I came across this problem in this book called Cambridge studies in advanced mathematics 97, page 217-218. I wad reading this line where the author starts the inductive steps and apply the inductive hypothesis and yields $$\Phi(x,x^{\frac{1}{u}}) = \frac{U\omega(U)x}{\ln(x)}+O\bigg(\frac{x}{\ln^2(x)}\bigg)+\sum_{x^{\frac{1}{u}}\leq p<x^{\frac{1}{U}}} \bigg(\frac{u_p\omega(u_p)x}{p\ln(\frac{x}{p})}+O\bigg(\frac{p}{\ln(p)}\bigg)+O\bigg(\frac{x}{p\ln^2(x)}\bigg)\bigg)$$ which indeed is $$\Phi(x,x^{\frac{1}{u}}) = \frac{U\omega(U)x}{\ln(x)}+O\bigg(\frac{x}{\ln^2(x)}\bigg)+O\bigg(\frac{x^{\frac{2}{U}}}{\ln^2(x)}\bigg)+O\bigg(\frac{x}{\ln^2(x)}\bigg)+\sum_{x^{\frac{1}{u}}\leq p<x^{\frac{1}{U}}} \frac{u_p\omega(u_p)x}{p\ln(\frac{x}{p})}$$ since the author wrote that the sum over $p$ of the first error term is $\ll x^{\frac{2}{U}}/\ln^2(x)$, and the sum over $p$ of the second is $\ll x/\ln^2(x)$. The author then estimate the contribution of the main term in the sum by writing the Prime Number Theorem in the form $\pi(t)=\mathrm{li}(t)+R(t)$, then apply Riemann-Stieltjes integration and yields
$$\sum_{x^{\frac{1}{u}}\leq p<x^{\frac{1}{U}}} \frac{\omega(\frac{\ln(x)}{\ln(p)}-1)x}{p\ln(p)}=\int_{x^{\frac{1}{u}}}^{x^{\frac{1}{U}}} \frac{\omega(\frac{\ln(x)}{\ln(t)}-1)x}{t\ln^2(t)}\,dt+f(t)R(t)\bigg|_{x^{\frac{1}{u}}}^{x^{\frac{1}{U}}}-\int_{x^{\frac{1}{u}}}^{x^{\frac{1}{U}}} R(t)\,df(t)$$ where $$f(t)=\frac{\omega(\frac{\ln(x)}{\ln(t)}-1)x}{t\ln(t)}$$ which in fact it is $$\sum_{x^{\frac{1}{u}}\leq p<x^{\frac{1}{U}}} \frac{\omega(\frac{\ln(x)}{\ln(p)}-1)x}{p\ln(p)}=\int_{x^{\frac{1}{u}}}^{x^{\frac{1}{U}}} \frac{\omega(\frac{\ln(x)}{\ln(t)}-1)x}{t\ln^2(t)}\,dt+\int_{x^{\frac{1}{u}}}^{x^{\frac{1}{U}}} \frac{\omega(\frac{\ln(x)}{\ln(t)}-1)x}{t\ln(t)}\,dR(t) $$ before applying Riemann-Stieltjes integration. I noticed that for large $t$, we have $$dR(t)=\bigg(\frac{\ln(t)-1}{\ln^2(t)}-\frac{1}{\ln(t)}\bigg)\,dt=-\frac{dt}{\ln^2(t)}$$ I am also reading this paper on Dickman function and noticed the similar situation $$\sum_{x^d<p\leq x^{u_1}} \rho\bigg(\frac{\ln(p)}{\ln(x)-\ln(p)}\bigg)\frac{x}{p}=x\int_{x^d}^{x^{u_1}} \rho\bigg(\frac{\ln(t)}{\ln(x)-\ln(t)}\bigg)\,dF(t)$$ where $F(t)=\sum_{p\leq t} \frac{1}{p}$, therefore I guess that $\frac{1}{p\ln(p)}$ could also be expressed as such and my conjecture would be $$\sum_{x^{\frac{1}{u}}\leq p<x^{\frac{1}{U}}} \frac{\omega(\frac{\ln(x)}{\ln(p)}-1)x}{p\ln(p)}=\int_{x^{\frac{1}{u}}}^{x^{\frac{1}{U}}} \frac{(\ln(t)-1)\omega(\frac{\ln(x)}{\ln(t)}-1)x}{t\ln^3(t)}\,dt$$ The difficulty that I am facing is to prove the conjecture I made above, but currently I have no idea how this can be achieved. Is this conjecture correct and if it is correct, then how do I show it? Therefore I hope someone could explain the things that I am missing.

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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Jan 28, 2021 at 9:54
  • $\begingroup$ Thank you so much for commenting, this is the first time that i am asking questions on MSE, i edit the question and add in context and further information, also some of my thoughts on the problem, but i am yet to figure out, maybe i am missing out something obvious. $\endgroup$ Jan 29, 2021 at 3:42
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    $\begingroup$ It's a fine question now. $\endgroup$ Jan 29, 2021 at 7:23
  • $\begingroup$ Some of the $=$ are $\sim\ $ (in particular in the last formula) $\endgroup$
    – reuns
    Jan 29, 2021 at 10:09
  • $\begingroup$ Yeap, you are right, i have made the assumption for large t, which in fact it's the $\pi(t)~\frac{t}{\ln(t)}$ $\endgroup$ Jan 29, 2021 at 10:25


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