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Questions on analytic representations of the Kronecker delta function $\delta(x-1)$ and the Moebius function $\mu(n)$

This question is related to analytic formulas for $a(n)$ where $f_a(x)$ and $F_a(s)$ defined in formulas (1) and (2) below are the summatory function and Dirichlet series associated with $a(n)$. $$...
Steven Clark's user avatar
  • 1,126
12 votes
1 answer
663 views

Error term when truncating series for $1/\zeta(s)$

Let $s=\sigma+it$, $0\leq \sigma\leq 1$, $|t|\geq 1$, say. Using Euler-Maclaurin, one can easily show that, for $x\geq |t|$, $$\zeta(s) = \sum_{n\leq x} \frac{1}{n^s} + \frac{x^{1-s}}{s-1} + O\left(\...
H A Helfgott's user avatar
  • 20.2k
2 votes
1 answer
1k views

On the convergence of Dirichlet series over the Mobius Mu function

It is known that if $\sum_{k=1}^{\infty} \frac{\mu(k)}{k^s} = \frac{1}{\zeta(s)}$ for $\Re(s) > 1/2$ then RH holds. My question is: Under RH why is it not $\sum_{k=1}^{\infty} \frac{\mu(k)}{k^s} = ...
Roupam Ghosh's user avatar
1 vote
1 answer
419 views

On a certain integral representation for Dirichlet L-functions

It is an ancient result of Jensen that $$(s-1)\zeta(s)=\frac{\pi}{2} \int_{-\infty}^{\infty} \frac{(1/2+it)^{1-s}}{\cosh^{2}\pi t} \mathrm{d}t$$ where $\zeta$ denotes the Riemann zeta function. Is ...
Q_p's user avatar
  • 1,019
1 vote
1 answer
261 views

Generalizing closed form representations related to conjectured analytic formulas for $f_a(x)=\sum\limits_{n=1}^x a(n)$

Consider the summatory function $f_a(x)$ defined in formula (1) below where the related Dirichlet series $F_a(s)$ defined in formula (2) below converges for $\Re(s)\ge 2$. $$f_a(x)=\sum\limits_{n=1}^...
Steven Clark's user avatar
  • 1,126
1 vote
1 answer
188 views

Can one show $h(x)=|2(\zeta'(x))^2-\zeta''(x)\zeta(x)|$ is a decreasing function for $x\in\mathbb{R}\cap [1,\infty)$?

This question is related to This question. When I tried to approach it I couldn't even proof that the LHS is a decreasing function on the given domain using regular methods. I have tried to write the ...
Haidara's user avatar
  • 178