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Let $H$ be a positive definite Hermitian matrix. How to prove

$$\int_{\mathbb R^n}\int_{\mathbb R^n}e^{-z^*Hz}dxdy=\frac{\pi^n}{\det H},$$ where $z=x+iy$ and $dx$, $dy$ denote the integration over real $n$ dimensional region?

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Would you mind explaining to me what is $z^*Hz$? –  Américo Tavares Aug 16 '11 at 20:03
If one defines the covariance between two complex-valued random column vectors $Z\in\mathbb{C}^n$, $W\in \mathbb{C}^m$ as the $n\times m$ matrix $E((Z-E(Z))(W-E(W)^*)$ where $^*$ is the conjugate-transpose, then when $Z$ and $W$ are the same random vector, this is like the variance. If $H$ is the variance, then with a Gaussian distribution, the density would be a constant times $e^{-z^*Hz/2}$, differing from what you've got only by that factor of $2$. This is probably ultimately reducible to $\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}$. –  Michael Hardy Aug 16 '11 at 20:07
@Américo: Note that $z \in \mathbb{C}^n$ and $H \in \mathbb{C}^{n \times n}$. The row vector $z^{\ast}$ is the conjugate transpose of $z$ and the product $z^{\ast}Hz$ is simply the matrix product (you can write this as the scalar product $\langle Hz, z \rangle$ with the standard Hermitian scalar product on $\mathbb{C}^n$ if you prefer). –  t.b. Aug 16 '11 at 20:49
@Theo: I was unsure about the order of $H$ and the meaning you clarified of $z^*Hz$. –  Américo Tavares Aug 16 '11 at 21:21
@Américo: Oh, sorry... No offense intended by further explaining and providing links. –  t.b. Aug 16 '11 at 21:23
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1 Answer

up vote 8 down vote accepted

Since $H$ is Hermitian, it admits Cholesky decomposition, i.e. $H = U^\dagger U$. Notice that $\det H = \vert \det U \vert^2$.

Then $-z^\ast H z = (U z)^\ast (U z)$. Introducing $w = U z$, and noticing that $\mathrm{d} x \mathrm{d} y = \frac{1}{\vert \det U \vert^2 } \mathrm{d}w_x \mathrm{d} w_y$ we have ( here $w_x = \mathop{Re}(U z)$ and $w_y = \mathop{Im}(U z)$):

$$ \int_{\mathbb{R}^n \times \mathbb{R}^n} \mathrm{d} x \mathrm{d} y \,\, \exp(-z^\ast H z) = \int_{\mathbb{R}^n \times \mathbb{R}^n} \frac{\mathrm{d} w_x \mathrm{d} w_y}{\vert \det U \vert^2} \,\, \exp(-w^\ast w) = \frac{\pi^n}{\det H }. $$

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