# In this prime counting identity, can this limit of a sum be expressed as integrals?

Here's the identity I'm working with.

$E_{0}(n,b) = 1$
$E_{k}(n,b) = \displaystyle\sum_{j=2}^{\lfloor n \rfloor}E_{k-1}(\frac{n}{j},b)-b \sum_{j=1}^{\lfloor \frac{n}{b} \rfloor}E_{k-1}(\frac{n}{j b},b)$ $\Pi(n)=li(n) - \log \log n - \gamma + \lim_{b \rightarrow 1^{+}}\displaystyle\sum_{k=1}^{\lfloor log_b n \rfloor}\frac{(-1)^{k-1}}{k}E_k(n,b)+\frac{1}{k}$
where $\Pi(n)=\displaystyle\sum_{j=2}^n \frac{\Lambda(j)}{\log j}$ is the Riemann Prime counting function, $li(n)$ is the logarithmic integral, and $\gamma=.57721566...$

My interest is in simplifying the function for $\Pi(n)$.

Specifically, I'm convinced the $\lim_{b \rightarrow 1^{+}}\displaystyle\sum_{k=1}^{\lfloor log_b n \rfloor}...$ term can be replaced by a piece-wise integral expression... somehow. If you graph it, it looks like this (for n = 10 and x = 1.015):

but I've come up short on how to go from the current expression to an piece-wise integral expression. Can anyone show me how to do that?

(an hand-wavy explanation, but not proof, of this approach can be found in section 8 of this pdf)

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