# How to prove a result related to prime number theorem in research paper of Rivoal and Zudilin

Question is ->I am studying research paper: A note on odd zeta values and I am unable to think how to deduce a result which the authors don't prove. This result has to be proved assuming the prime number theorem and it's on Page 12 of the paper :

Prove that $$\lim_ {n\to\infty} \frac{\log(\Phi_n) } {n} =\int_0^{1} \rho_0 (t) d(\psi(t) + 1/t)$$, where $$\psi(t)$$ = $$\frac {\Gamma'(t) } {\Gamma(t) }$$.

where $$\Phi(n)$$ and $$\rho(n)$$ are described in this image: Can someone please tell how to prove this result ?

I shall be really thankful.

• This question already had an accepted answer. Why have you bumped it with an edit? Jan 6 at 0:41

If we start by partitioning the range of summation into intervals on which $$\rho_0(\frac np)$$ is constant, we obtain \begin{align*} \log \Phi_n &= \sum_{2\sqrt n where $$\theta(x) = \sum_{p\le x} \log p \sim x$$ by the prime number theorem. Thus \begin{align*} \log \Phi_n &\sim \sum_{k=6}^{3\sqrt n-1} \rho_0\big( \tfrac k6 \big) \big( \tfrac{6n}k - \tfrac{6n}{k+1} \big) \sim 6n \sum_{k=6}^\infty \frac{\rho_0(k/6)}{k(k+1)}. \end{align*} A similar partitioning calculation on the integral will hopefully lead to the same result.
• in the fifth line of your answer how did you change $\sum_{2√n<p$\leq$n$ = $\sum_{k=6}^{3√n-1}$ $\sum_{6n/(k+1) p$\leq$6n/k) . Can you please elaborate why it is right? Mar 20, 2020 at 5:36 • There was a typo (now fixed). We're just splitting up the interval of summation from the first line into$3\sqrt n-1$consecutive intervals of summation. Mar 20, 2020 at 7:33 • On the first line, every prime between$2\sqrt n$and$n$appears exactly once, and smaller or larger primes don't appear at all. On the second line, do any primes smaller than$2\sqrt n$or larger than$n$appear? Given a prime$p$between$2\sqrt n$and$n$, for how many values of$k$will$p\$ appear in the inner sum? Apr 16, 2020 at 15:30
• We're just splitting up the range of summation into intervals on which the function is simpler. A strong analogy would be the computation $$\int_0^N \lfloor x\rfloor^2\,dx = \sum_{k=0}^{N-1} \int_k^{k+1} \lfloor x\rfloor^2\,dx = \sum_{k=0}^{N-1} k^2 \int_k^{k+1} 1\,dx.$$ Apr 16, 2020 at 15:32