A corrected form of the question asks to show that $\int_{\mathbb R^n} e^{-x^tAx}\;dx\;=\; \pi^{n/2}/\sqrt{\det A}$ for symmetric $n$-by-$n$ $A$ with positive-definite real part. First, for $A$ *real* (positive-definite), there is a (unique) positive-definite square root $S$ of $A$, and the change of variables $x=S^{-1}y$ gives the result, as the questioner had noted.

The trick here, as in many similar situations asking for extension to complex parameters of a computation that succeeds simply by change of variables in the purely real case, is invocation of the Identity Principle from complex analysis. That is, if $f,g$ are holomorphic on a non-empty open $\Omega$ and $f(z)=g(z)$ for $z$ in some subset with an accumulation point, then $f=g$ throughout $\Omega$. This can be iterated to apply to several complex variables, in various manners. In the case at hand, this gives an extension from symmetric *real* matrices to symmetric *complex* matrices (with the constraint of positive-definiteness on the real part, for convergence of everything).

To be sure, the complex span (in the space of $n$-by-$n$ matrices) of real symmetric matrices is complex *symmetric* matrices, not $n$-by-$n$ complex matrices with arbitrary imaginary part.

EDIT: To discuss meromorphy in each of the entries, observe that if $A$ is symmetric with positive-definite real part, then so is $A+z\cdot (e_{ij}+e_{ji})$ for sufficiently small complex $z$, where $e_{ij}$ is the matrix with $ij$-th entry $1$ and otherwise $0$. Without attempting to describe the precise domain, this allows various proofs of holomorphy of both sides of the asserted equality. To prove connectedness of whatever that domain (for fixed $i>j$) is, it suffices to observe that it is *convex*: if $A$ and $B$ are symmetric complex with positive-definite real part, then the same is true of $tA+(1-t)B$ for real $t$ in the range $0\le t\le 1$.