This has been giving me nothing but a headache:
Let the Tchebychev Function, $\psi (x)$ be defined:
$$\psi (x) = \sum_{p^m \le x}\log p \space \space \space , \space \space \space p \in \mathbb P$$
Prove that the integrated Tchebychev Function, $\psi_1 (x)$, is as follows:
$$\psi_1 (x) = \frac{x^2}2 - \sum_{\rho}\frac{x^{\rho + 1}}{\rho(\rho +1)} - E(x)$$
Where $\rho$ is a zero of the Riemann Zeta Function in the critical strip and:
$$E(x) = \left(\frac{\zeta ' (0)}{\zeta(0)}\right)x + \left(\frac{\zeta ' (-1)}{\zeta(-1)}\right) + \sum_{k=1}^{\infty}\frac{x^{1-2k}}{2k(2k-1)}$$
Can anyone point me to the right direction? Not really sure where to begin. Thanks!
I'd suggest using the following theorem:
$$ \psi_1(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \frac{x^{s + 1}}{s(s+1)} \left( \frac{-\zeta'(s)}{\zeta(s)} \right) \mathrm{d}s$$
where $c > 1$. A proof of this equality can, for example, be found in Complex Analysis by Stein and Shakarchi. It is on page 191 being proposition 2.3 of chapter 7. Once you have this, you can see the same/similar question which I also linked to in my comment.