Proof
\begin{align*}
E[(X-\mu)^2] & = \int_{-\infty}^{\infty} (x-\mu)^2 \cdot \frac{1}{\sigma \sqrt{2 \pi}} e^{ - \left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx\\
& = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \sigma^2 z^2 \cdot e^{ -z^2 / 2} dz\\
& = \frac{\sigma^2}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} z \cdot z e^{ -z^2 / 2} dz\\
& = \frac{\sigma^2}{\sqrt{2 \pi}} \cdot \big [ -z e^{-z^2 / 2} \big |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} e^{ -z^2 / 2} dz \big ]\\
& = \frac{\sigma^2}{\sqrt{2 \pi}} \cdot \big [ 0 + \sqrt{2 \pi} \big ]\\
& = \sigma^2
\end{align*}
using integration by parts. So, the use of \(\sigma\) as the other parameter in the normal probability function is also warranted.