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I would like to understand how to derive (2) from (1) below.

Problem: If $\psi_1$ is the integrated Tchebychev function below $$\psi_1(x)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}\frac{x^{s+1}}{s(s+1)}(\frac{-\zeta '(s)}{\zeta(s)})ds \quad(1)$$ then $$\psi_1(x)=\frac{x^2}{2}-\sum_\rho\frac{x^{\rho}}{\rho(\rho+1)}-E(x)\quad(2)$$ where the sum is taken over all zeros of zeta function in the critical strip. Which error term will be $$E(x)=c_1x+c_0+\sum_{k=1}^{\infty}\frac{x^{1-2k}}{2k(2k-1)}$$ where $c_1=\frac{\zeta '(0)}{\zeta(0)}$ and $c_0=\frac{\zeta '(-1)}{\zeta(-1)}$.

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I don't see a question. –  daniel Jan 18 '13 at 13:27
    
@daniel I forgot to determine the $\psi$. –  Hoseyn Heydari Jan 18 '13 at 13:49
    
@daniel I want to prove (2) and which come after it with (1). –  Hoseyn Heydari Jan 18 '13 at 14:19
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2 Answers

up vote 3 down vote accepted

According to Harold Edwards in his book Riemann's Zeta Function there are two methods for evaluating the indefinite integral (1) to derive (2). He outlines each method on page 73 of his text. I will not reproduce them in full here but if the following suggestions do not get you all the way you can easily find the rest in Edwards.

Method 1. Use the expansion $-\zeta'(s)/\zeta(s)= \sum\Lambda(n)n^{-s}.$ Then (1) is $$\sum_{n=1}^{\infty}\Lambda(n)\frac{1}{2\pi}\int_{a-i\infty}^{a+i\infty}\frac{x^{s+1}ds}{n^ss(s+1)}. $$ Expand $\frac{1}{s(s+1)} $using partial fractions.

Method 2. Expand $-\zeta'(s)/\zeta(s)$ as $\frac{s}{s-1}-\sum_{\rho}\frac{s}{\rho(s-\rho)}+\sum_{n=1}^{\infty}\frac{s}{2n(s+2n)}- \frac{\zeta'(0)}{\zeta(0)}.$ Again you would use partial fractions, but there are algebra problems which Edwards covers in his book.

Edit: In case Edwards is not ready to hand and since it's not trivial, at least for me:

The integrand in (1) becomes:

$$\frac{-\zeta'(s)x^{s+1}}{\zeta(s)}(\frac{1}{s}-\frac{1}{s+1})= \frac{x^{s+1}}{s-1}- \sum_{\rho}\frac{x^{s+1}}{\rho(s-\rho)}+\sum_n \frac{x^{s+1}}{2n(s+2n)}-\frac{\zeta'(0)x^{s+1}}{\zeta(0)s}- \frac{x^{s+1}}{2(s-1)}+\sum_\rho \frac{x^{s+1}}{(\rho+1)(s-\rho)}- \sum_{n}\frac{x^{s+1}}{(2n-1)(s+2n)}+\frac{\zeta'(-1)x^{s+1}}{\zeta(-1)(s+1)}.$$

You can see, for example, that the first and fifth terms when integrated will combine to give you $\frac{x^2}{2}$... and so on.

Hopefully this is enough detail.

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As an alternate approach, you can view your (1) as defining $\psi_1(x^{-1})/x$ to be an inverse Mellin Transform in sense of http://en.wikipedia.org/wiki/Mellin_inversion_theorem $$ \psi_1(x^{-1})/x=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-s}\phi(s)\, ds $$ where $$ \phi(s)=\frac{-\zeta^\prime(s)}{s(s+1)\zeta(s)}. $$ To prove your identity, it suffices to show that the right side of (2), after swapping $x$ for $x^{-1}$ and dividing by $x$, has the same Mellin transform $\phi(s)$. This can be done by integrating term by term.

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