Consider the integral

$$ \int_{\mathbb{R}^n}dx\,e^{-\frac12 x^TAx}=\frac{(2\pi)^{n/2}}{\sqrt{\det A}} $$

where $A=A^T$ is a symmetric $n\times n$ complex matrix with positive definite real part.

Question: can we explicitly calculate this integral (for complex $A$) without using analytic continuation?

Motivation: the standard proof of the above result starts off with a real $A$ and uses Cholesky decomposition to decouple the integral into $n$ one-dimensional Gaussian. (Diagonalizing $A$ with an orthogonal matrix with Jacobian $J=1$ essentially does the same.) Then one argues that, as long as the real part of $A$ remains positive definite, both sides are holomorphic and by analytic continuation the integral must have the value of the right hand side even for complex $A$, (see a good discussion on this here).

My question is motivated by the observation that for $n=1$ everything is scalar, $\det A=A$, and one can prove the above result for $\Re A>0$ using Cauchy theorem and contour integration with complex Jacobian $J=\sqrt{A}$, $\arg\sqrt{A}\in(-\pi/4,\pi/4)$. There is no need for analytic continuation (unless of course you want to go to $\Re A<0$), see the proof here.

So I wonder if there exists a direct proof for $n>1$ using some variant of Cauchy theorem in $\mathbb{C}^n$? Or some other way of integration with substitution using complex Jacobians, without having to rely on analytic continuation?

  • $\begingroup$ Yes, Integration by substitution helps it here. First step: Substitute $x\to Ux$ where $U$ is a unitary matrix such that $U^tAU$ is diagonal. Second step: Use the fact that the diagonal is non-negative and make a second substitution. $\endgroup$ – uniquesolution Sep 13 '15 at 11:58
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    $\begingroup$ @uniquesolution thanks for the feedback! Well, you have to be careful here. $A=B+iC$, where $B$ and $C$ are real symmetric and $B$ is positive definite; there is no further constraint on $C$. $B$ and $C$ are both diagonalisable but there is no guarantee that the same orthogonal matrix will do that for me (simultaneous diagonalizability) unless they commute, which they do not in general. So, you cannot - in general - diagonalize the whole (complex) quadratic form in the exponent and the integral will not factor into the product of 1d Gaussians over the new real/complex(?) coordinates... $\endgroup$ – Andras Vanyolos Sep 13 '15 at 12:40

Proposition. Given a symmetric complex $n\times n$ matrix $S=S^T\in {\rm Mat}_{n\times n}(\mathbb{C})$ with real part ${\rm Re}(S)>0$ positive definite, then (i) the matrix $S$ is invertible, and (ii) the Gaussian integral is $$ I~:=~\int_{\mathbb{R}^n} \! d^nx ~e^{-\frac{1}{2}x^T Sx} ~=~\sqrt{\frac{(2\pi)^n}{\det S }}. \tag{1}$$

Remark. Note that the condition$^1$ $${\rm Re}(S)~>~0\qquad\Leftrightarrow\qquad \forall x~\in~\mathbb{R}^n\backslash\{0\}:~~x^T{\rm Re}(S) x~>~0 \tag{2}$$ ensures that the integrand (1) is Lebesgue integrable.

Remark. We will below give a proof of the proposition that does not rely on analytic continuation of the $S$ matrix, as requested by OP.

Induction proof of the proposition. The case $n=1$ is e.g. proven in my Phys.SE answer here. Next consider the induction step. Write $$ S~=~\begin{pmatrix} a & b^T \cr b & C \end{pmatrix}, \qquad a~\in~\mathbb{C}, \quad b~\in~\mathbb{C}^{n-1}, \quad C~\in~{\rm Mat}_{n-1 \times n-1}(\mathbb{C}). \tag{3}$$ It follows that $${\rm Re}(a)~\stackrel{(2)+(3)}{>}~0, \tag{4}$$ which implies that $a\neq 0$. Define $$ \gamma~:=~\alpha + i \beta~:=~b/a~\in~\mathbb{C}^{n-1}, \qquad \alpha , \beta~\in~\mathbb{R}^{n-1}, \tag{5}$$ $$ S^{\prime}~:=~C-\gamma a \gamma^T ~\in~{\rm Mat}_{n-1 \times n-1}(\mathbb{C}), \tag{6}$$ and $$ \widetilde{S}^{\prime}~:=~S^{\prime}-\beta a \beta^T ~\in~{\rm Mat}_{n-1 \times n-1}(\mathbb{C}). \tag{7}$$ Then $$ S~\stackrel{(3)+(5)+(6)}{=}~\begin{pmatrix} 1 & 0 \cr \gamma & \mathbb{1} \end{pmatrix} \begin{pmatrix} a & 0 \cr 0 & S^{\prime} \end{pmatrix} \begin{pmatrix} 1 & \gamma^T \cr 0 & \mathbb{1} \end{pmatrix} ~\stackrel{(9)}{=}~\begin{pmatrix} 1 & 0 \cr \alpha & \mathbb{1} \end{pmatrix} \widetilde{S} \begin{pmatrix} 1 & \alpha^T \cr 0 & \mathbb{1} \end{pmatrix}, \tag{8}$$ where $$ \widetilde{S}~:=~\begin{pmatrix} 1 & 0 \cr i\beta & \mathbb{1} \end{pmatrix} \begin{pmatrix} a & 0 \cr 0 & S^{\prime} \end{pmatrix} \begin{pmatrix} 1 & i\beta^T \cr 0 & \mathbb{1} \end{pmatrix} ~\stackrel{(7)}{=}~\begin{pmatrix} a & i\beta^T \cr i\beta & \widetilde{S}^{\prime} \end{pmatrix} . \tag{9}$$ Eq. (8) implies that ${\rm Re}(\widetilde{S})>0$. Eq. (9) then implies that ${\rm Re}(\widetilde{S}^{\prime})>0$. Eq. (4) implies that ${\rm Re}(\beta a \beta^T)\geq 0$. Eq. (7) then implies that ${\rm Re}({S}^{\prime})>0$. [Here we have used that (semi)positive-definite matrices form a convex cone.] By induction we then know that ${S}^{\prime}$ is invertible. Eq. (8) then implies that $S$ is invertible. The integral becomes $$ \begin{align}I~\stackrel{(1)+(8)}{=}&~\int_{\mathbb{R}^n} \! d^nx ~e^{-\frac{1}{2}x^T \widetilde{S}x}\cr ~\stackrel{(9)}{=}~&\int_{\mathbb{R}^{n-1}} \! d^{n-1}x^{\prime} ~e^{-\frac{1}{2}x^{\prime T} S^{\prime}x^{\prime}} \int_{\mathbb{R}+i\beta^Tx^{\prime}} \! dz~e^{-\frac{1}{2}az^2}\cr ~=~&\int_{\mathbb{R}^{n-1}} \! d^{n-1}x^{\prime} ~e^{-\frac{1}{2}x^{\prime T} S^{\prime}x^{\prime}} \int_{\mathbb{R}} \! dz~e^{-\frac{1}{2}az^2}\cr ~=~&\sqrt{\frac{(2\pi)^{n-1}}{\det S^{\prime}}} \sqrt{\frac{2\pi}{a}}\cr ~\stackrel{(8)}{=}~&\sqrt{\frac{(2\pi)^n}{\det S}} . \end{align}\tag{10}$$ To perform the 1-dimensional $z$-integral, we have used the fact that it is possible to shift the horizontal integration contour back to the real axis $\mathbb{R}$ without changing the integral value, cf. Cauchy's integral theorem. $\Box$


$^1$In this answer, the notation $M>0$ means a positive-definite matrix $M$; not a matrix with positive entries.

  • $\begingroup$ Very nice direct proof, thanks! $\endgroup$ – Andras Vanyolos Dec 16 '19 at 22:42

Here is a proof that requires neither analytic continuation nor the Cauchy Theorem. Instead, it depends on ambiguities permitted in the standard SVD.

Note that the SVD of a complex positive definite matrix must necessarily take the form $$\underline {\overline {\bf{A}} } = \underline {\overline {\bf{U}} } \,\underline {\overline {\bf{\Lambda }} } \,{\underline {\overline {\bf{U}} } ^ + }$$ where $\underline {\overline {\bf{U}} }$ is some unitary matrix and $\underline {\overline {\bf{\Lambda }} }$ is the diagonal singular value matrix with positive real entries. However, any decomposition of the form $$\underline {\overline {\bf{A}} } = \left( {\underline {\overline {\bf{U}} } \,\underline {\overline {\bf{\Phi }} } } \right)\,\underline {\overline {\bf{\Lambda }} } {\left( {\underline {\overline {\bf{U}} } \,\underline {\overline {\bf{\Phi }} } } \right)^ + } = \underline {\overline {\bf{U}} } \,\underline {\overline {\bf{\Phi }} } \,\underline {\overline {\bf{\Lambda }} } \,{\underline {\overline {\bf{\Phi }} } ^ + }{\underline {\overline {\bf{U}} } ^ + }$$ where $$\underline {\overline {\bf{\Phi }} } = \left[ {\begin{array}{*{20}{c}} {{e^{j{\varphi _1}}}}& \cdots &0\\ \vdots & \ddots & \vdots \\ 0& \cdots &{{e^{j{\varphi _N}}}} \end{array}} \right]$$ is an equally valid SVD since $\underline {\overline {\bf{\Phi }} }$ commutes with $\underline {\overline {\bf{\Lambda }} }$.

Now consider the transformation $$\underline {\bf{y}} = \underline {\overline {\bf{\Phi }} } {\left( {\underline {\bf{x}} } \right)^ + }{\underline {\overline {\bf{U}} } ^ + }\underline {\bf{x}}$$ where, for each individual value of $\underline {\bf{x}}$, $\underline {\overline {\bf{\Phi }} } \left( {\underline {\bf{x}} } \right)$ is chosen to insure $\underline {\bf{y}}$ remains real with the same sign pattern as $\underline {\bf{x}}$.

Then the transformation remains unitary, with a Jacobian determinant of one, but the transformed integral is now completely real by construction. QED.

The sign pattern requirement guarantees that the integration range of $\underline {\bf{y}}$ remains the same as that of $\underline {\bf{x}}$. There are some secondary subtleties addressing elements in $\underline {\bf{x}}$ that are precisely zero, but the proof is immune to them, as such subsets are of zero measure.

  • $\begingroup$ Nice approach, thanks! It made me look into SVD theorem. But you implicitly assume $A$ is positive definite and hence $det A= det \Lambda >0$. Would this derivation hold if I assume that only the real part of $A$ is positive definite? $\endgroup$ – Andras Vanyolos Jul 1 '18 at 12:03
  • $\begingroup$ If A is "symmetric", which I take to mean being Hermitian, and the real part is positive definite, the whole matrix must be positive definite, since the imaginary part must be antisymmetric. The only way this argument doesn't work is if the assumption of symmetry applies only to the real part, but then the integral itself would have a complex value. $\endgroup$ – John Polcari Jul 1 '18 at 13:43
  • $\begingroup$ Sorry for late reply. The integral can be complex. Even in 1d it is complex when we use Cauchy and move contours (see second to last paragraph of my original question). Btw, if A is Hermitian the imaginary part (being antisymmetric) does not contribute to the integral since it cancels in the quadratic form; x is a real n-tuple. So I am still interested in the case where A is a complex symmetric matrix with pos def real part. $\endgroup$ – Andras Vanyolos Jul 18 '18 at 19:45

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