Consider $F(a)=\int_{0}^{+\infty}{\frac{1}{1+x^a(\ln(1+x)^2f(x)}} \, \mathrm{d}x$ Let $f$ be $T$-periodic, continuous, such that $T=1$, $f(0)=0$, $f'(0)>0$, and $f(x)>0$ for $x\in(0,1)$.
$$F(a) = \int_{0}^{+\infty} {\frac{1}{1+x^a(\ln(1+x))^2f(x)}} \, \mathrm{d}x$$
Consider the domain, continuity, and differentiability of $F(a)$.
This is a question of Unified National Graduate Entrance Examination (China). I don't know how to do this problem. So I want to get some help and hint. If I make progress, I will edit it soon.
Thanks. Any help would be greatly appreciated :-)
My attempt
I think there exists a contradiction.
$$f(0)=f(1)=0,f'(1)=f'(0)>0$$
So $\exists \xi\in(0,1),f(\xi)<0$
This is the picture of question.

 A: Let me first assume that $f$ is continuous, non-negative and $1$-periodic. To study convergence of $F(a)$, we focus on the integral of the form
$$ I_n(a) = \int_{0}^{1} \frac{\mathrm{d}x}{1 + n^a \left( \log (n+1) \right)^2 f(x)}. $$
This is related to $F(a)$ by the inequality $\sum_{n=1}^{\infty} I_n(a) \leq F(a) \leq 1 + \sum_{n=1}^{\infty} I_n(a) $, and so, it suffices to investigate the convergence of $\sum_{n=1}^{\infty} I_n(a)$ instead. To this end, we make extra assumptions:


*

*There exist constants $0 < c_1 < C_1$ and $\delta > 0$ such that $c_1 |x| \leq f(x) \leq C_1|x|$ near $x = 0$.

*$f(x) > 0$ for all $x \in (0, 1)$.
Notice that the first condition is more general than the assumption that both the right-derivative $f'(0^+)$ and the left-derivative $f'(0^-)$ exist and are positive. Also, it is easy to check that these two assumptions are equivalent to the following single condition:


*

*There exist constants $0 < c_2 < C_2$ such that $c_2|x| \leq f(x) \leq C_2|x|$ for all $|x| \leq \frac{1}{2}$.


From this, we obtain bounds
$$
I_n(a)
\geq 2\int_{0}^{1/2} \frac{\mathrm{d}x}{1 + n^a \left( \log(n+1) \right)^2 C_2x}
= \frac{2\log\left(1 + (C_2/2) n^a \left( \log(n+1) \right)^2 \right)}{n^a \left( \log(n+1) \right)^2}
$$
and
$$
I_n(a)
\leq 2\int_{0}^{1/2} \frac{\mathrm{d}x}{1 + n^a \left( \log(n+1) \right)^2 c_2x}
= \frac{2\log\left(1 + (c_2/2) n^a \left( \log(n+1) \right)^2 \right)}{n^a \left( \log(n+1) \right)^2}
$$
From this, we find that there exist constants $0 < c_3 < C_3 $, possibly depending only on $a$, such that
$$ \frac{c_3}{n^a \log(n+1)} \leq I_n(a) \leq \frac{C_3}{n^a \log(n+1)}. $$
Therefore $F(a)$ converges if and only if $a > 1$.
Once the convergence is established, continuity of $F$ on $(1, \infty)$ is an easy consequence of the dominated convergence theorem. Differentiability is a bit more cumbersome to establish, but we can indeed prove that $F$ is differentiable on $(1, \infty)$.
