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Quite generally I thought that every holomorphic function defined in some domain can be analytically continued into the entire complex plane. However the dedekind eta function seems to not have such an analytic continuation to the lower half as is stated here Dedekind Eta Function.

So what is the problem here precisely? As an example the zeta function defined via the dirichlet series is only convergent for s>1. Probably it also converges for s=1 if Im(s) is not equal zero. Still it can be continued.

Now the dedekind eta (or similar functions as the jacobi theta) are only defined for Im(t)>0. Again I would assume it to converge on the real axis if t is not an integer. So what prevents it of being continued?

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  • $\begingroup$ The logarithm would be an easier function. There you always need a branch cut and so there isn't even a meromorphic continuation like for the zeta function. $\endgroup$ – Paul K Sep 27 '18 at 19:27
  • $\begingroup$ See this: math.stackexchange.com/questions/1379758/… . There are simply many functions that are holomorphic on some open set of the complex plane, but that fail to be analytic at some points. Recall that being analytic is a local condition, so there is not reason why a function that is analytic somewhere has to be so everywhere. $\endgroup$ – user2520938 Sep 27 '18 at 19:41
  • $\begingroup$ Well, the logarithm is missing one single line typical for a branch cut, but not half the complex plane as in the example above. $\endgroup$ – Diger Sep 27 '18 at 19:43
  • $\begingroup$ So in the specific example above of the eta function and looking at your link I would assume in accordance that for every $a\in {\mathbb R}$ and $r>0$ $$ \sup_{t \in U(a,r)} |\eta(t)| = \infty $$ where $$ U(a,r) = D(a,r) \cap {\mathbb H} \, ? $$ Is that the case? $\endgroup$ – Diger Sep 27 '18 at 20:12
  • $\begingroup$ So just from looking at plots of the eta function it seems that it behaves pretty chaotic on the real line. From the product formula it vanishes at the rationals. There are finite values at e.g. $\exp(-\pi)$ and it seems to diverge at $\exp(1)$. $\endgroup$ – Diger Sep 28 '18 at 9:37

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