My problem is to show that $$ \sum_{p\leq \sqrt x}\frac{x}{p\log(x/p)}=x\int_{2}^{\sqrt{x}}\frac{1}{u\log(x/u)\log u}\,\mathrm{d}u+O(x/\log x) $$ using Abel's summation formula and PNT. It is understood that $p$ is prime here. Letting $\phi(n)=\frac{x}{n\log(x/n)}$, we have $$ \sum_{p\leq \sqrt x}\frac{x}{p\log(x/p)}=\pi(\sqrt x)\phi(\sqrt x)-\int_{2}^{\sqrt{x}}\pi(t)\phi'(t)\,\mathrm{d}t, $$ but determining $\phi'(t)$, the integral becomes too big.


It's not super nice, but it's not so bad, actually. Write $\phi(t) = \psi_1(t)\cdot \psi_2(t)$ with

$$\psi_1(t) = \frac{x}{t}\quad\text{and}\quad \psi_2(t) = \frac{1}{\log (x/t)} = \frac{1}{\log x - \log t}.$$

Then we have

$$\psi_1'(t) = - \frac{x}{t^2}\quad\text{and}\quad \psi_2'(t) = -\frac{1}{(\log x - \log t)^2}\cdot \biggl(-\frac{1}{t}\biggr) = \frac{1}{t(\log (x/t))^2},$$


\begin{align} \phi'(t) &= \psi_1'(t)\psi_2(t) + \psi_1(t)\psi_2'(t) \\ &= -\frac{x}{t^2\bigl(\log (x/t)\bigr)} + \frac{x}{t^2\bigl(\log (x/t)\bigr)^2}. \end{align}

For $t \leqslant \sqrt{x}$, the second term is in modulus smaller than the first by a factor between $\frac{1}{\log x}$ and $\frac{1}{\log \sqrt{x}} = \frac{2}{\log x}$, so in the integral the first term is the dominant one, and the second can [a priori, only probably, we need to check that this is the case] be subsumed in the error term.

Using $\pi(t) = \frac{t}{\log t} + O\bigl(\frac{t}{(\log t)^2}\bigr)$ we first get

$$\pi(\sqrt{x})\phi(\sqrt{x}) = \frac{\sqrt{x}}{\log \sqrt{x}}\cdot\frac{x}{\sqrt{x}\log (x/\sqrt{x})}\cdot\Bigl(1 + O\bigl((\log x)^{-1}\bigr)\Bigr) \in O\biggl(\frac{x}{(\log x)^2}\biggr),$$

so this can be subsumed in the error term. As the main contribution to the integral we have

$$-\int_2^{\sqrt{x}} \pi(t)\psi_1'(t)\psi_2(t)\,dt = x\int_2^{\sqrt{x}} \frac{dt}{t(\log t)\bigl(\log(x/t)\bigr)} + x\int_2^{\sqrt{x}} O\biggl(\frac{1}{(\log t)^2}\biggr)\frac{dt}{t\bigl(\log (x/t)\bigr)},$$

where the first term is exactly the integral we want, and the second is smaller due to the additional $\log t$ factor in the denominator. With $\frac{1}{2}\log x \leqslant \log \frac{x}{t} \leqslant \log x$ for $1 \leqslant t \leqslant \sqrt{x}$, we can estimate the second integral by a multiple of

$$\frac{x}{\log x} \int_2^{\sqrt{x}} \frac{dt}{t(\log t)^2},$$

and since $\int_2^{\infty} \frac{dt}{t(\log t)^2} < +\infty$, that is $O(x/\log x)$. For the remaining

$$-\int_2^{\sqrt{x}} \pi(t)\psi_1(t)\psi_2'(t)\,dt,$$

we again pull out $\frac{1}{(\log (x/t))^2}$ and use $\pi(t) \in O(t/\log t)$ to get

$$\Biggl\lvert\int_2^{\sqrt{x}} \pi(t)\psi_1(t)\psi_2'(t)\,dt\Biggr\rvert \leqslant C\cdot\frac{x}{(\log x)^2} \int_2^{\sqrt{x}} \frac{dt}{t(\log t)} \leqslant \tilde{C} \cdot\frac{x\log \log x}{(\log x)^2},$$

so that belongs to $o(x/\log x)$.

Altogether, we have seen that indeed

$$\sum_{p \leqslant \sqrt{x}} \frac{x}{p\log (x/p)} = x\int_2^{\sqrt{x}} \frac{dt}{t(\log t)\bigl(\log (x/t)\bigr)} + O\biggl(\frac{x}{\log x}\biggr).$$

(Note: that yields $\pi_2(x) \sim \frac{x\log \log x}{\log x}$, where $\pi_2(x)$ is the number of semiprimes not exceeding $x$.)

  • $\begingroup$ You write that this yields that $\pi_2(x) \sim \frac{x \log \log x }{\log x}$. But how are you able to see that from the above calculations? $\endgroup$ – slowpoke Dec 17 '20 at 1:56
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    $\begingroup$ @slowpoke Which part are you asking about, the last integral being asymptotically $\frac{\log \log x}{\log x}$, or the sum being essentially $\pi_2(x)$? Neither of those is totally obvious. On the other hand, neither is deep. For the integral, the substitution $u = \log t$ and a partial fraction decomposition lead to an explicit evaluation. For the sum, note that grouping semiprimes by their smaller prime factor leads to $$\pi_2(x) = \sum_{p \leqslant \sqrt{x}} \biggl(\pi\biggl(\frac{x}{p}\biggr) - \bigl(\pi(p)-1\bigr)\biggr)$$ (remove the "${}-1$" if you don't want to count squares of primes) $\endgroup$ – Daniel Fischer Dec 17 '20 at 14:30
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    $\begingroup$ and the second part of that sum is asymptotically negligible. Then approximating $\pi(y)$ with $\frac{y}{\log y}$ leads to the sum in question. (Of course one needs to verify that this approximation doesn't change the asymptotics.) $\endgroup$ – Daniel Fischer Dec 17 '20 at 14:30
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    $\begingroup$ @slowpoke Again I'm not quite sure what you're asking, so I may be off in answering. The "second part" refers to $S_2(x) := \sum_{p \leqslant \sqrt{x}} \bigl(\pi(p) - 1\bigr)$, the first part of the sum giving the value of $\pi_2(x)$ is $S_1(x) := \sum_{p \leqslant x} \pi\bigl(x/p\bigr)$. In order to obtain an asymptotic expression for $\pi_2(x)$, there are several steps to carry out (of course the way can be modified). A) Find that indeed $\pi_2(x) = S_1(x) - S_2(x)$. B) Show that $S_2(x)$ is of smaller order than $S_1(x)$, so that $\pi_2(x) \sim S_1(x)$ follows. $\endgroup$ – Daniel Fischer Jan 5 at 14:02
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    $\begingroup$ C) Show that $$S_1(x) = \sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr) \sim \sum_{p \leqslant \sqrt{x}} \frac{x}{p\log \frac{x}{p}}\,.$$ D) Find the asymptotic behaviour of that last sum. That is here done by writing it as an integral per Abel's formula, and then evaluating/estimating the integral. In each step, except A), one needs to show that some approximation one makes to simplify doesn't affect the overall asymptotics. $\endgroup$ – Daniel Fischer Jan 5 at 14:02

Taking $\phi\left(z\right)=1/\left(z\log\left(x/z\right)\right) $ in the Abel's formula we have $$S=x\sum_{p\leq\sqrt{x}}\frac{1}{p\log\left(x/p\right)}=\frac{2\pi\left(\sqrt{x}\right)}{\log\left(x\right)}-x\int_{2}^{\sqrt{x}}\frac{\pi\left(t\right)\left(1-\log\left(x/t\right)\right)}{t^{2}\log^{2}\left(x/t\right)}dt $$ $$=\frac{2\pi\left(\sqrt{x}\right)}{\log\left(x\right)}-x\int_{2}^{\sqrt{x}}\frac{\pi\left(t\right)}{t^{2}\log^{2}\left(x/t\right)}dt+x\int_{2}^{\sqrt{x}}\frac{\pi\left(t\right)}{t^{2}\log\left(x/t\right)}dt. $$ Now using the estimation $$\pi\left(z\right)=O\left(\frac{z}{\log^{2}\left(z\right)}\right) $$ and the obvious inequality $$\frac{1}{\log\left(x/t\right)}\leq\frac{2}{\log\left(x\right)},\, t\in\left[2,\sqrt{x}\right]\tag{1} $$ we get $$\int_{2}^{\sqrt{x}}\frac{\pi\left(t\right)}{t^{2}\log^{2}\left(x/t\right)}dt=O\left(\int_{2}^{\sqrt{x}}\frac{dt}{t\log^{2}\left(x/t\right)\log\left(t\right)}\right) $$ $$=O\left(\frac{1}{\log^{2}\left(x\right)}\int_{2}^{\sqrt{x}}\frac{dt}{t\log\left(t\right)}\right)=O\left(\frac{\log\left(\log\left(\sqrt{x}\right)\right)}{\log^{2}\left(x\right)}\right). $$ Using the PNT in the form $$\pi\left(z\right)=\frac{z}{\log\left(z\right)}+O\left(\frac{z}{\log^{2}\left(z\right)}\right) $$ in the other integral we get $$\int_{2}^{\sqrt{x}}\frac{\pi\left(t\right)}{t^{2}\log\left(x/t\right)}dt.=\int_{2}^{\sqrt{x}}\frac{1}{t\log\left(x/t\right)\log\left(t\right)}dt+O\left(\int_{2}^{\sqrt{x}}\frac{1}{t\log\left(x/t\right)\log^{2}\left(t\right)}dt\right) $$ and using again $(1)$ we have $$ \int_{2}^{\sqrt{x}}\frac{1}{t\log\left(x/t\right)\log^{2}\left(t\right)}dt=O\left(\frac{1}{\log\left(x\right)}\right) $$ and finally $$\frac{2\pi\left(\sqrt{x}\right)}{\log\left(x\right)}=O\left(\frac{\sqrt{x}}{\log^{2}\left(x\right)}\right) $$ hence $$x\sum_{p\leq\sqrt{x}}\frac{1}{p\log\left(x/p\right)}=\int_{2}^{\sqrt{x}}\frac{1}{t\log\left(x/t\right)\log\left(t\right)}dt+O\left(\frac{x}{\log\left(x\right)}\right) $$ as wanted.


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