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I am trying to solve this integral:

$H(z)=\int_{\mathbb{C}}{p(z)\log_2{p(z)}dz}$,

where $z$ is a complex number with complex normal distribution $p(z)$, and $\mathbb{C}$ denoted the complex plain. $H(z)$ is called entropy function of variable $z$.

Now, it is clear that probability distribution $p(z)$ is a real function, but how one calculates this integral? $H(z)$ is also a physical quantity with real values (it shows how much uncertainty is in $z$).

Let me also express the pdf function $p(z)$ above. A circular symmetric complex normal pdf (wich is a common assumption for gaussian noise model in engineering) of a complex vector $\mathbf{z}$ with length $n$, is equivalent to pdf of a real vector with $2n$ elements (of real and imaginary parts) which are jointly guassian. In this problem,

$p(z) = \frac{1}{c\pi}\sum_{x_1}\exp(-|z-ax_1|^2)$,

where $x_1$ is complex and $a,c$ are real constants.

To evaluate the probability density, I choose complex points on a 2D grid whose span is the support of $p(z)$, and put into $p(z)$ relation above. The question is that how $dz$ should be treated, that is, should it be transformed to $dxdy$ or $dx+jdy$. The second form leads to a complex value for the integral, which is what makes me confused.

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    When $u = (x_1, \ldots, x_n)\in \mathbb{R}^n$, it is pretty common to write $\mathrm{d}u$ for $\mathrm{d}x_1\mathrm{d}x_2\cdots \mathrm{d}x_n$.2012-09-12

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