$\int^x_2 (\log u)^{-2} du\ll x(\log x)^{-2}$

Question

I came across this estimation in a book:

$$ \int^x_2 (\log u)^{-2} du \ll x(\log x)^{-2}. $$

I tried to prove it by first integrating by parts:

$$ \int^x_2 (\log u)^{-2} du = x(\log x)^{-2}-2(\log 2)^{-2} + 2\int^x_2 (\log u)^{-3} du, $$

but was then stuck. Could someone show me how to establish the estimation?

By $f\ll g$ the book meant that there is a constant $C>0$ such that $f(x)\le C g(x)$, in some domain of $x$. Here, I actually wanted for all $x\ge2$. — Tapioka, May 21 '20 at 05:40

score 1 · Accepted Answer · answered May 21 '20 at 05:27

1

By L'Hospital's rule $$ \mathop {\lim }\limits_{x \to + \infty } \frac{{\int_2^x {\frac{{du}}{{\log ^2 u}}} }}{{\frac{x}{{\log ^2 x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{{\log ^2 x}}}}{{\frac{1}{{\log ^2 x}} - \frac{2}{{\log ^3 x}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{1 - \frac{2}{{\log x}}}} = 1. $$ Thus we in fact have $$ \int_2^x {\frac{{du}}{{\log ^2 u}}} \sim \frac{x}{{\log ^2 x}}. $$

answered May 21 '20 at 05:27

Gary

31,845

Thanks for the answer. But actually I wanted for all $x\ge2$. – Tapioka May 21 '20 at 05:41
@Tapioka That is a consequence of this. – Gary May 21 '20 at 05:55
@ Gary: At first glance, it seemed that only for sufficiently large $x$ was proved. – Tapioka May 21 '20 at 05:57
The ratio tends to $1$. Thus there is an $X>2$ so that for $x>X$ the ratio is at most $2$, say. For $2\leq x\leq X$, the ratio has a maximum value $K$ since it is a continuous function. Thus, the ratio is at most $\max(2,K)$ for all $x\geq 2$. – Gary May 21 '20 at 06:06
Thanks for your clear explanation! – Tapioka May 21 '20 at 06:12
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@Tapioka If you appreciate an answer then you should upvote it and give it the $\color{green}{\checkmark}$ – gen-ℤ ready to perish May 21 '20 at 07:02

$\int^x_2 (\log u)^{-2} du\ll x(\log x)^{-2}$

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