Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Problem: If $\psi_1$ is the integrated Tchebychev function below $$\psi_1(x)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}\frac{x^{s+1}}{s(s+1)} \left(\frac{-\zeta '(s)}{\zeta(s)}\right) \, ds \tag 1$$ where $c$ > 1, then $$\psi_1(x)=\frac{x^2}{2}-\sum_\rho\frac{x^{\rho+1}}{\rho(\rho+1)}-E(x)\quad(2)$$ 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 $c_0=\frac{\zeta '(-1)}{\zeta(-1)}$.

share|improve this question
1  
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
add comment

2 Answers

up vote 3 down vote accepted

Note that the usual definition of $\psi_1$ is as an integral of the Chebyshev function. See Wikipedia on the Chebyshev function for further information. There they use $\psi$ to denote what we write as $\psi_1$, and they also refer to the latter as the second Chebyshev function (and also, the smoothing function).

According to Harold Edwards in his book Riemann's Zeta Function, there are two methods for evaluating the indefinite integral (1). The first method demonstrates the validity of (1), and the second is used to derive (2). He outlines each method on pages 72-74 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 $$\sum_{n=1}^{\infty}\Lambda(n)\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{x^{s+1} \,ds}{n^ss(s+1)},$$ where $\Lambda$ is the von Mangoldt function. 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.

share|improve this answer
    
Small typo: in method 1, there should be an $i$ in the fraction next to the integral. i.e. it should be $\frac{1}{2\pi i}$. Also, I think that the $a$ in the limits of integration is the $c$ in the question. –  Andrew Kelley Apr 18 at 17:30
    
@AndrewKelley: thanks, will look at this. –  daniel Apr 18 at 18:53
    
The book was checked out from my library, and so a librarian suggested I look for it on googlebooks. Hence, the link. –  Andrew Kelley Apr 21 at 15:07
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.