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Let $\Lambda$ be the von Mangoldt function. I am interested in understanding the average $$\sum_{n=1}^x \Lambda(n)^2.$$ By partial summation and the prime number theorem one can prove that this is $$ x(\log x) -x +O\left(\frac{x}{(\log x)^A}\right)$$ for any fixed $A>0$. But my question is whether one can see this directly from the residue of some Perron style formula? Presumably the Dirichlet series of $\Lambda^2$ has a double pole at $s=1$ but I am not aware of any nice Dirichlet series for $\Lambda^2$.

Replacing $\Lambda^2(n)$ by $(\log n) \Lambda(n)$ would change almost nothing in any asymptotics and $(\log n) \Lambda(n)$ is nicer to work with as it comes up naturally as the derivative of $\zeta'/\zeta$. The equality
$$\left(\frac{\zeta'}{\zeta}\right)'= \frac{\zeta"}{\zeta} -\left(\frac{\zeta'}{\zeta}\right)^2 = \frac{\zeta"}{\zeta'} \frac{\zeta'}{\zeta} -\left(\frac{\zeta'}{\zeta}\right)^2 $$ is equivalent to $$ -\Lambda \cdot \log =f\ast \Lambda - \Lambda \ast \Lambda ,$$ where $\cdot$ is pointwise multiplication, $\ast$ is the Dirichlet convolution and $f$ is the coefficient of $-\zeta"/\zeta'$. But $f$ looks a bit mysterious!

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    $\begingroup$ note that since $\Lambda(n)(\log n-\Lambda(n))$ is supported on prime powers at least $2$, the difference between $\sum \Lambda(n)^2 n^{-s}$ and $\left(\frac{\zeta'}{\zeta}\right)'$ is a Dirichlet series convergent for $\Re s >1/2$ so indeed the asymptotics and the function to apply Perron too is $\left(\frac{\zeta'}{\zeta}\right)'$ $\endgroup$ Commented Oct 3 at 21:30
  • $\begingroup$ Yes indeed! But $\Lambda(n) (\log n)$ also seems a bit unnatural. In fact I need it in a different setting where partial summation does not work... $\endgroup$ Commented Oct 4 at 7:31
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    $\begingroup$ I don't see the difficulty: just displace the contour to the left for $(\zeta'/zeta)'$. The analysis is not any harder. There are contexts in which the identity you've given is the first step towards something (since $\zeta''/\zeta$ is more amenable to a sieve approach, and $\Lambda\ast \Lambda$ gives rise to a "type II" (bilinear) sum), but this is not one of them. $\endgroup$ Commented Oct 5 at 6:19
  • $\begingroup$ Have you any thoughts, Dr. Pi, on the answer posted some time ago by Steven Clark? $\endgroup$ Commented Dec 12 at 21:53

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The asymptotic for

$$b(x)=\sum\limits_{n\le x} \Lambda(n)\, \log(n)\tag{1}$$ is $$b(x)\approx x\, (\log(x)-1)\tag{2}$$

which follows from the explicit formula

$$b_o(x)=\lim\limits_{\epsilon\to 0} \left(\frac{b(x-\epsilon)+b(x+\epsilon)}{2}\right)\\=x\, (\log(x)-1)+\frac{\pi^2}{12}-\gamma^2-2 \gamma_1+\sum\limits_{\rho} \frac{1-\rho \log(x)}{\rho^2}\, x^{\rho}\\+\sum\limits_{n=1}^{\infty} \frac{1+2 n \log(x)}{4 n^2}\, x^{-2 n}\\=x\, (\log(x)-1)+\frac{\pi^2}{12}-\gamma^2-2 \gamma_1+\sum\limits_{\rho} \frac{1-\rho \log(x)}{\rho^2}\, x^{\rho}+\frac{1}{4} \left(\text{Li}_2\left(\frac{1}{x^2}\right)-2 \log\left(1-\frac{1}{x^2}\right) \log(x)\right),\quad x\ge 1\tag{3}$$.

Formula (3) above is equivalent to

$$b_o(x)=\lim\limits_{\epsilon\to 0} \left(\frac{b(x-\epsilon)+b(x+\epsilon)}{2}\right)\\=x\, (\log(x)-1)+1+\sum\limits_{\rho} \frac{x^{\rho} (1-\rho \log(x))-1}{\rho^2}\\+\sum\limits_{n=1}^\infty \frac{x^{-2 n} (1+2 n \log(x))-1}{4 n^2}\\=x\, (\log(x)-1)+1+\sum\limits_{\rho} \frac{x^{\rho} (1-\rho \log(x))-1}{\rho^2}+\frac{1}{4} \left(\text{Li}_2\left(\frac{1}{x^2}\right)-2 \log(x) \log\left(1-\frac{1}{x^2}\right)-\frac{\pi^2}{6}\right),\quad x\ge 1\tag{4}$$

which makes it more clear that $b_o(1)=0$.

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