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Can provide us a linear convex combination of $Li(x)=\int_2^x\frac{1}{\log(t)}dt$ and $\frac{x}{\log(x)}$ a better approximation for $\pi(x)$, the prime counting function? Or not, is better $Li(x)$ when we consider the ratio $\frac{\pi(x)}{Li(x)}$ for large real numbers $x$? I know that $Li(x)$ is better than $\frac{x}{\log x}$.

Let a fixed $0<\theta<1$, for such linear convex combination $L_\theta(x)$ we can compute using Prime Number Theorem $$\lim_{x\to\infty}\frac{L_\theta(x)}{\pi(x)}=1,$$ but it does improve the approximation that provide us the logarithmic integral $Li(x)$?

Question. Let $\theta$ a real number such that $0<\theta<1$, does improve $$\theta\int_2^x\frac{1}{\log(t)}dt+(1-\theta)\frac{x}{\log(x)}$$ the aproximation of Prime Number Theorem? I say the asymptotic $\sim 1$ for $\pi(x)$ given by $Li(x)$ or $x/\log x$. Thanks in advance.

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  • $\begingroup$ I don't know if has sense $(Li(x))^{\theta}\cdot(x/\log(x))^{1-\theta}$, for a fixed real $0<\theta<1$. I don't know how use PNT easily to test this possibility as $x\to\infty$. $\endgroup$ – user243301 Dec 9 '15 at 20:34
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    $\begingroup$ No, $x/\log x$ is much worse an approximation to $\pi(x)$ than $\mathrm{Li}(x)$, and taking linear convex combinations does not improve the estimate except when $\theta = 1$. See my answer to a related question here: mathoverflow.net/questions/70713/… $\endgroup$ – Peter Humphries Dec 9 '15 at 21:13
  • $\begingroup$ Then as you see, if you want edit an useful answer for this MSE you are welcome @PeterHumphries . I try understand your reference. Very thanks much $\endgroup$ – user243301 Dec 9 '15 at 23:49

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