# How do I calculate the values of $\zeta(0.5+ie^x)$ for large $x$ ?

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

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.
• 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. – davidlowryduda Jun 27 '15 at 0:22