# How to prove: for some $c>0,x>2 , c,x\in \mathbb R , \, \int_2^x \frac{\mathrm dt}{\log t}-\frac{x}{\log x} \leq \frac{cx}{(\log x)^2}$

How to prove: for some $c>0,x>2 , c,x\in \mathbb R$

$$\int_2^x \frac{\mathrm dt}{\log t}-\frac{x}{\log x} \leq \frac{cx}{(\log x)^2}$$

I have tried my textbook, notes and also tried to find something similar on the internet, if someone could please help.

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## 1 Answer

Hint: Integrate by parts. Let $u=\dfrac{1}{\log t}$ and let $dv=dt$. Then $du=-\dfrac{1}{t\log^2 t}$ and we can take $v=t$. When we go through the process, there will be a main term of $\dfrac{x}{\log x}$, and a couple of other terms (one of them an integral) that are not hard to bound.

Remark: The integration by parts "trick" that we used is in fact a standard and useful method for producing estimates of integrals.

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Would computing the integral be sufficient, being that the questions asks for a proof? –  Dick Feb 20 '13 at 12:47
When you do the integration by parts, you will get the main term, minus $\frac{2}{\log 2}$, plus $\int_2^x \frac{dx}{\log^2 x}$. You need to prove that these two terms together are bounded by some $\frac{cx}{\log^2 x}$, so there is something to prove. You will not be able to find an explicit expression for $\int_2^x \frac{dx}{\log^2 x}$. –  André Nicolas Feb 20 '13 at 12:54
Thanks for your help. I am very grateful. –  Dick Feb 20 '13 at 22:02
You are welcome. I take it the rest went just fine. –  André Nicolas Feb 20 '13 at 22:05