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I've been exploring recovering Riemann's prime-power counting function $J[y]$ and the second Chebyshev function $\psi(y)$ from functions of $\zeta[s]$ via relationships such as the following.
$$J(y)=\frac{1}{2\pi i}\int_{a-\infty\ i}^{a+\infty\ i}\log\zeta[s]\ y^s\frac{ds}{s}$$ $$\psi(y)=\frac{1}{2\pi i}\int_{a-\infty\ i}^{a+\infty\ i}(-\frac{\zeta'[s]}{\zeta[s]})\ y^s\frac{ds}{s}$$

I've noticed these relationships seem to evaluate to twice the step-size of the associated prime counting function, and I'm trying to understand if there's an error in these relationships or an error in my evaluation method.

I'm primarily interested in investigating these relationships in the context of using Fourier series representations of $J'[x]$ and $\psi'[x]$ to recover $\log\zeta[s]$ and $-\frac{\zeta'[s]}{\zeta[s]}$ respectively (see Illustrations of Fourier Series for Prime Counting Functions), but below I illustrate the problem I'm encountering in a simpler context.

For example, the second relationship above can be evaluated using term-wise integration as follows. This approach is based on section 3.2 of Edward's book "Riemann's Zeta Function".
$$-\frac{\zeta'[s]}{\zeta[s]}=\int_{0}^{\infty}\psi'[x]\ x^{-s}dx=\sum_{n=2}^\infty \Lambda[n]\ n^{-s}$$ $$\psi(y)=\sum_{n=2}^\infty \Lambda[n]\frac{1}{2\pi i}\int_{a-\infty\ i}^{a+\infty\ i}(\frac{y}{n})^s\ \frac{ds}{s}$$

I believe the integral above evaluates as follows.
$$\int(\frac{y}{n})^s\ \frac{ds}{s}=Ei[s\log\frac{y}{n}]$$

This leads to the following formula to recover $\psi[y]$. $$\psi(y)=\frac{1}{2\pi i}\sum_{n=2}^N \Lambda[n]\ (Ei[(a+i\ M)\log\frac{y}{n}]-Ei[(a-i\ M)\log\frac{y}{n}]),\quad N,M\to\infty$$

The formula above evaluates with an offset which I'm less concerned about and which can can be taken care of by subtracting off the evaluation at y=1. But as I noted earlier, the formula above seems to evaluate to twice the step size of $\psi[y]$ which I'm more concerned about.

So my question is:
Are the relationships above correct or is there an error in my evaluation method? For example, are these relationships conditionally convergent based on assumed relationships between the relative values of the evaluation parameters N, a, and M?

Another discrepancy which I've noticed is the following integral is supposed to evaluate to zero for $0<z<1,$ and it seems to evaluate to $-1$ for $0<z<1$, which is most likely related to the evaluation problems I'm having with the relationships above.
$$\frac{1}{2\pi i}\int_{a-\infty\ i}^{a+\infty\ i} z^s\frac{ds}{s}$$

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  • $\begingroup$ $Ei(z)$ is a complicated function with some branch points. You are supposed instead to use the residue theorem or the inverse Fourier/Laplace transform for saying that if $a > 0$ and $t \in \mathbb{R}$ then $\frac{1}{2i \pi}\int_{a-i\infty}^{a+i\infty} e^{st} \frac{ds}{s} = 1_{t > 0}$ $\endgroup$ – reuns Dec 16 '16 at 23:09
  • $\begingroup$ And note that $F(s) = \frac{1}{s}$ is the bilateral Laplace transform of $f(t) =1_{t > 0}$ and $g(t) = -1_{t < 0}$ in the same time, depending on the domain of convergence ($Re(s) > 0$ or $Re(s) < 0$) $\endgroup$ – reuns Dec 16 '16 at 23:11
  • $\begingroup$ @user1952009 The integral $\frac{1}{2\pi i}\int_{a-\infty\ i}^{a+\infty\ i} z^s\frac{ds}{s}$ is supposed to evaluate to zero for $0<z<1$ and to one for $z>1$. The integral evaluates correctly for $z>1$ which seems to be the situation you've addressed in your comment, but the integral evaluates incorrectly for $0<z<1$ (i.e. it evaluates to -1 versus zero). I'm using a positive value for $a$, so $Re[s]>0$. $\endgroup$ – Steven Clark Dec 16 '16 at 23:55
  • $\begingroup$ @user1952009 I've explored this integral a bit more in Mathematica, and apparently I'm supposed to be using three different formulas to evaluate the integral for the three different conditions $0<z<1$, $z=1$, and $z>1$. $\endgroup$ – Steven Clark Dec 17 '16 at 0:20
  • $\begingroup$ You don't care of the special case $x/n=1$. And $1_{t > 0} = \begin{cases} 1 \text{ if } t > 0\\ 0 \text{ if } t < 0 \end{cases}$ (and again we don't care of $t=0$) $\endgroup$ – reuns Dec 17 '16 at 0:50

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