To prove that the univariate Gaussian distribution is normalized, we will first show that it is normalized for a zero-mean Gaussian and extend that result to show that $\mathcal{N}$($x$$\mid$$\mu$, $\sigma$$^{2}$) is normalized.

The pdf of the zero-mean Gaussian distribution is given by:

\begin{align} \varphi(x)= \frac{1}{\sqrt{2\pi\sigma^2}}\, \exp\left(-\frac{1}{2\sigma^2}x^2\right)\,\,\,\,\,\,\,-\infty < x < \infty. \end{align} To prove that the above expression is normalized, we have to show that \begin{align} \int_{-\infty}^{\infty} \exp\left(-\frac{1}{2\sigma^2}x^2\right)dx = \sqrt{2\pi\sigma^2} \end{align} Proof. Let \begin{align} I = \int_{-\infty}^{\infty} \exp\left(-\frac{1}{2\sigma^2}x^2\right)dx \end{align} Squaring the above expression, \begin{align} I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\left(-\frac{1}{2\sigma^2}x^2-\frac{1}{2\sigma^2}y^2\right)dx\,dy \end{align} To integrate this expression we make the transformation from Cartesian coordinates (x, y) to polar coordinates (r, $\theta$), which is defined by \begin{align} x = r \, cos\,\theta\\ y = r \, sin\, \theta \end{align} and using the trigonometric identity $cos^2\,\theta + sin^2\, \theta = 1$, we have $x^2 + y^2 = r^2$. Also the Jacobian of the change of variables is given by, \begin{align*} \dfrac{\partial \left(x, y\right)}{\partial \left(r, \theta\right)} \,\,&= \,\,\,\begin{vmatrix} &\dfrac{\partial \left(x \right)}{\partial \left(r \right)} &\dfrac{\partial \left(x \right)}{\partial \left(\theta \right)}&\\ \\ &\dfrac{\partial \left(y \right)}{\partial \left(r \right)} &\dfrac{\partial \left(y \right)}{\partial \left(\theta \right)}& \end{vmatrix}\\ \\ &= \,\,\, \begin{vmatrix} &cos \,\theta &-r\,sin \, \theta& \\ &sin \,\theta &r\,cos\,\theta& \end{vmatrix}\\ &=\,\,\, r\, cos^2\,\theta + r\, sin^2\,\theta \\ &=\,\,\, r \end{align*} using the same trigonometric identity $cos^2\,\theta + sin^2\, \theta = 1$. Thus equation (4) can be rewritten as \begin{align} I^2 \,\,&=\,\,\, \int_{0}^{2\pi}\int_{0}^{\infty} \exp\left(-\frac{r^2}{2\sigma^2}\right) r\,dr\,d\theta \\ &= \,\,\,2\pi\int_{0}^{\infty} \exp\left(-\frac{r^2}{2\sigma^2}\right) r\,dr \\ &= \,\,\,2\pi\int_{0}^{\infty} \exp\left(-\frac{u}{2\sigma^2}\right) \frac{1}{2}\,du \\ &=\,\,\,\pi\left[\exp\left(-\frac{u}{2\sigma^2}\right)\,\left(-2\sigma^2\right)\right]_0^\infty \\ &=\,\,\,2\pi\sigma^2 \end{align} where we have used the change of variables $r^2 = u$. Thus $$I = \left(2\pi\sigma^2\right)^{1/2}.$$ Finally to prove that $\mathcal{N}$(x$\mid$$\mu$, $\sigma$$^{2}$) is normalized, we make the tranformation $y = x - \mu$ so that, \begin{align*} \int_{-\infty}^{\infty} \mathcal{N}(\textit{x}\mid\mu, \sigma^{2})\,dx \,\,&= \,\,\,\frac{1}{\left(2\pi\sigma^2\right)^{1/2}} \int_{-\infty}^{\infty} \exp\left(-\frac{y^2}{2\sigma^2}\right)\,dy\\ &=\,\,\,\frac{I}{\left(2\pi\sigma^2\right)^{1/2}}\\ &=\,\,\,1 \end{align*} as required.