Compute multivariate complex Gaussian integral

Question

I don't know how to work out the homework of Leib&Loss P121, Ex4(b), in which we need to compute the following $$ \int_{\mathbb{R}^n}\exp(-x^tAx)dx=\pi^{n/2}/\sqrt{\det A} $$ where $A=A^t$ is a symmetric (thank Paul, see the comments) complex matrix with positive definite real part.

It hints to use something like continuous extension, but I don't know how to do this?

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UPDATE

Since it is easy to show in case $A$ is real, I try to show that $$ F(t)=\int_{\mathbb{R}^n}\exp(-x^t(A+tBi)x)dx-\pi^{n/2}/\sqrt{\det (A+tBi)} , $$ is independent to $t$, the DCT make us differentiate under the integral, but I can't show that $F'(t)=0$.

Answer 1

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 $0\le t\le 1$.

Answer 2

The formula given is false if $\sqrt{\cdot}$ is interpreted as meaning the principal branch. There is the following counterexample: let $n \geq 5$ and let $A = e^{\frac{2\pi i}{n} } I_{n \times n}$ where $I_{n \times n}$ is the $n\times n $ identity matrix ($n \geq 5$ ensures the real part of $A$ is positive definite). Then the integral is equal to $- (2\pi )^\frac{n}{2}$ however the formula gives $(2\pi )^\frac{n}{2}$, an error by a sign.

On the other hand the formula is true if you say the left hand side is $\pm$ the right hand side. You can show this by squaring both sides and showing that both sides are holomorphic/analytic in $A$, agree for real pos. def. matrices, and the set of matrices which have positive definite real part is path connected (analytic continuation). The fact that the matrix is symmetric rather than Hermitian is important for analyticity (because a Hermitian matrix has a nonzero antiholomorphic derivative with respect to $\overline{A_{ij}}$).

The problem with making a similar argument without squaring is that for the right hand side to be analytic, you need to exclude matrices $A$ that have a negative determinant $\det{A} < 0$ (if the negative real line is your branch cut). However the set of matrices with positive definite real part and non-negative determinant is not a connected set, so you cannot analytically continue.

Admittedly I do find it strange that $\int_{\mathbb{R}^n} e^{-\frac{1}{2} x^{\mathrm{T}}A x } dx$ is holomorphic with no branch cuts, yet on some open set it is equal to an analytic function with branch cuts.