# Integral bound for square of log derivative

I am currently facing the following problem:

Given a polynomial $$f(x) = \sum_{s \in S_f} u_s x^s$$, $$f(0)\neq 0$$, $$\lvert S_f \rvert \leq t$$ (i.e. $$f$$ is $$t$$-sparse) with $$u_s$$ coming as samples from i.i.d. $$\mathcal{N}(0,1)$$-distributed variables, bound

$$\int_0^1 \bigg(\frac{f'(x)}{f(x)}\bigg)^2 dx = \int_0^1 \bigg(\frac{d}{dx} \log(\lvert f(x) \rvert)\bigg)^2 dx$$

in terms of the coefficients $$u_s$$ and the sparsity $$t$$, but not in terms of $$\deg(f)$$. It is not too difficult to bound $$\int_0^1 \frac{f'(x)}{f(x)} dx = \log(\lvert f(x) \rvert) \rvert_0^1 = \log(\lvert f(1) \rvert) - \log(\lvert f(0) \rvert)$$ if $$f(0) \neq 0$$, as we can plug in upper and lower bounds for $$\log$$. This makes me hopeful a bound of the squared integrand should exist too. In the worst case, a bound of the expectation of the integral with respect to the $$u_s$$ would also suffice, i.e. a bound for

$$\mathbb{E}_{u_s, s\in S_f} \bigg[\int_0^1 \bigg(\frac{f'(x)}{f(x)}\bigg)^2 dx \bigg].$$

It would take too long to explain where this comes from - I arrived at this problem looking at zero distributions of certain polynomials.

Thank you for all your ideas!

If the cardinality of the set $$S_f$$ is $$\ge1$$, then the value of the integral $$I:=\int_0^1\Big(\frac{f'(x)}{f(x)}\bigg)^2 dx$$ will be $$\infty$$ with nonzero probability, because with nonzero probability the polynomial $$f$$ will have a non-multiple root in the interval $$[0,1]$$. This will then also imply that the expectation of $$I$$ is $$\infty$$.
• @AzadTasan : Without squaring $f'/f$, the integral will in general exist only in the principal value en.wikipedia.org/wiki/Cauchy_principal_value sense; consider e.g. $f(x)\equiv x-1/2$. When you do square $f'/f$, even the principal value may be infinite. Commented Apr 14, 2021 at 20:04