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If I have a function $f(x)$ that is $C^2$ and I know that $\int_{-\infty}^{+\infty}f(x)\mathrm{d}x=1$ and $\int_{-\infty}^{+\infty}x f(x)\mathrm{d}x=\mu$ in real axis $x$, what can I say about this integration in the real axis $$ G(a)=\int_{b'}^{b''}f(x)\ln(1+a x)\mathrm{d}x $$ knowing that $b',b'',a,\mu$ are real?

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Nothing. $G(a)$ can be more or less everything as your assumptions on $f$ don't tell you anything about $f$'s behaviour on the intervall $[b',b'']$. – martini Mar 23 '12 at 12:16
f is always finite and continuous in al the real axis. – emanuele Mar 23 '12 at 12:34
The problem is, that the integrals of $f$ and $x\mapsto xf(x)$ over the whole real axis don't tell us anything about $f$ in the interval $[b',b'']$. One can give examples where $f$ is constant zero or constant one in all of $[b',b'']$. – martini Mar 23 '12 at 12:37
f is a well behaved function. – emanuele Mar 23 '12 at 18:44
Even if $f$ is $C^\omega$, you cannot infer its behaviour on $[-b,b]$ from these two integrals. – martini Mar 23 '12 at 21:00
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The Taylor series expansion of $\log(1+ax)$ around the point $b'$ (assuming $b'$ is chosen in a valid range) is $\log(1+ab') + \frac{a}{1+ab'}(x-b')+O(x^{2})$. If we assume $b''$ is close enough that a linear approximation works and we plug this in then: $$ \int_{b'}^{b''}f(x)\log(1+ax)dx$$ $$\approx \log(1+ab')\cdot\biggl(F(b'')-F(b')\biggr) + \frac{a}{1+ab'}\cdot{}\biggl(\int_{b'}^{b''}xf(x)dx - b'[F(b'')-F(b')]\biggr)$$ $$ = \biggl[ \log(1+ab') - \frac{ab'}{1+ab'}\biggr]\cdot{}\biggl(F(b'')-F(b') \biggr) + \frac{a}{1+ab'}\int_{b'}^{b''}xf(x)dx,$$

where $F(x)$ is the CDF coming from the known density $f(x)$.

Like others have mentioned, you won't be able to squeeze much out of the last term without making significant assumptions of the form of $f(x)$ on $[b',b'']$. If you're willing to explore various assumptions about whether the random variable $X$ that has $f(x)$ as its density is bounded or non-negative, then you can probably make use of Bennett's inequality and/or Markov's inequality to get some inequality constraints.

Another approach, which I may try to flesh out tomorrow, would be to look at any bounds that the moment generating function of $X$ yields. Your integral is a part of the expected value of $\log(1+aX)$, so if you find its moment generating function, you can relate one part of the integral to other parts.

But again, you'll require strict assumptions on $f(x)$. Even in a Gaussian family, I can shift the mean so far to the left of $b'$ that this integral is as small as desired. Without relationships between the parameters you list, inequality bounds won't be too helpful.

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$f(x)$ is an $\alpha$-stable distribution. – emanuele Mar 24 '12 at 18:38
... and $(-1 \lt b',b'' \ge +\infty]$. – emanuele Mar 24 '12 at 18:44
Even if $b'>-1$, that doesn't mean $\log(1+ab')$ is well-behaved. I think you have a backward inequality as well. I'm not sure that $\alpha$-stability buys you anything, because you're only ever working with one copy of the distribution. You're not using any central limit theorem type properties, and so all of the suggestions above about how there can be arbitrary probability density on $[b',b'']$ still applies. There's just not much you can possibly say without stronger assumptions. – EMS Mar 24 '12 at 18:54
...yes but my doubt is, if $x \gt \log(1+x)$ in $(-1,+\infty]$ and i know that $\int_{-1}^{+\infty}f(x)x d x=\mu$, why i can't say that $\int_{-1}^{+\infty}f(x)\log(1+x) d x\lt +\infty$ ? – emanuele Mar 26 '12 at 7:15
The space of functions that is integrable with respect to a given probability measure $\nu=f(x)dx$ is $L^{1}(\mathbb{R},\nu)$. The function $\log(1+x)$ doesn't decay towards $0$ as $x\to\infty$. This is a problem because then it's much harder to determine if $\log(1+x)\in L^{1}(\mathbb{R},\nu)$. If $f(x)$ does not die off to zero sufficiently quickly, then this function won't be integrable. You should look up some standard results in integrability theory to see why your stated conditions are not sufficient to determine the integrability of $\log(1+ax)$. – EMS Mar 26 '12 at 15:59
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