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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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    f is a well behaved distribution. you can imagine it as a gaussian.2012-03-24

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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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    As an aside, I had a fun time assuming that $X$ was exponentially distributed with $\lambda=1$ and then computing the mean of $\log(1+X)$. It turns out to be [Gompertz constant](http://mathworld.wolfram.com/GompertzConstant.html); I wonder if this coincidence has been noted before.2012-03-26