Gaussian Characteristic Function

Since the Gaussian PDF is

$$p(t) \isdef \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(t-\mu)^2}{2\sigma^2}}$$

and since the Fourier transform of $p(t)$ is

$$P(\omega) = e^{-j\mu \omega} e^{-\frac{1}{2}\sigma^2\omega^2}$$

It follows that the Gaussian characteristic function is

$$\Phi(\omega) = \overline{P(\omega)} = e^{j\mu \omega} e^{-\frac{1}{2}\sigma^2\omega^2}.$$