\begin{align*} &\int^\infty_{-\infty}e^{-\alpha x^2}dx\int^\infty_{-\infty}e^{-\alpha y^2}dy = \int^\infty_{-\infty}\int^\infty_{-\infty}e^{-\alpha(x^2+y^2)}dxdy = \int^{2\pi}_0d\theta\int^\infty_0re^{-\alpha r^2}dr \\ &= 2\pi\int^\infty_0\frac{-1}{2\alpha }e^{-\alpha r^2}d(-\alpha r^2)=\frac{-\pi}{\alpha }\int^\infty_0 d(e^{-\alpha r^2}) = \frac{-\pi}{\alpha }[e^{-\alpha r^2}]^\infty_0=\frac{\pi}{\alpha }\\ &\text{Therefore, } \bigg[\int^\infty_{-\infty}e^{-\alpha x^2}dx\bigg]^2=\frac{\pi}{\alpha } \Rightarrow \int^\infty_{-\infty}e^{-\alpha x^2}dx = \sqrt{\frac{\pi}{\alpha }} \end{align*}

$$\mathcal{N}(x;\mu,\sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2 }}e^{-\frac{(x-\mu)^2 }{2\sigma^2 }}$$

\begin{align*} &\frac{1}{\sqrt{2\pi\sigma^2} }\int^\infty_{-\infty}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx = \frac{1}{\sqrt{2\pi\sigma^2 }}\int^\infty_{-\infty }e^{-\alpha y^2 }dy \\ &= \frac{1}{\sqrt{2\pi\sigma^2} }\sqrt{\frac{\pi}{\alpha}} = \sqrt{\frac{2\pi\sigma^2 }{2\pi\sigma^2 }} =1 \end{align*}