# Twin prime conjecture implies $\limsup_{n\to\infty}\frac{\sigma(n)\pi(n)}{n^2}\left(\pi(\log n)-\frac{\pi_2(\log n)}{2C_2}\right)=e^{\gamma}$?

Let $\sigma(n)$ the sum of positive divisor function, $\pi(x)$ is the prime counting function, $\pi_2(x)$ is the twin prime counting function (we will assume that Twin prime conjecture holds), $C_2$ is the twin prime constant and $\gamma$ is Euler's constant.

My question is a proof verification of:

a symbolic use of $\limsup_{n\to\infty}$, an exercise about the growth of $\sigma(n)$, Gronswall's theorem, a use of the method of integration by parts and if my use of the equivalences $\sim$ between prime counting functions and its corresponding logarithmic integral functions, in the statements of Prime Number Theorem and Twin Prime conjecture, holds. My goal is encourage to study more mathematics with nice topics and refresh my mathematics with easy computations.

Question. Assuming that the Twin prime conjecture holds it is possible to prove that $$\limsup_{n\to\infty}\frac{\sigma(n)\pi(n)}{n^2}\left(\pi(\log n)-\frac{\pi_2(\log n)}{2C_2}\right)=e^{\gamma},$$ by the way that I've sketched below? Thanks in advance.

PLease, if there are mistakes you are welcome to say it to me.

Integration by parts provide us of following statement (I've read in [1], page 5 as Remark 4) when you take $dv=dt$, $u=1/\log t$ thus $du=\frac{-1/t}{\log^2 t}dt$ and this gives $$\int_2^x \frac{dt}{\log t}=\Big(\frac{t}{\log t}\Big|_2^x+Li_2(x),$$

where we will take $Li(x)=\int_2^x \frac{dt}{\log t}$ and $Li_2(x)=\int_2^x \frac{dt}{\log^2 t}$.

Thus when we made the substitution $x$ by $\log x$, and after we multiply previous identity by $\frac{\sigma(x)}{x\log x}$ we obtain if there were not mistakes (labelling $x=n$, since $\sigma(n)$ is an arithmetic function)

$$\limsup_{n\to\infty}\frac{\sigma(n)}{n\log n}\left(Li(\log n)-Li_2(\log n)\right)=e^\gamma,$$ where I've used Gronswall's theorem, that states $\limsup \sigma(n)/(n\log\log n )=e^\gamma$ (see this site or MathWorld [3]), and this exercise $\sigma(n)\leq n(1+\log n)$ (see [2] in page 10 and page 214, for which reference I can not provide us a free access).

Now as subterfuge to try compute more and put on game more statements, I've asked to me if I can use equivalences to change all continuos functions that involves previous limit, I say if previous computations are rigth, by arithmetic functions. Thus Prime Number Theorem provide us of the asymptotic equivalences $\pi(n)\sim Li(n)$ and $\pi(n)\sim\frac{n}{\log n}$ and the Twin prime conjecture of this $\pi_2(n)\sim 2C_2Li_2(n)$, where $C_2$ is the twin prime constant (see [1], page 1 and page 5). And the use that leave implicit in the statement of my question, finish the sketch of my proof.

Thus I require details if the use of this statements is wrong, if it isn't possible this use. The statements and its symbolic use as I've detailed: substitutions, limits, bounds and use of asymptotic equivalences as is implicit. If it is possible write without problems, tell me as your answer. Very thanks much.

References:

[1] Sebah and Gourdon, Introduction to twin primes and Brun's constant computation, http://numbers.computation.free.fr/Constants/Primes/twin.pdf

[2] Ram Murthy, Problems in Analytic Number Theory, GTM RIM 206, Springer 2008.

[3] MathWorld, Gronswall's Theorem, http://mathworld.wolfram.com/GronwallsTheorem.html

• The identity couldn't a special meaning, I am looking made computations, only to encourage to me to study mathematics.
– user243301
Nov 7, 2015 at 21:22
• I ran a loop for n between $10^2$ and $10^4$, and found these results. Nov 8, 2015 at 12:15

I think your compuations are right. Note that $$\textrm{Li}\left(x\right)=\int_{2}^{x}\frac{dt}{t}=\left.\frac{t}{\log\left(t\right)}\right|_{2}^{x}+\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}$$ hence $$\textrm{Li}\left(\log\left(n\right)\right)-\textrm{Li}_{2}\left(\log\left(n\right)\right)=\frac{\log\left(n\right)}{\log\left(\log\left(n\right)\right)}-\frac{2}{\log\left(2\right)}\sim\frac{\log\left(n\right)}{\log\left(\log\left(n\right)\right)}$$ then $$\frac{\sigma\left(n\right)\pi\left(n\right)}{n^{2}}\left(\pi\left(\log\left(n\right)\right)-\frac{\pi_{2}\left(\log\left(n\right)\right)}{2C_{2}}\right)\sim\frac{\sigma\left(n\right)}{n\log\left(n\right)}\left(\textrm{Li}\left(\log\left(n\right)\right)-\textrm{Li}_{2}\left(\log\left(n\right)\right)\right)=$$ $$=\frac{\sigma\left(n\right)}{n\log\left(n\right)}\left(\frac{\log\left(n\right)}{\log\left(\log\left(n\right)\right)}-\frac{2}{\log\left(2\right)}\right)\sim\frac{\sigma\left(n\right)}{n\log\left(\log\left(n\right)\right)}$$ then $$\limsup_{n\rightarrow\infty}\frac{\sigma\left(n\right)\pi\left(n\right)}{n^{2}}\left(\pi\left(\log\left(n\right)\right)-\frac{\pi_{2}\left(\log\left(n\right)\right)}{2C_{2}}\right)=e^{\gamma}$$ and those are, essentially, your calculations.