Question 13.3(a) Tom Apostol Introduction to Analytic number theory

Question

I am self studying analytic number theory from Tom M Apostol and got struck on question 3(a) of Chapter-13.

It's image:

In (a) I thought of using Perron's Formula (statement here)

by taking identity $log\zeta(s) =\sum_{n=2}^{\infty}\frac{\Lambda(s) }{n^s log n}$ .

But the problem is that I am not sure what will occur on the RHS ie I am not able to simplyfy $\sum_{n=2}^{\infty}\frac{\Lambda(n) }{n^s log n}$.

I thought of Perron's Formula due to the setting in LHS in which question was asked.

So, can you please tell how to simplify RHS if question can be simplified by Perron's Formula?

Answer 1

To solve (a), you are correct to use the Perron's formula. Namely, we get that

\begin{align*} \frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty}\log(\zeta(s))\frac{x^s}{s}ds&=\frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty}\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s\log(n)}\frac{x^s}{s}ds\\ &=\sum_{n<x}\frac{\Lambda(n)}{\log(n)} \end{align*}

We now note that there are $\pi\left(x^{1/k}\right)$ powers $p^k$ of primes, and at powers $p^k$ we have

$$\frac{\Lambda(n)}{\log(n)}=\frac{\log(p)}{\log(p^k)}=\frac{1}{k}$$

Summing over all $k$ and counting the number of elements of each group yields the result.

To solve (b), we note that $\pi(x)<x$ and so since $x^{1/\log_2(x)}=2$ we get that

\begin{align*} \sum_{k=2}^{\infty}\frac{\pi(x^{1/k})}{k}&=\sum_{k=2}^{\log_2(x)}\frac{x^{1/k}}{k}\\ &\leq \frac{1}{2}\log_2(x)\sqrt{x}\\ &=o(x/\log(x)) \end{align*}

and thus this will only be assymptotic to $x/\log(x)$ if $\pi(x)$ is which completes our proof.

Answer 2

Whenever $n$ is not a prime power, we have $\Lambda(n)=0$, but when $n=p^k$, we have

$$ {\Lambda(n)\over\log n}={\log p\over\log p^k}=\frac1k $$

Hence, we have

$$ \sum_{n\le x}{\Lambda(n)\over\log n}=\sum_{p^k\le x}\frac1k=\sum_{k\ge1}\frac1k\sum_{p\le\sqrt[k]x}=\sum_{k\ge1}{\pi(\sqrt[k]x)\over k} $$

Consequently, by Perron's formula and the fact that $\log\zeta(s)=\sum_{n\ge2}{\Lambda(n)\over n^s\log n}$, we deduce

$$ {1\over2\pi i}\int_{c-i\infty}^{c+i\infty}\log\zeta(s)x^s{\mathrm ds\over s}=\sum_{n\le x}{\Lambda(n)\over\log n}=\pi(x)+{\pi(x^{1/2})\over2}+{\pi(x^{1/3})\over3}+\cdots $$