# What is the value of the Dirichlet Eta Function at s=1/2?

Although $$\eta(1)$$ is known to be $$\ln(2)$$, I have not seen an analytically calculated value for $$\eta(\frac{1}{2});$$

$$\eta\left(\frac{1}{2}\right) = \sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{\sqrt{n}}$$

A web calculator gives the value to be 0.6, which seems to be right.

Isn't just $$\eta\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{\sqrt{n}}=\left(1-\sqrt{2}\right) \zeta \left(\frac{1}{2}\right)\approx 0.6048986434$$

Edit

Remember the general relation $$\eta\left(s\right)=\left(1-2^{1-s}\right) \zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $$0 \leq s \leq 1$$, you could use $$\eta\left(s\right)=\frac 12+\left( \log (2)-\frac{1}{2}\right)\, s^{0.895}$$

A careful computation shows that the numerical value is $$0.6048986434216303702472...$$ which is not $$0.6$$. One should be aware that the above series converge really slowly.

As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $$1/2$$. However, as far as I know, there is no analytic formula of $$\zeta(1/2)$$, so this is why you haven't seen an "analytically calculated value for $$\eta(1/2)$$".

EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$\gamma/2 + \pi/4 - (1/2 + \sqrt{2})\log(2) + \log(\pi)/2,$$where $$\gamma$$ is the Euler-Mascheroni constant, is $$\eta'(1/2)/\eta(1/2)$$ not $$\eta(1/2)$$.

• You are very correct ! We start a no-end loop. By the way $\to +1$ – Claude Leibovici Nov 13 '18 at 10:51
• It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though. – leftaroundabout Nov 13 '18 at 16:01
• @leftaroundabout Do you have a reference, I searched, but I didn't find it. – Josué Tonelli-Cueto Nov 13 '18 at 18:47
• @user3059799 Corrected, editing from the phone is hard – Josué Tonelli-Cueto Nov 13 '18 at 20:11
• The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $\zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise. – user98602 Nov 14 '18 at 1:18

The numerical value of 0.604898... is provided in http://oeis.org/A113024 .