, $y = r \sin \theta$
We calculate the Jacobian to change integration variables.
$$\begin{align} &\left| \frac{\partial (x,y)}{\partial (r,\theta)} \right | = \left | \begin{array}{cc} \frac{\partial x}{\partial r}& \frac{\partial y}{\partial r}\\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta}\\ \end{array} \right | \\ &=\left | \begin{array}{cc} \cos \theta & \sin \theta\\ -r \sin \theta & r \cos \theta \\ \end{array} \right | = r \cos^2 \theta + r \sin^2 \theta = r \end{align}$$

$$\begin{align} &\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{x^2 + y^2}{2}} dx dy \\ = &\int_{0}^{\infty} \int_{0}^{2 \pi} \left| \frac{\partial (x,y)}{\partial (r,\theta)} \right | e^{-\frac{r^2}{2}} dr d\theta \\ = &\int_{0}^{\infty} \int_{0}^{2 \pi} r e^{-\frac{r^2}{2}} dr d\theta \\ & = 2 \pi [-e^{-\frac{r^2}{2}}]^{\infty}_0 = 2 \pi \end{align}$$
so that
$$\begin{align} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} = \sqrt{2 \pi} \end{align}$$
i.e.
$$\begin{align} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} dx = 1 \end{align}$$

Moments of a Normal Distribution.
$$\begin{align} \mathbb{E}[X] = & \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x e^{-\frac{x^2}{2}} dx \\ =& \frac{1}{\sqrt{2 \pi}} \left[-e^{-\frac{x^2}{2}}\right]^{\infty}_{-\infty} = 0 \end{align}$$
This is also obvious by symmetry. $p(x)$ is a symmetric function i.e. $p(x) = p(-x)$ so multiplying p(x) by an odd function $f(x)$ i.e. $f(x) = -f(-x)$ and calculating $\int_{-\infty}^{\infty} p(x) f(x) dx = 0$.
Since $\int_{-\infty}^{0} p(x) f(x) dx = -\int^{\infty}_{0} p(x) f(x) dx$. Substitute $y = -x$ to see this.
$$\begin{align} \mathbb{E}[X^2] = &\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2}} dx \\ &=\frac{1}{\sqrt{2 \pi}} \left ( [- x e^{-\frac{x^2}{2}}]^{\infty}_{-\infty} + \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} dx \right) = 1 \end{align}$$