Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.