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I'm trying to provide a full-fledged proof of the estimate

$$\Psi(x,x^{1/u}) = x \, \rho(u)+ O\left(\frac{x}{\log x}\right) \qquad (*)$$

via the sly inductive approach commonly attributed to A. Granville. (For any $U\geq 0$ the formula holds uniformly for $u \in [0,U]$ and all $x\geq 2$; the $\rho$ in it denotes the Dickman function.) As many of you know, Prof. Granville begins his proof by noticing that $(*)$ holds for every $u \in(0,2]$ and every $x\geq 2$ and by assuming afterwards that it also holds for every $u \in (0,N]$ and every $x\geq 2$. Then, resorting to the Buchstab identity

$$\Psi(x,y) = 1 + \sum_{p \leq y} \Psi \left(\frac{x}{p},p\right),$$

he obtains that

$$\Psi(x, x^{1/u}) = \Psi(x,x^{1/N})- \sum_{x^{1/u}<p\leq x^{1/N}}\Psi\left(\frac{x}{p},p\right). \qquad (**)$$

At this point, the idea is to invoke the induction hypothesis and rewrite every term in the sum on the right-hand side of $(**)$. Given that

$$p = \left(\frac{x}{p}\right)^{\log p / \log(x/p)},$$

it follows that if $p>x^{1/u}$ and $u \in (N,N+1]$, then

$$\frac{\log(x/p)}{\log p} = \frac{\log x}{\log p}-1 < u-1 \leq N$$

and whence

\begin{eqnarray*} \Psi(x,x^{1/u}) &=& x \, \rho(N) + O\left(\frac{x}{\log x}\right) - \sum_{x^{1/u}<p\leq x^{1/N}}\left[\frac{x}{p} \, \rho\left(\frac{\log x}{\log p}-1\right) + O\left(\frac{x/p}{\log(x/p)}\right)\right]\\ &=& x \, \rho(N) + O\left(\frac{x}{\log x}\right) + O\left(\frac{x}{\log x}\sum_{x^{1/u}<p\leq x^{1/N}} \frac{1}{p}\right) - x\sum_{x^{1/u}<p\leq x^{1/N}}\frac{1}{p} \, \rho\left(\frac{\log x}{\log p}-1\right)\\ &=& x \, \rho(N) + O\left(\frac{x}{\log x}\right) - x\sum_{x^{1/u}<p\leq x^{1/N}}\frac{1}{p} \, \rho\left(\frac{\log x}{\log p}-1\right).\\ \end{eqnarray*}

Notice that in going from the next-to-last line to the last one, we resorted to the famed second theorem of F. Mertens. Having said all this, my question is the following one:

What's in your experience the most clear-cut way to deal with the remaining sum in the last line of the math environment preceding the previous sentences?

By recurring to Merten's second theorem (once again) and the Abel summation formula, I am able to recover the expected main term but I don't see a way to sense in advance if proceeding thus one is to obtain a suitable error term (I eventually obtain it; yet, one usually prefers to find out whether he/she is on track before actually getting there, right?).

I would greatly appreciate any insights, suggestions, or comments you may want to leave for me below.

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    $\begingroup$ Use standard estimates on the prime counting function (with $\log^2$ saving at least in the error term) instead of Mertens' theorem. $\endgroup$ – js21 Feb 8 '17 at 9:58
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    $\begingroup$ Are you able to put an explicit constant times $x/\log x$ in place of $O(x/\log x)$? That would help me tremendously on some current research. Gerhard "Still Looking For Explicit Estimates" Paseman, 2017.04.28. $\endgroup$ – Gerhard Paseman Apr 29 '17 at 5:54
  • $\begingroup$ If I am not mistaken, in order to do so one would need to apply, instead of the asymptotic formulas for the sum of the reciprocal of the primes up to $x$ and the prime-counting function, some of the known sharp upper bounds for both functions (like those by J. B. Rosser and L. Schoenfeld). $\endgroup$ – José Hdz. Stgo. Apr 30 '17 at 7:26

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