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While self studying analytic number theory from Introduction to sieve methods and it's applications by M Ram Murthy and Alina Carmen,I have a doubt in text.

Authors write that Assuming Chebysheff theorem whose statement is - There exist positive constants A and B such that Ax < $\theta(x) $ < Bx.

Now authors says by partial summation, this implies the bound on π(x) ie π(x) = O($\frac {x} { log x } $ ) .

What I have done - > In Abel Summation formula I take a(n) = b(n) / log (x) , b(n) = 1 if n is prime and 0 otherwise. f(n) = 1/logx . So RHS becomes $\theta(x) $ / log x + $\int_{2}^x \frac{\theta(t) } { t log^2(t) } dt $ . Now, using $\theta(x) $ ~ x as x ->$\infty $ . And writing $\int_{2}^x$ = $\int_{2}^y$ + $\int _{y}^x$ and y - >$\infty$ .

But now on RHS I have O( x/ logx) + O ( $\int_{2}^{\infty} \frac{1} { log^2(t) } dt $ )- M ×$\int_ {x} ^{\infty} \frac{1} {log^2(t) } dt $ .

Now there are two problems about which I am unable to think 1. How to prove $\int_{2}^{\infty} \frac {1} {log^2(t) } dt $ is convergent? 2.How to evaluate integral $\int_{x}^{\infty} \frac {1} { log^2(t) } dt $ ?

Can someone please explain!!

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You don't need anything else than $$\pi(x)= \pi(x)-\pi(x^{1/2})+O(x^{1/2})$$ $$\frac{\theta(x)-\theta(x^{1/2})}{\log x}\le \pi(x)-\pi(x^{1/2})\le \frac{\theta(x)}{\log x^{1/2}}$$ to get $$\frac{\theta(x)}{x}\in [A,B]\implies \frac{\pi(x)}{x/\log x}\in [a,b]$$

Once we know that $\theta(x)=O(x)$ then the partial summation is $$\pi(x)=\sum_{n\le x}\frac{\theta(n)-\theta(n-1)}{\log n}=\frac{\theta(x)}{\log x}+\sum_{n\le x-1} \theta(n)(\frac1{\log n}-\frac1{\log (n+1)})$$ $$ = \frac{\theta(x)}{\log x}+\sum_{n\le x} O(n \frac{\log(n)-\log(n+1)}{\log^2 n})=\frac{\theta(x)}{\log x}+\sum_{n\le x} O(n \frac{1/n}{\log^2 n})$$ $$=\frac{\theta(x)}{\log x}+O(\frac{x}{\log^2 x})$$

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  • $\begingroup$ can you please tell how after the partial summation is ..you changed sum of $\theta(n-1)$ / logn to $\theta(n)$/ log(n+1) ? $\endgroup$ – Ben Jan 16 '20 at 6:53
  • $\begingroup$ Yes, I made trivial calculations. $\endgroup$ – reuns Jan 16 '20 at 6:54
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    $\begingroup$ No, this is completely trivial. $\endgroup$ – reuns Jan 16 '20 at 7:13
  • $\begingroup$ yes , i was overthinking . $\endgroup$ – Ben Jan 16 '20 at 7:15
  • $\begingroup$ can you please look at this question of mine if you have some spare time .math.stackexchange.com/questions/3510920/… $\endgroup$ – Ben Jan 16 '20 at 7:16

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