How to prove it without using the expression of $F'(x)$ , using the mean value theorem two times

Question

$\displaystyle F(x) = \int_{x}^{x^2} \dfrac{t-1}{\ln^2t} \,dt$ and $F(1)=\ln(2)$ defined for all $x >1$

I have to prove that : for all $x >1 $: $\quad \dfrac{3}{2} \le \dfrac{F(x)-F(1)}{x-1} \le x^2\dfrac{x+2}{2}$

using the mean value theorem two times. The problem is that I tried using two ways: the first is to use the theorem to the function $F$ between $[1;x]$ or to the antiderivative of the function inside the integral between $[x,x^2]$ but it seems to not work.

I managed to prove it but I had to use the expression of $F'(x)$. How can I prove it wihout using it?

Answer 1

This is not an answer. Just an attempt of clarification of this interesting but puzzling question. Here is a graphical representation of function $F$ :

together with a representation of the line segment $[A,X]$ (where $A(1,$ln$(2))$ and $X=(x,F(x))$) whose slope is

$$s_x=\dfrac{F(x)-F(1)}{x-1}$$

One can find also on this figure point $M(x_0,F(x_0))$ such that, for a given x, one has a form of mean value theorem :

$$F'(x_0)=\dfrac{F(x)-F(1)}{x-1}$$

Assuming that $F$ is convex (to be established), we have a graphical proof of the first equality (the slope is always greater than the slope in $A$).

How to apply (twice !) mean value theorem to obtain in particular the RHS $x^2\frac{x+2}{2}$ as a bounding value.

However, I have an idea : the integrand can be written under the form :

$$\frac{t-1}{ln^2(t)-ln^2(1)}$$

which can be "bracketed" by using mean value theorem...

Final remark : You say you have a solution using $F'(x)$ that does not use mean value theorem. Can you say briefly what it is ?

Besides, I have a problem with $F'(x)$ : I find $F'(x)=\frac{(x-1)(x^2-x+2)}{2 ln(x)^2}$ but it does not match the (limit) value $3/2$ it should have for $x=1$.

Answer 2

Consider the function $f:(0, \infty) \to\mathbb {R} $ defined by $$f(t) =\frac {t-1}{\log ^2t}-\frac{1}{t-1},t\neq 1,f(1)=1$$ then $f$ is continuous on $(0,\infty)$. The function $F$ of the question can be expressed as $$F(x) =\int_{x} ^{x^2}f(t)\,dt+\int_{x}^{x^2}\frac{dt}{t-1},\,x\neq 1,x>0$$ or $$F(x) =\log(1+x)+\int_{x}^{x^2}f(t)\,dt$$ and thus $F(x) \to\log 2$ as $x\to 1$. In this manner the value $F(1)=\log 2$ ensures that $F$ is continuous on $(0,\infty) $.

Now we can see that $$F(x) - F(1)=\log\frac{1+x}{2}+\int_{x}^{x^2}f(t)\,dt$$ The integral on right can be expressed as a difference of two integrals over $[1,x^2]$ and $[1,x]$ respectively and in the first integral we can put $t=u^2$ to get $$F(x) - F(1)=\log\frac{1+x}{2}+\int_{1}^{x}\{2tf(t^2)-f(t)\}\,dt$$ Now $$2tf(t^2)-f(t)=\frac{(t-1)^2(t+2)}{2\log^2t}-\frac{1}{t+1}$$ and hence $$F(x) - F(1)=\frac{1}{2}\int_{1}^{x}\left(\frac{t-1}{\log t} \right)^2(t+2)\,dt$$ We can now use the inequality $$1\leq\frac{t-1}{\log t} \leq t$$ and prove the desired result.