# Values of the completed Riemann $\xi(1+it)$ for small t?

I'm editing this question heavily for clarity:

I am looking for methods to compute $\zeta(1+it)$, or the (partially) completed Riemann zeta function $$\pi^{-s/2}\Gamma(s/2)\zeta(s)$$ along the line Re$(s)=1$. Certainly as I have left out the $s(s-1)$ factor, there is a pole at $s=1$. I would like to know the behaviour of the function above and below that point.

In particular, are there known mean value estimates, or even exact formulas for the argument of the functions above, preferably for small values of $t$?

• In general, the absolute value decreases, but it is not monotone.
– joro
Oct 15, 2015 at 8:08
• Wolfram alpha will do this, too. The Gamma function is available. Oct 15, 2015 at 13:02
• @Stopple, Yes, I realized this afterwards, and it was helpful. But I'd like to get more precise information and be able to control for the height with greater freedom. Also, having now some heuristic I'm more inclined to have a theoretical approach to estimating these values at real and imaginary parts. Oct 15, 2015 at 14:08
• Down vote with no explanation? I got the numerical heuristic I was looking for on my own, so modified the question to ask for a more theoretical answer. Oct 18, 2015 at 18:56

I'm not really sure what you're looking for. Mathematica computes the Laurent series expansion at $s=1$: $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\frac{1}{s-1}+\left(\gamma -\frac{\log (\pi )}{2}+\frac{\psi \left(\frac{1}{2}\right)}{2}\right)+O\left(s-1\right)$$ where $\gamma$ is Euler's constant, and $\psi$ is the logarithmic derivative of the Gamma function. The next term, the coefficient of $s-1$, is $$\frac{1}{16} \left(-16 \gamma _1+\pi ^2+2 \log ^2(\pi )-8 \gamma \log (\pi )+8 \gamma \psi \left(\frac{1}{2}\right)+2 \psi\left(\frac{1}{2}\right)^2\\ -4 \log (\pi ) \psi \left(\frac{1}{2}\right)\right)$$
• Maybe worth noting additionally that numerically these coefficients seem to approach $\pm1$ very quickly: \begin{align*} -&0.9769042910338789661\\ &1.0002481555681051493\\ -&0.9997501717181822006\\ &1.0000033534484987277\\ -&0.9999983031937065084\\ &1.0000000241807483700\\ -&0.9999999918023337510\\ &1.0000000001183020012\\ -&0.9999999999697778090\\ &1.0000000000004334114\\ -&0.9999999999999110273\\ &1.0000000000000012570\\ -&0.9999999999999997840\\ &1.0000000000000000030 \end{align*} Oct 16, 2015 at 7:11