My attempt to calculate the characteristic function $\mathbb{E}\left(e^{it^T X}\right)$ of multivariate normal distributed $X \sim \mathcal{N}_d(\mu,\Sigma)$, finding a lecture note that writes as follows, but I am confused about the $\overset{?}{=}$ step.
\begin{align*} \phi_X(t) &= \mathbb{E}\left(e^{it^T X}\right) \\ &= \int_{-\infty}^{\infty} e^{it^T X} f_X(x) dx \\ &= \int_{-\infty}^{\infty} e^{it^T X} \frac{1}{(2\pi)^{d/2}|\Sigma|^{1/2}} e^{-\frac{1}{2}(x-\mu)^T \Sigma^{-1} (x-\mu)} dx \\ &= \int_{-\infty}^{\infty} \frac{1}{(2\pi)^{d/2}|\Sigma|^{1/2}} \exp{\left(-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T x\right)} dx \\ &\overset{?}{=} \exp{\left(it^T \mu - \frac{1}{2} t^T \Sigma t\right)} \\ &= e^{it^T \mu} e^{-\frac{1}{2}t^T\Sigma t} \end{align*}
Can you show me more details about that step?
What they do is completing the square:
\begin{align}-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T x & = -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T (x - \mu) + it^T \mu \\ & = -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T (x - \mu) +\frac{1}{2} t^T \Sigma t - \frac{1}{2} t^T \Sigma t\\ & \;\;\;\; +\ it^T \mu \\ & = -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)+it^T \Sigma \Sigma^{-1} (x - \mu) +\frac{1}{2} t^T \Sigma \Sigma^{-1}\Sigma t \\ & \;\;\;\; - \frac{1}{2} t^T \Sigma t +\ it^T \mu \\ & = -\frac{1}{2}(x-\mu-i\Sigma t)^T\Sigma^{-1}(x-\mu-i\Sigma t) - \frac{1}{2} t^T \Sigma t +\ it^T \mu \end{align}
Then, the two last terms factorize out with the exponential, while the first gives rise to a reparametrized Gaussian. So that integral is 1.