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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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You shouldn't write it as $\mathrm{d}z$: this makes you want to think of $z$ as a complex number and the integral the complex line integral. But since you are integrating over the entire complex plane, the "area" element should be $\mathrm{d}x\mathrm{d}y$ where $(x,y)$ is a point in $\mathbb{R}^2$. ($\mathrm{d}x + j\mathrm{d}y$ has "unit" of "length", which doesn't make sense if you are integrating over the entire plane, where the unit should be that of "area") –  Willie Wong Sep 12 '12 at 11:14
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Hang on, now I am not quite sure what you mean. How does $H(z)$ depend on $z$? The right hand side is an integral of a function of $z$ in $\mathrm{d}z$. This should give you a constant, not a function of $z$. –  Willie Wong Sep 12 '12 at 11:21
    
Yes, you're right and many thanks. It is $H(z|x_1)$, so that it shows an expectation over parameter $z$, but in fact as you noted it will be a function of $x_1$ and not $z$. I was not sure if treating it with d$x$d$y$ is ok, because my original problem only contained a single integration with d$z$. –  Mona Hajimomeni Sep 12 '12 at 11:39
    
When $u = (x_1, \ldots, x_n)\in \mathbb{R}^n$, it is pretty common to write $\mathrm{d}u$ for $\mathrm{d}x_1\mathrm{d}x_2\cdots \mathrm{d}x_n$. –  Willie Wong Sep 12 '12 at 11:47

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