Exactly how do you derive Riemann's explicit formula correctly? I know the functional equation etc and I have seen a complete proof of von Mangoldt's version of the formula via a free paper online but I would like to read a direct proof.
You know, the formula that says 
$$J(x)=li(x)-\sum\limits_{\rho} li(x^{\rho})-\log(2)+\int\limits_x^{\infty}\frac{1}{t(t^2-1)\log t}dt$$
Please give links that are free or at least free to read online or something like that.
 A: Let $\Lambda(n) = \log p$ if $n=p^k$ is a prime power, $\Lambda(n)=0$ otherwise. Then $$\frac{-\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \Lambda(n) n^{-s} = s \int_1^\infty (\sum_{n \le x} \Lambda(n)) x^{-s-1}dx, \qquad \Re(s) > 1$$
Thus, by inverse Mellin/Laplace/Fourier transform
$$\sum_{n \le x} \Lambda(n) = \frac{1}{2i \pi} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{-\zeta'(s)}{\zeta(s)} \frac{x^s}{s}ds, \qquad \Re(s) > 1$$
Then the Riemann explicit formula follows by applying the residue theorem
$$\frac{1}{2i \pi} \int_{\sigma-iT}^{\sigma+iT}+\int_{\sigma+iT}^{-\infty+iT}+\int_{-\infty+iT}^{-\infty-iT}+\int_{-\infty-iT}^{\sigma-iT} \frac{-\zeta'(s)}{\zeta(s)} \frac{x^s}{s}ds $$
$$=\sum_\beta Res(\frac{-\zeta'(s)x^s}{\zeta(s)s},\beta) = x-\log2\pi -\sum_{k=1}^\infty \frac{x^{-2k}}{-2k}-\sum_{|\Im(\rho)|< T} \frac{x^\rho}{\rho}$$
And carefully letting $T \to \infty$,  using the density of zeros to show that $\int_{\sigma+iT}^{-\infty+iT}+\int_{-\infty+iT}^{-\infty-iT}+\int_{-\infty-iT}^{\sigma-iT} \frac{-\zeta'(s)}{\zeta(s)} \frac{x^s}{s}ds  \to 0$ and hence 
$$\sum_{n \le x} \Lambda(n) = x-\log2\pi -\sum_{k=1}^\infty \frac{x^{-2k}}{-2k}-\sum_{\rho} \frac{x^\rho}{\rho}$$
Finally the explicit formula for $J(x)= \sum_{n \le x} \frac{\Lambda(n)}{\log n}$ follows by differentiating, multiplying by $\frac{1_{x \ge 2}}{\log x}$ and integrating, transferring the formula for $\frac{\zeta'(s)}{\zeta(s)}$ to $\log \zeta(s)$.
