In wolfram alpha the values of $$\zeta(0.5+ie^x)$$ closed to zero then How

do I know the real values of $\zeta(0.5+ie^x)$ for large real number $x$ ?

Thank you for any help


Choose your favorite analytic continuation and calculate it. That's one way.

For instance, you might use that $$ \zeta(s) = (1 - 2^{1 - s})^{-1} \eta(s)$$ where $$ \eta(s) = \sum_{n \geq 1} \frac{(-1)^{n - 1}}{n^s},$$ which simply converges at values $s = \frac 12 + it$.

If you're asking how others go about it, many use a so-called Approximate Functional Equation (or series accelerations of it or the $\eta$ function). See this MO question for a bit more about the approximate functional equation.

  • $\begingroup$ @thank you for your answer, is n(s) able to calculate :n(0.5+ie^x) for large x ? $\endgroup$ – zeraoulia rafik Jun 26 '15 at 23:48
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    $\begingroup$ Yes, of course. What I call $t$, you call $e^x$. I'm not promising that it converges quickly, but it converges all the same. $\endgroup$ – davidlowryduda Jun 27 '15 at 0:22
  • $\begingroup$ but should be know that w.alpha also used n(s) but didn't up to calculate it $\endgroup$ – zeraoulia rafik Jun 27 '15 at 0:23
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    $\begingroup$ Ok. What does that have to do with anything? $\endgroup$ – davidlowryduda Jun 27 '15 at 0:26

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