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This is I think an algebra confusion about an equality of Littlewood,

$$\frac{\psi(x) - x}{\sqrt{x}} = -2\sum_{1}^{\infty}\frac{\sin( \gamma_n\log x)}{\gamma_n} + O(1).\hspace{20mm}(1)$$

He refers the reader to "equivalent formulas" in Landau, and for this one the equivalent formula is the explicit formula,

$\psi(x) = x - \frac{1}{2}\log(1 - 1/x^2)-\log 2\pi -\sum\frac{x^\rho}{\rho}.\hspace{20mm}(2)$

I infer that $\sum\frac{x^{\rho}}{\rho} \sim 2\sqrt{x}\sum\frac{\sin\gamma_n\log x}{\gamma_n}\hspace{40mm}(3)$

Now if we assume RH we can show that (edit: for the real part of $\sum \frac{x^{\rho}}{\rho}$)

$Re\left(\sum \frac{x^{\rho}}{\rho}\right) = \sqrt{x}\sum \frac{\cos(\gamma_n \log x) +2\gamma_n\sin(\gamma_n\log x) }{1/4+\gamma_n^2}.\hspace{35mm}(4)$

Littlewood says he is assuming RH, so I would think (for the real parts)

$$2\sum_{2}^{\infty}\frac{\sin( \gamma_n\log x)}{\gamma_n} = \sum \frac{\cos(\gamma_n \log x) +2\gamma_n\sin(\gamma_n\log x) }{1/4+\gamma_n^2}\hspace{20mm}(5)$$

So (5) is a check on my inferences and I haven't been able to make the algebra work. I think (3) is perhaps wrong but then I don't know what equivalence Littlewood meant to suggest. Any help with inferences and/or algebra appreciated. Thank you.

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This previous answer may help. – Raymond Manzoni Jul 21 '13 at 17:41
@RaymondManzoni: Well for starters I do see that your (1) and Littlewood's formula--(1) above-- are the same. – daniel Jul 21 '13 at 17:55
To continue : in $(3)$ you probably consider only the non trivial roots (so that your later equalities should include an error term). The idea is then to take the nontrivial zeros by pairs $\rho=\frac 12 \pm i\gamma$. This is what I did while going from $(5)$ to $(6)$ (look only at the terms at the right if you find the left part confusing) so that you should get $\frac{2\;\sqrt{x}\;\cos(\gamma\ln x-\arg{\rho})}{|\rho|\;\ln x}$ that you may rewrite using the approximation from my $(2)$ ($\ |\rho|\approx \gamma\ $ and $\ \arg\,\rho\approx \frac{\pi}2$). – Raymond Manzoni Jul 21 '13 at 18:57
@RaymondManzoni: Thanks for this link, which will take me some time to work through! – daniel Jul 22 '13 at 2:26
Well you don't really need to go through the three explicit formulas (from $\,\pi(x)=R(x)-\sum\cdots\;$ to $\,\Pi(x)=\operatorname{li}(x)-\sum\cdots\;$ to your $\,\psi(x)=x-\cdots\:$ keeping only the first term(s) like I had to) to solve your specific problem (Greg Martin's answer and my last comment should suffice). Anyway I find the subject itself very interesting so "Excellent Journey !" – Raymond Manzoni Jul 22 '13 at 8:24
up vote 2 down vote accepted

The formula (5) is not correct as an equality. But it is correct up to a $O(1)$ term, which is all you need for proving (1). To see this, note that $$ \bigg| \sum \frac{\cos(\gamma_n \log x)}{1/4+\gamma_n^2} \bigg| \le \sum \frac{|\cos(\gamma_n \log x)|}{1/4+\gamma_n^2} \le \sum \frac1{\gamma_n^2}, $$ which is known to converge (it follows, for example, from the fact that there are only $O(\log T)$ zeros between $T$ and $T+1$). Note also that $$ \bigg| \sum \frac{2\sin(\gamma_n\log x) }{\gamma_n} - \sum \frac{2\gamma_n\sin(\gamma_n\log x) }{1/4+\gamma_n^2} \bigg| = \bigg| \sum \frac{2\sin(\gamma_n\log x)}{\gamma(1/4+\gamma_n^2)} \bigg| \le \sum \frac1{2\gamma_n^3} $$ which converges even faster. Therefore the two sides of (5) differ by a bounded amount.

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This is a gloss on a perfectly good answer, in case someone is interested in where this comes up in an early source. At page 284 of v. Mangoldt's paper ( book really) on Riemann's paper, he has

$$\lim_{h\to \infty} \frac{1}{2\pi i}\int_{a-ih}^{a+ih}\frac{2(s+r-1/2)}{(s+r-1/2)^2+ \gamma_n^2}\cdot\frac{x^s~ds}{s} $$

$$= \frac{2(r-1/2)}{(r-1/2)^2+\gamma_n^2} -2\cdot x^{-r+1/2}\cdot \frac{(r-1/2)\cos(\gamma_n\log x) -\gamma_n\sin(\gamma_n\log x)}{(r-1/2)^2+ \gamma_n^2}.\hspace{10mm}(1) $$

At page 293, summing over (1) he re-writes it as three sums, the dominant being $$2x^{-r+1/2}\sum_{n=1}^{\infty}\frac{\sin(\gamma_n\log x)}{\gamma_n.}\hspace{30mm}(2)$$

Taking $r=0$ and summing over $\gamma_n$ the last term in (1) is just the r.h.s. of (5) in the OP.

So while Littlewood cites Landau for (1) in the OP a better cite for the uninitiated would have been von Mangoldt. Peeling the term on the r.h.s. of (1) in the OP out of the r.h.s. of (5) in OP requires not only some tedious algebra but the motivation to do so.

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