An integral related to Gaussian distribution I am trying to evaluate the integral
$$\int_{\mathbb{R}^d} e^{i \langle z,x \rangle}  e^{- \langle x,A^{-1}x \rangle /2} dx $$
where $A$ is a positive-definite symmetric matrix. 
As a first step (which is quite pointless, but just to remove the inverse sign) I did
$$ |\det A|\int_{\mathbb{R}^d} e^{i \langle z, Ax \rangle}  e^{- \langle A x,x \rangle /2} dx . $$
What I really want to do now is make some orthogonal change of variables so that all the matrices inside the integral are diagonal. However, I cannot make this calculation work out, regardless of what I try.
Note that it is possible to diagonalize $A$ using an orthogonal matrix, given the hypotheses on $A$.
Many thanks for your help.
EDIT:
Let
$$f_X(x) = \frac{1}{\sqrt{(2\pi)^d\operatorname{det}(A)}}\exp\left(-\frac{1}{2}\langle x, A^{-1}x\rangle\right).$$
We calculate as follows (as per the suggested substitution):
\begin{align}
 \int_{\mathbb{R}^d} e^{i \langle z,x \rangle} f_X(x)dx &= \frac{1}{\sqrt{(2\pi)^d\operatorname{det}(A)}} \int_{\mathbb{R}^d} e^{i \langle U^T z,y \rangle}  e^{\langle y, D^{-1}y \rangle /2} dy \\
&= \frac{1}{\sqrt{(2\pi)^d\operatorname{det}(A)}} \prod_{n=1}^{d} \int_{\mathbb{R}} e^{i (U^T z)_n y_n} e^{-y_n^2/2d_n} dy_n
\end{align}
where, for a vector $x \in \mathbb{R}^d$ I write $x = (x_1,...,x_d)$, and where the $d_n$ for $n \in \{1,...,d\}$ are the diagonal entries of $D$.
Continuing from above, using the integral from the $1$-D case, we get
\begin{align}
&= \frac{1}{\sqrt{(2\pi)^d\operatorname{det}(A)}} \prod_{n=1}^{d} \exp \left( -\frac{1}{2} d_n (U^T z)^2_n \right) (2 \pi d_n)^{1/2}\\
&=\prod_{n=1}^{d} \exp \left( -\frac{1}{2} d_n (U^T z)^2_n \right)\\
&= \exp \left( -\frac{1}{2} \langle z, Az \rangle \right),
\end{align}
as expected.
 A: Hint:
$$
\langle x,A^{-1}x\rangle = 
x^T A^{-1}x = 
x^T A^{-1/2}A^{-1/2}x = 
\langle A^{-1/2}x,A^{-1/2}x \rangle
$$
from there, take $u = A^{-1/2}x$.
Alternatively, to use orthogonal matrices: note that if $A = UDU^T$ (with $U$ orthogonal), we have
$$
\langle x,A^{-1}x\rangle = 
\langle x,UD^{-1}U^Tx\rangle = 
\langle U^Tx,D^{-1}U^Tx\rangle = 
\langle [U^Tx],D^{-1}[U^Tx]\rangle
$$
from there, make the substitution $y = U^T x$.
A: Let $X = (X_1, X_2, \ldots, X_d)$ denote a vector of $d$ zero-mean jointly Gaussian random variables with invertible covariance matrix $A$. Then, the joint density is
$$f_X(x) = \frac{1}{\sqrt{(2\pi)^d\operatorname{det}(A)}}\exp\left(-\frac{1}{2}\langle x, A^{-1}x\rangle\right)$$
and so we recognize the given integral as having value 
$cE\left[e^{i\langle z,X\rangle}\right] = cE\left[e^{-iZ}\right] 
= c\Psi_Z(1)$ where
$c = \sqrt{(2\pi)^d\operatorname{det}(A)}$, and
$\Psi_Z(\cdot)$ denotes the characteristic function of the random variable 
$Z = -\langle z, X\rangle = -\sum_i z_i X_i$. But, linear combinations
of jointly Gaussian random variables are Gaussian random variables, and
so we conclude that $Z$ is a zero-mean Gaussian random variable with
 variance $\langle z, Az\rangle$. Thus,
$\Psi_Z(\omega) = \exp\left(-\frac{1}{2}\langle z, Az\rangle\omega^2\right)$,
giving the value of the integral as
$$\sqrt{(2\pi)^d\operatorname{det}(A)}\cdot\exp\left(-\frac{1}{2}\langle z, Az\rangle\omega^2\right).$$
I would be interested in knowing what value you obtained via the approach
pointed out by @Omnomnomnom.
