Upper bound for an integral.

Question

Let F and G be functions defined on $[1,\infty[$ by : $$F(x)=\int_1^x \frac{\ln^2(t)}{t(t-1)}dt$$ and $$G(x)=\int_1^x \frac{\ln^2(t)}{t^2}dt$$ I don't know if this is related or not but we first proved that $G(x)=2-\frac{1}{x}(\ln^2(x)+2\ln(x)+2)$ and it followed that $\lim_{x\to \infty} G(x) = 2$

Now, suppose $\lim_{x\to \infty} F(x)=l$ is a real number and prove that $0\lt l \lt 4+F(2)$

If you don't mind me bothering you, I would greatly appreaciate some help.

I tried everything I could think of but failed, thanks in advance.

Answer 1

For $t \ge 2$, $t-1 \ge t/2$ and so $\frac1{t-1} \le \frac2 t$ $$F(x) = F(2) + \int_{2}^{x}\frac{\ln^2t}{t(t-1)}\mathrm d t \le F(2) + 2\int_2^x \frac{\ln^2t}{t^2}\mathrm d t \le F(2) + 2G(x)- 2G(1)$$

As $F(2) \le F(x)$ for $x \ge 2$, then $$0 < F(2) \le l \le F(2) + 4 - 2G(2) < 4 + F(2)$$

Answer 2

This is clearly not the spirit of the question, but $G(+\infty), F(2)$ and $F(+\infty)$ all have simple closed forms. As you noticed, $G(+\infty)=2$, and

$$ F(+\infty)=\int_{1}^{+\infty}\left(\frac{1}{t-1}-\frac{1}{t}\right)\log^2(t)\,dt\stackrel{t\mapsto 1/u}{=}\int_{0}^{1}\frac{\log^2(t)}{1-t}\,dt=2\sum_{n\geq 1}\frac{1}{n^3}=2\,\zeta(3)$$ while $$ F(2) = \int_{0}^{1}\left(\frac{1}{t}-\frac{1}{t+1}\right)\log^2(t+1)\,dt=-\frac{\log^3(2)}{3}+2\sum_{n\geq 1}\frac{(-1)^{n+1}H_n}{(n+1)^2}=-\frac{\log^3(2)}{3}+\frac{\zeta(3)}{4} $$ by integration by parts and alternating Euler sums with weight $3$. On the other hand $\log(t+1)\leq \sqrt{t+1}-\frac{1}{\sqrt{t+1}}$ by the Cauchy-Schwarz inequality, hence $$ F(2) \leq \int_{0}^{1}\frac{(t+1)+\frac{1}{t+1}-2}{t(t+1)}\,dt =\log(2)-\frac{1}{2}\leq \frac{1}{\sqrt{2}}-\frac{1}{2}.$$ This is much sharper than the requested bound.