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Nov 19, 2017 at 15:44 comment added François Brunault Another approach would be to take the constant term in the asymptotic expansion of $\sum_{|a|,|b| \leq N} \log (a^2+b^2) $ when $N$ tends to infinity.
Nov 19, 2017 at 2:16 comment added Vesselin Dimitrov This amounts again to the Kronecker limit formula. See, for example, this paper of Quine and Stong (A double Stirling formula): ams.org/journals/proc/1993-119-02/S0002-9939-1993-1164151-5 .
Nov 18, 2017 at 23:53 comment added AHusain ncatlab.org/nlab/show/… with $\tau=i$
Nov 18, 2017 at 23:33 history edited john mangual CC BY-SA 3.0
Remove log
Nov 18, 2017 at 22:02 comment added reuns In general to see if $F'(0) $ is well-defined where $F(s) =\sum_{n=1}^\infty a_n n^{-s}$ you need to find a convergent series $g(x)= \sum_{k=0}^\infty c_k x^{\alpha_k} (\log x)^{r_k}$ such that $\sum_{n=1}^\infty a_n e^{-nx}-\sum_{k=0}^\infty c_k x^{\alpha_k} (\log x)^{r_k}=o(1/\log^3 x)$ as $x \to 0$ ($-\alpha_k$ are the poles of $F(s)\Gamma(s)$)
Nov 18, 2017 at 21:58 comment added reuns Your question is unclear. $ 4 \zeta(s) L(s,\chi_4) = 4\zeta_{\mathbb{Q}(i)}(s) = \sum_{(a,b) \in \mathbb{Z}^2 \setminus (0,0)} (a^2+b^2)^{-s} = \frac{1}{(2\pi)^{-s}\Gamma(s)}\int_0^\infty x^{s-1}(\theta(ix)^2-1)dx$ where $\theta(z)^2 = \sum_{(a,b) \in \mathbb{Z}^2} e^{2i \pi z(a^2+b^2)})= (z/2i)^{-1} \theta(-4/z)^2$.
Nov 18, 2017 at 21:58 comment added john mangual @reuns I'm trying to write in a style that's typical of what I'm seeing in the hep-th section of arXiv. I'm asking what happens if you set $s=0$ in the zeta functions, and what happens if $z \in \mathbb{R}$ for the theta functions. With the formulas you've provided we can regularize the products in question.
Nov 18, 2017 at 21:04 history edited john mangual CC BY-SA 3.0
Product not sum
Nov 18, 2017 at 20:54 history asked john mangual CC BY-SA 3.0