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juan
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Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?
With mpmath I check also this zero. I think now that my parenthesis "(with some effort)" contains an error. :-/ What it is clear is that the real zeros are those of the function given. But the behavior of this function for t complex is really not simple. I will try to do an X-ray of this function. Later we will try to post it. If I know how to do it.
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A measure theory question
Yes, the constant $e$ is only to make $x \log(e/x)$ monotone on $[0,1]$. In fact $H_f = H_g$ if $f$ and $g$ coincide on an interval $[0,\varepsilon]$, therefore there is some liberty in choosing $f$.
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A measure theory question
Perhaps I must explain that the existence of $A\subset{\bf R}$ with $0<H_f(A)<1$ is well known and due to Dvoretzky: Dvoretzky, A. A note on Hausdorff dimension functions. Proc. Cambridge Philos. Soc. 44, (1948). 13–16.
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Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?
added 76 characters in body; deleted 7 characters in body; added 8 characters in body
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Is this sum of reciprocals of zeta zeros correct?
In fact the two sums are equal $\sum_\rho(\rho^{-1}+(1-\rho)^{-1})= \sum_\rho (\rho^{-1}+\overline{\rho}^{-1})$. But I (or Davenport) am speaking about $\zeta(s)$. The case of $L(\chi,s)$ is a little more complicate. In this case the zeros are symmetric with respect to the line $\sigma=\frac12$ but not respect the real axis. The determination of the constant $B(\chi)$ is recent, due to Vorhauer in 2006. Davenport do not include it. You must see the book by Montgomery and Vaughan, Chapter 10. In the sum you must consider the zeros of $L(s,\chi)$ and those of $L(s,\overline{\chi})$.
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Is this sum of reciprocals of zeta zeros correct?
In Lagarias formula the sum is over the non trivial zeros repeated according to its multiplicity. In fact the formula is the Mittag-Leffler expansion of the meromorphic function $\frac{\hat\zeta'(s)}{\hat\zeta(s)}$ that has poles just at the non-trivial zeros and at $0$ and $1$
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