This question concerns the book Riemann's Zeta Function, H.M. Edwards (Dover, 1974). On pages 25 to 36, Edwards provides a detailed proof of Riemann's explicit formula for the prime counting function. I'll first show two key equations in that proof.
For $x > 1$, Edwards derives a formula for $J(x)$: \begin{equation*} J(x) = \operatorname{li}(x) -\sum_{Im(\rho) > 0} \left[\operatorname{li}(x^{\rho}) + \operatorname{li}(x^{1-\rho})\right] + \int_x^{\infty} \frac{dt}{t (t^2 - 1) \log t} -\log2, \end{equation*} where the last two terms of $J(x)$ are small and do not grow with $x$.
Let $\pi(x)$ be the prime counting function and $\mu(n)$ be the Möbius function. For any fixed $x \geq 2$ choose the smallest integer $N$ such that $x < 2^N$. Edwards shows: \begin{equation*} \pi(x) = \sum_{n=1}^{N} \frac{\mu(n)}{n} J(x^{1/n}). %\quad\quad \text{ for } x < 2^N.\quad N = \min\LR{N\in\Nbb, x < 2^N} \end{equation*}
Disregarding the last three terms of $J(x)$, we have \begin{equation*} \pi(x) \sim \sum_{n=1}^{N} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}), \end{equation*} which has proved to be a good approximation of $\pi(x)$.
Edwards then gives (page 35) the following formula for the error in the above approximation: \begin{equation*} \pi(x) - \sum_{n=1}^{N} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}) = \sum_{n=1}^{N}\sum_{Im(\rho)>0} \left[ \operatorname{li}(x^{\rho/n}) + \operatorname{li}(x^{(1-\rho)/n}) \right] + \text{ lesser terms}. \end{equation*}
Now to my question. Edwards error formula does not look correct to me. Shouldn't the formula be \begin{equation*} \pi(x) - \sum_{n=1}^{N} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}) = -\sum_{n=1}^{N} \frac{\mu(n)}{n} \sum_{Im(\rho)>0} \left[ \operatorname{li}(x^{\rho/n}) + \operatorname{li}(x^{(1-\rho)/n}) \right] + \text{ lesser terms}. \end{equation*}
As an aside, I searched the Internet for errata on the Edwards book and found nothing. Is anyone aware of a listing of errata for his book?
(Edit - Added an Example).
Fix $x = 15 = 2^4 - 1$. Note that $\mu(1) = 1$, $\mu(2) = -1$, $\mu(3) = -1$ and $\mu(4) = 0$. Let \begin{equation*} S(n) = \sum_{Im(\rho)>0} \left[ \operatorname{li}(x^{\rho/n}) + \operatorname{li}(x^{(1-\rho)/n}) \right]. \end{equation*} Using $x=15$, we now show the two formulae for the approximation error: \begin{align*} \pi(15) - \sum_{n=1}^{4} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}) &= S(1) + S(2) + S(3) + S(4) + \text{ lesser terms},\\ \pi(15) - \sum_{n=1}^{4} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}) &= -S(1) + \frac{S(2)}{2} + \frac{S(3)}{3} + \text{ lesser terms}. \end{align*} Edwards calls his expression ``an exact analytical formula for the error''. That suggests that either: (1) the two expressions above are equal (which seems unlikely), or (2) one of the two expressions is incorrect.