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?


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 $F'(t)=0$.

  • $\begingroup$ What does "positive definite real part" mean? Never heard of that before . . . $\endgroup$ – user14717 Jan 10 '15 at 14:42
  • $\begingroup$ It means $A=B+C i$, $B$ is positive definite, i.e., all its eigenvalue is positive. $\endgroup$ – van abel Jan 10 '15 at 14:45
  • $\begingroup$ So the only condition on $C$ is that it has real entries? $\endgroup$ – user14717 Jan 10 '15 at 15:15
  • $\begingroup$ @NickThompson Yes. $\endgroup$ – van abel Jan 10 '15 at 15:16
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    $\begingroup$ The imaginary part must be skew-symmetric, as @NickThompson's (counter-) example shows. Then the identity principle in complex analysis gives the general result. $\endgroup$ – paul garrett Jan 10 '15 at 15:43

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 $0\le t\le 1$.

  • $\begingroup$ It seems not easy to show that both sides are holomorphic, in fact use the complex one variable theory, it is not clear what is the domain space of $z_{ij}=a_{ij}+b_{ij}i$. $\endgroup$ – van abel Jan 11 '15 at 1:19
  • $\begingroup$ One need not be too precise about domains to prove holomorphy of both sides in each entry... the main thing is to be sure of connectedness, which follows from convexity of the cone of positive-definite real matrices. I can fill in further details tomorrow, ... $\endgroup$ – paul garrett Jan 11 '15 at 4:48
  • $\begingroup$ It seem in order to show that $f(z)=\sqrt{\det A(z)}$, where $A(z)=A+z(e_{ij} +e_{ji} )$, is holomorphic, I need to show that $C∖\{\det(A(z))|Re(A(z))>0\}$ can be cutted by a path from $0$ to $\infty$ , but what I can show is only that $f(z)$ is nonvanishing near $z=0$. $\endgroup$ – van abel Jan 12 '15 at 11:30
  • $\begingroup$ To define a square root of a non-zero function on a region it is only required that the region be simply connected, which a convex region certainly is. $\endgroup$ – paul garrett Jan 12 '15 at 17:09
  • $\begingroup$ I can only show the nonvanishing near $z=0$, when $z$ is large, certainly there are zeros of $\det(A(z))$. $\endgroup$ – van abel Jan 13 '15 at 5:41

Take \begin{align} A:= \begin{pmatrix} 1 & -2i \\ i & 1 \end{pmatrix} \end{align} Then $\sqrt{\det(A)} = \pm i$, and your formula predicts that the value of the integral is $\pm i\pi$. However, $\mathbf{x}^{T}A\mathbf{x} = x^2 + y^2 - ixy$ and \begin{align} \int_{\mathbb{R}^2} \exp(-x^2 - y^2 + ixy) \, \mathrm{d}x \mathrm{d}y = \frac{2\pi}{\sqrt{5}}. \end{align}


In the case that A is complex, $\sqrt{detA}$ has two values. The problem should involve which one to choose. There's ambiguity to say this $\int_{\mathbb{R}^2} e^{-x^{t}Ax}dx=\pi^{n/2}\sqrt{detA}$.

One way to consider this problem is to diagonalize $A$ (possible because A Hermitian) and change coordinate.

The integral is then $\int_{\mathbb{R}^2}e^{-(a_1+i b_1){x_1}^{2}-(a_2+ib_2){x_2}^{2}}dx_1dx_2$ with $a_1,a_2>0$.

So this problem is tranformed to a one dimension complex Gaussian. Then one can consider a complex contour which are two sectors closed by real axis and a complex ray in compex plane. We want the integral over the real axis and the integral over the complex ray to be the same. This requirement forces the angle between the complex ray and the real axis to be less or equal to $\frac{\pi}{4}$. This yields which value of $\sqrt{det A}$ one should choose.

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    $\begingroup$ $A$ isn't hermitian though $\endgroup$ – Spine Feast Nov 6 '16 at 15:25
  • $\begingroup$ complex symmetric is hermitian. $\endgroup$ – Dai Nov 6 '16 at 22:41
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    $\begingroup$ I'm pretty sure it isn't. $\endgroup$ – Spine Feast Nov 7 '16 at 19:11
  • $\begingroup$ en.wikipedia.org/wiki/Symmetric_matrix $\endgroup$ – Dai Nov 8 '16 at 4:54
  • $\begingroup$ When talking about "symmetric", the background is there is a metric such that $<Ax,y>=<x,A*y>$. In the complex case, the natural setting is Hermitian. $\endgroup$ – Dai Nov 8 '16 at 4:57

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.


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