Equality involving $\sum_n \sin(\gamma_n \log x)/\gamma_n$ 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.  

 A: 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.
A: 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.   
