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$\Omega$ is a convex open set in $\mathbb {C^n}$ and $f$ is an analytical function Edit: without zero point on $\Omega$, then can we define an analytical branch of $\ln {f}$ on $\Omega$ ?

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sorry, I have corrected it. – Hezudao Oct 22 '11 at 23:58
No problem. However, I made your change in the question clearly visible, in order for my answer to still make sense. – t.b. Oct 23 '11 at 0:08
up vote 11 down vote accepted

I am happy to report that, yes, you can define $\log f$ even in much greater generality than in your situation.
Namely, if you have a real differentiable manifold $M$ and a function $f\in \mathcal C^\infty (M)$ never zero on $M$, then $f$ has a $\mathcal C^\infty $ logarithm as soon as the first De Rham cohomology group of $M$ vanishes: $H^1_{DR}(M, \mathbb R)=0$.

The definition of the logarithm is straightforward: fix a point $x_0\in M$ and define $$(\log f)(x)= \int_\gamma \frac {df}{f} $$
where $\gamma$ is a differentiable path joining $x_0$ to $x$, along which we can integrate the closed $1$-form $\frac {dg}{g}$.
The vanishing cohomology hypothesis ensures that the value of $\log f$ at $x$ does not depend of the path $\gamma$ chosen.

If $M$ happens to be a complex holomorphic manifold and if $f\in \mathcal O(M)$ is holomorphic, then the logarithm of $f$ will automatically be holomorphic: $\log f\in \mathcal O(M)$.
This applies to your case since a simply connected manifold -and a fortiori a convex set in a vector space- has zero first De Rham cohomology group.

Finally, just for old times' sake, let me sum this up in the language of classical physics :
$$\text{Every conservative vector field has a potential}$$

A variant
Specialists in complex manifolds are addicted to the exact sequence of sheaves on the complex manifold $X$: $$0\to 2i\pi \mathbb Z\to \mathcal O_X \stackrel {exp}{\to} \mathcal O^\ast_X \to 0 $$

A portion of the associated cohomology long exact sequence is $$ \Gamma(X,\mathcal O_X) \stackrel {exp}{\to} \Gamma(X,\mathcal O^\ast_X) \to H^1(X,\mathbb Z) $$
which shows again that every nowhere vanishing holomorphic function on $X$ is an exponential (in other words: has a logarithm) as soon as $H^1(X,\mathbb Z)=0$.

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What is $g$? Where does it come from? – scineram Oct 23 '11 at 12:01
@scineram: I wrote $g$ instead of the correct $f$. Thanks a lot for noticing this typo, now corrected. – Georges Elencwajg Oct 23 '11 at 12:05
Why require the region convex? – Phosphorus Feb 16 '14 at 9:51
Dear @Ren: because Hezuado asked about a convex open set in his question! – Georges Elencwajg Feb 16 '14 at 10:02
@GeorgesElencwajg, are you available to answer some questions about your answer? I am in doubt if I should consider $df = \partial f + \bar{\partial} f$ and how could I use this to show that $exp(log(f)) = f$. – Marra May 13 '14 at 14:30

Edit: I misread the question and only address the case $n=1$.

If you allow the function $f$ to have roots, then the answer is clearly no. There is no analytic branch of the logarithm of $z$ in any neighborhood of $0$.

If you exclude roots, the answer is yes.

More generally:

Let $f: G \to \mathbb{C}$ be holomorphic with $0 \notin f(G)$ and assume that $G$ is a simply connected domain (in particular $G$ convex suffices). Then $f$ has a holomorphic logarithm $g: G \to \mathbb{C}$, that is $f(z) = e^{g(z)}$ for all $z \in G$.

Fix $z_0 \in G$ arbitrarily, choose $g(z_0)$ in such a way that $f(z_0) = \exp{g(z_0)}$ and put $g(z) = g(z_0) + \int_{z_0}^{z} \frac{f'(w)}{f(w)}\,dw$. Check that $g$ is well-defined, holomorphic, and that $\frac{d}{dz} [f \cdot e^{-g}] = 0$, hence $f = e^g$.

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Your answer is definitely right when applied to one variable, but how can we generalize it to the situation of multiple one? – Hezudao Oct 23 '11 at 0:16
Oh I'm sorry, I only addressed the one-dimensional case. I missed that you were talking about $\mathbb{C}^n$. I don't know off the top of my head. I need to think a little. – t.b. Oct 23 '11 at 0:21
I don't have the expertise to say anything at the moment, I'm not feeling comfortable enough with several complex variables and it's getting late. I'll leave the answer here for the moment, hoping that it's useful in some way. – t.b. Oct 23 '11 at 0:41
Yess, well definedness of the integral above has to see with the fact that in a simply-connected domain, the integral is independent of path. – gary Oct 23 '11 at 1:29
@gary: Are you asking? Anyway: yes, of course. – t.b. Oct 23 '11 at 1:34

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