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I came across this series of following analytic functions $$ f_n\ :\ z\mapsto\frac{1}{\Big(\Pi_{k=0}^{n-1}\Gamma(1+ e^{2i\frac{k}{n}\pi}z)\Big)^{\frac{1}{n}}} $$
One can get easily a local (around zero) expansion (as an exponential, hence my question) with convergence radius $R=1$ and I wonder whether it can be continued as an entire function. Is it linked to some multiple gamma function defined by Barnes ?

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    $\begingroup$ $\Gamma(1+z)\Gamma(1-z)=\dfrac{\pi z}{\sin\pi z}$ $\endgroup$ – Nosrati Aug 20 '18 at 14:27
  • $\begingroup$ (+1) and thanks. I wanted to give a simplified version of it ! $\endgroup$ – Duchamp Gérard H. E. Aug 20 '18 at 14:57
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Using a product representation of the gamma function your term is holomorphic in $\mathbb{C}$ without $-z\in\mathbb{N}$.

For $n\ge 2$ and on the right side $\,|z|<1\,$ (and $z=1)$ your term is equivalent to:

$$\left(\prod\limits_{k=1}^\infty\left(1+(-1)^{n-1}\left(\frac{z}{k}\right)^n\right)\right)^{\frac{1}{n}}=\exp\left(-\sum\limits_{k=1}^\infty\frac{(-z)^{nk}\zeta(nk)}{nk}\right)$$

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  • $\begingroup$ (+1) ... and then has a radius of convergence one, I presume. Am I right ? $\endgroup$ – Duchamp Gérard H. E. Aug 21 '18 at 16:59
  • $\begingroup$ Because, for n=2 already, the function of the question is NOT extendable to an entire function. $\endgroup$ – Duchamp Gérard H. E. Aug 21 '18 at 17:19
  • $\begingroup$ @DuchampGérardH.E.: It depends what you use for the Gamma function. The product on the left side of the equation is valid for $z\in\mathbb{C}$ , because for $\bigg|\frac{z}{k_0}\bigg|<1$ we have $\bigg|\frac{z}{k}\bigg|<1$ for every $k>k_0$ which is trivial. The value range of the right side is mentioned in the answer. The problem is, if the left side is differentiable in $\mathbb{C}$ . If we use $\ln(1+(z/k)^m)=\sum\limits_{k=1}^\infty\frac{1}{k}(z/(k+z))^{mk}$ for $|z/k|\geq 1$ we can extend the differentiation to $\mathbb{C}$ . But you are right: We get a problem with $z=-1,-2,-3,...$ . $\endgroup$ – user90369 Aug 21 '18 at 17:44
  • $\begingroup$ @DuchampGérardH.E. : It follows: The product is not an entire function in $\mathbb{C}$. $\endgroup$ – user90369 Aug 21 '18 at 17:50
  • $\begingroup$ In fact, for $n\geq 2$ it is never true (I'll add this to my answer). I accept your answer on the ground of interaction (it helped me a lot :) $\endgroup$ – Duchamp Gérard H. E. Aug 21 '18 at 18:42
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This is the problem of n-th root of an entire function within the space of entire functions. For $n\geq 2$, there is no solution as soon as the original function has (at least) a simple zero. Considering $$ g_n\ :\ z\mapsto\frac{1}{\Big(\Pi_{k=0}^{n-1}\Gamma(1+ e^{2i\frac{k}{n}\pi}z)\Big)} $$ we have $g_n=f_n^n$ for $|z|<1$. The function $g_n$ is entire and has zeroes at the points such that $$ 1+ e^{2i\frac{k}{n}\pi}z\in \mathbb{Z}_{\leq 0} $$ for some $0\leq k<n$, this is the following disjoint union (each term of the union being the set of - simple - zeroes of the corresponding factor with the same $k$) $$ Z_n=\cup_{0\leq k\leq n-1} \{ne^{2i\pi\frac{(n-k)}{n}}\}_{n\leq -1} $$ hence the zeroes are simple whence the impossibility that $f_n$ be entire.

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