# Reference request for the explicit formula for $\sum_{n\leq x} \Lambda(n)n^{-s}$

Denote by $$\Lambda(n)$$ th e von Mangoldt function, which is equal to $$\log p$$ if $$p\geq 2$$ is a prime, and $$0$$ otherwise. Let $$\rho$$ denote a complex zero of the Riemann $$\zeta$$-function. If i recall well, i once heard sometime ago that

$$\sum_{n\leq x} \Lambda(n)n^{-s} = -\frac{\zeta'}{\zeta}(s) + \frac{x^{1-s}}{1-s} - \sum_{|Im \rho| \leq x} \frac{x^{\rho-s}}{\rho-s} + O(\log^{2}x)$$ for $$s\neq 1, s\neq \rho$$ and $$s\neq -2k, k\in \mathbb{N}$$.

Does anyone have a reference for this result ?

• @Sylvain von Mangoldt's work came 30 years after Riemann's. Jul 29, 2020 at 12:07
• As stated the asymptotic is only valid in the regime where $s$ is fixed and $x$ is sent to infinity (or to put it another way, the implied constant in $O(\log^2 x )$ will depend on $s$). At this level of non-uniformity the $-\frac{\zeta'}{\zeta}(s)$ term can be absorbed into the error. There are more uniform versions of this formula (see e.g., Exercise 27 of terrytao.wordpress.com/2014/12/09/… ) but the restriction $|\mathrm{Im} \rho| \leq x$ should be replaced by $|\mathrm{Im} (\rho-s)| \leq x$. Aug 16, 2020 at 16:20
• In my notes I referred to this sort of formula as a "truncated Landau explicit formula" but after looking through the 1911 and 1912 papers of Landau on the subject it does not appear that this formula appears in this exact form, though many related formulae of this type do. Aug 16, 2020 at 16:21
• @TerryTao The non-truncated version is indeed due to Landau. I needed this formula recently and a colleague (J. Merikoski) pointed out to me that it is used in H. L. Montgomery's original paper on pair correlation in number theory, "The Pair correlation of zeros of the zeta function" (Analytic Number Theory, Proc. Sympos. Pure Math. 24, St. Louis Univ. Missouri 1972, 181-193, 1973). There, a precise reference is given to a book of Landau: page 353 of "Handbuch der Lehre von der Verteilung der Primzahlen", Teubner, Berlin, 1909. Dec 23, 2022 at 22:44
• @OfirGorodetsky Thanks for clearing up the reference! I don't recall where I first found this formula attributed to Landau but it is good to actually have a concrete citation to back it up. Dec 24, 2022 at 2:02

I don't know a reference off-hand, but here is a sketch of the proof (the details need to be checked carefully, and I have not done it). One can start from $$\sum_{n\leq x}\frac{\Lambda(n)}{n^s}=\frac{1}{2\pi i}\int_{(\sigma)}-\frac{\zeta'(z)}{\zeta(z)}\cdot\frac{x^{z-s}}{z-s}\,dz,\qquad\sigma>\max(1,\mathrm{Re}\,s),$$ which is a variant of Theorem 5.1 in Montgomery-Vaughan: Multiplicative number theory I, and can be proved in the same way. As in the theorem, the RHS is understood as a Cauchy principal value, while in case of $$x\in\mathbb{N}$$ the term corresponding to $$n=x$$ in the LHS is counted with weight $$1/2$$.

The integrand is meromorphic with simple poles at $$z=s$$, $$z=1$$, and $$z=\rho$$. The corresponding residues are $$-\zeta'(s)/\zeta(s)$$, $$x^{1-s}/(1-s)$$, and $$-m_\rho\cdot x^{\rho-s}/(\rho-s)$$, where $$m_\rho$$ is the multiplicity of $$\rho$$. So one can derive the OP's display by performing the following steps:

1. Truncate the integral on the RHS to $$|\mathrm{Im}\,z|\leq x$$ and estimate the error introduced. Perturb $$x$$ slightly if it is very close to some $$\mathrm{Im}\,\rho$$.
2. Extend the truncated contour (which is a vertical line segment) to a rectangle containing the points $$s=1$$ and $$s=0$$, hence all the $$\rho$$'s with $$|\mathrm{Im}\,\rho|\leq x$$. By the Residue Theorem, the integral weighted by $$1/(2\pi i)$$ equals the sum of corresponding residues listed above.
3. Estimate the contribution of the horizontal line segments of the rectangular contour, as well as of the vertical line segment to the left of $$s=0$$.
4. The LHS equals the sum of residues listed in item 2, up to the error terms listed in items 1 and 3.

The non-truncated version of this estimate is due to E. Landau, see page 353 of his book "Handbuch der Lehre von der Verteilung der Primzahlen", Teubner, Berlin, 1909. It states that for $$s \neq 1$$ for which $$\zeta(s) \neq 0$$ one has $$(\star)\, {\sum_{n \le x}}'\frac{\Lambda(n)}{n^s} = \frac{x^{1-s}}{1-s}-\frac{\zeta'}{\zeta}(s)-\sum_{\rho}\frac{x^{\rho-s}}{\rho-s} + \sum_{k=1}^{\infty} \frac{x^{-2k-s}}{2k+s},$$ where the $$\prime$$ in the sum indicates that the last term is counted with weight $$1/2$$ if $$x$$ is a positive integer. Here the sum is over non-trivial zeros of $$\zeta$$, as usual.
One should also say that $$(\star)$$ holds for $$s=1$$ as well, if one interprets $$x^{1-s}/(1-s) - \zeta'(s)/\zeta(s)$$ as its limit at $$s=1$$, which is $$\log x -\gamma$$.
A truncated version has been used many times in the literature (e.g. Lemma 6 of the paper "On integers free of large prime factors" by A. Hildebrand and G. Tenenbaum, Trans. Am. Math. Soc. 296, 265-290, 1986) and is known to experts. Apart from the blog post Terry Tao mentioned in the comments, and the proof sketch given in GH from MO's answer, a similar proof can also be found in the appendix of the arXiv preprint 2211.08973. It states that uniformly for $$\Re s \ge 0$$, $$x \ge 4$$ and $$T \ge 2 +3 |\Im s|$$ we have $${\sum_{n \le x}}'\frac{\Lambda(n)}{n^s} = \frac{x^{1-s}}{1-s}-\frac{\zeta'}{\zeta}(s)-\sum_{|\Im(\rho-s)|\le T}\frac{x^{\rho-s}}{\rho-s} + \sum_{k=1}^{\infty} \frac{x^{-2k-s}}{2k+s} + R(x,T)$$ if $$\zeta(s)\neq 0$$, where $$R$$ satisfies the bound $$R(x,T) \ll (\log x) (x-1)^{-\Re s}\min\left\{ 1, \frac{x}{T\langle x \rangle}\right\} + \frac{\log^2 (xT)}{T}\left( 2^{\Re s } x^{1-\Re s}+\frac{2^{-\Re s}}{\log x}\right)$$ with an absolute implied constant. For $$s=0$$ this recovers the usual explicit formula (Theorem 12.5 from Montgomery--Vaughan), and its proof is the same. Taking $$T \to \infty$$ recovers $$(\star)$$. Here $$\langle x \rangle$$ is as in the statement of Theorem 12.5: it's the distance of $$x$$ to the nearest prime power not equal to $$x$$.
• I'm curious why you use both $\rho+s$ and $\rho-s$ this formula; it would have been more aesthetic to me to just use $\rho-s$ (though in the regime you indicate where $\Im s$ is small compared to $T$, it makes little difference). Dec 24, 2022 at 2:07
• @TerryTao Oh, that's an embarrassing mathematical sign mistake on my part that I didn't catch until now, thanks for pointing it out. I'll update the sign in the next revision of the preprint. (Fortunately, as you say, this sign mistake is inconsequential -- in the considered range, the error incurred by replacing a sum over $|\Im(\rho+s)|<T$ with a sum over $|\Im(\rho-s)|<T$ can be absorbed in the bound for $R(x,T)$.) Dec 25, 2022 at 15:15
• Need to add $1_{x >1}$ in the explicit formulas. The inverse Mellin transform of $1/(z-\rho)$ on $\Re(z)=\sigma > \Re(\rho)$ is $1_{x >1} x^\rho$. Dec 26, 2022 at 2:28
• @reuns Note that I already assume $x \ge 4$. (This could be relaxed to $x>c$ for fixed $c>1$, as in the statement of Theorem 12.5 in Montgomery--Vaughan.) Dec 26, 2022 at 12:45