Estimates of $f(x) = \int_x^{x^2} \dfrac{dy}{\ln y}$

Question

I'm trying to solve this question:

Let $f$ be the function defined for $x \in (0;1): f(x) = \int_x^{x^2} \dfrac{1}{\ln y}\,dy$.

Show that

$$\forall x \in (0;1), x^2 \ln2 \le \ f(x) \leq x \ln2$$

Probably it's nothing difficult, but I'm completely blocked. Because I'm not able to compute that integral, I've find the $\int_x^{x^2} \dfrac{dy}{y \ln y}$, an what was my surprise that this integral equal $\ln 2$. However, there I've finished. I have no idea how to use this result, or if it's possible to use it somehow.

I wanted to make some estimates of $\dfrac{1}{y}$, but it didn't make any sense.

If you had some little hint for help me, I would be so happy.

Answer 1

You are on the right track. Simply write

$$\int_x^{x^2}\frac{1}{\log y}\,dy=\int_x^{x^2}\frac{y}{y\log y}\,dy$$

Then, we have for $x\in (0,1)$, $0<x^2<x<1$. Therefore,

$$x^2\int_x^{x^2}\frac{1}{y\log y}\,dy\le\int_x^{x^2}\frac{y}{y\log y}\,dy\le x\int_x^{x^2}\frac{1}{y\log y}\,dy$$

Can you finish now?

Answer 2

If we set $y=x^{\alpha}$ we have $dy = \log(x)\,x^{\alpha}\,d\alpha$ and: $$ f(x)=\int_{x}^{x^2}\frac{dy}{\log y} = \int_{1}^{2}\frac{x^{\alpha}}{\alpha}\,d\alpha. $$ By applying the Cauchy-Schwarz inequality we have an extremely tight upper bound: $$ \forall x\in(0,1),\qquad f(x)\leq\frac{x}{2}\sqrt{\frac{1-x^2}{-\log x}} \tag{1}$$ and a not-so-bad lower bound comes from $x^\alpha = e^{\alpha\log x} = x e^{(\alpha-1)\log x} \geq x\left(1+(\alpha-1)\log x\right)$: $$ \forall x\in(0,1),\qquad f(x)\geq x\log(2)+(1-\log 2)\,x\log x.\tag{2} $$ The difference between the RHS of $(1)$ and $(2)$ is at most $0.04$ for $x\in (0,1)$.