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Gauss sums are analogous to the Gamma function:

Fix a complex number $s$ with real part $>0$. Then we have a multiplicative character $\chi_s :\mathbb R^{\times}_{>0} \to \mathbb C^\times$ given by $x \mapsto x^s$. We also have an additive character $\lambda: \mathbb R \to \mathbb C^\times$ given by $x \mapsto e^{-x}$. The inner product of these two characters over $\mathbb R^{\times}_{>0}$, $$\langle \chi_s, \lambda \rangle : = \int_0^\infty x^s e^{-x} \frac{dx}{x}$$ is the Gamma function.

Remark that $dx/x$ is the Haar measure on $\mathbb R^{\times}_{>0}$.

For Gauss sums, say over the finite field $\mathbb F_p$, we take the inner product of a Dirichlet character $\chi: \mathbb F_p^\times \to \mathbb C^\times$ with the additive character $\lambda : \mathbb F_p \to \mathbb C^\times$ given by $x \mapsto e^{2\pi i x/p}$:

$$\eta(\chi) = \sum_{x \in \mathbb F_p^\times} \chi(x) \lambda(x).$$

With this notation, the Gamma function 'is' $\eta(\chi_s)$.

There are many analogies between Gauss sums and the Gamma function (Jacobi sums vs Euler beta function, Gauss multiplication formula vs Hasse–Davenport product relation...).

But I haven't seen a Gauss sum equivalent of the relation $\eta(\chi_{s+1}) = s\eta(\chi_s)$. Is there one, and if so what could it be?

Bruno Joyal
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    The usual proof of the Gamma function recurrence relation is (of course) via integration by parts; in a real sense, it's a consequence of the relation $x^s=x\cdot x^{s-1}$. AFAIK there's no corresponding 'parametrized' relation among different Dirichlet characters $\bmod p$, but if there were then you could presumably use a 'summation by parts' approach to derive the corresponding relation... – Steven Stadnicki Oct 27 '13 at 21:34
  • The Gross-Koblitz formula expresses Gauss sums in terms of the $p$-adic $\Gamma$-function which has the analog of the functional equation for "usual" $\Gamma$, so in principle something can be said, but... – მამუკა ჯიბლაძე Mar 09 '17 at 09:03
  • @BrunoJoyal Your gamma definition using Gauss sums is amazing:) Could you tell your reference? – user1062 Aug 16 '21 at 02:17

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