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Suppose we have the following function: $$ f(b)=\sum_{n=0}^{\infty} e^{-b \sqrt n}$$ Is there a closed form expression for the analytic continuation of $f(b)$ to $f(-b)$?

sigma
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  • for understanding the singularity at $x = 0$ of $g(x) = \sum_{n=1}^\infty e^{- n^{1/2} x }$ I look at $\int_0^\infty g(x) x^{s-1}dx = \Gamma(s) \zeta(s/2)$ whose poles are all of order $1$ : at $s=2$ and $s = -k$, so as $x \to 0$ : $g(x) \sim \Gamma(1/2) x^{-2} + \sum_{k=0}^\infty \frac{(-1)^k}{k!} \zeta(-k/2) x^{k}$. The residue theorem should say this is indeed a series representation of $g(x)$ valid for $|x|$ small enough – reuns Jul 19 '16 at 13:59
  • How does the residue theorem lead to the last expression for $f(x)$? – sigma Jul 19 '16 at 14:04
  • inverse Mellin transform (just the Laplace transform with a change of variable) : $g(x) = \frac{1}{2 i \pi} \int_{\sigma-i\infty}^{\sigma+i\infty} x^{-s} \Gamma(s) \zeta(s/2) ds$ and you can apply the residue theorem to this, using the functional equation for $\zeta(s)$ – reuns Jul 19 '16 at 14:05
  • How do you apply the residue theorem to that form? Assuming we have $\sigma=0$, it seems like there is only one singularity on the imaginary axis at $x=0$. Where did you get the sum, and the factor of $\Gamma (1/2) x^{-2}$ ? – sigma Jul 19 '16 at 14:57
  • $\sigma > 2$ and told you where are the poles of $\Gamma(s) \zeta(s/2)$ : at $s = 2$ and $s = -k$ – reuns Jul 19 '16 at 15:36
  • Where did you get the factor of $\Gamma (1/2)$ from? – sigma Jul 19 '16 at 15:39
  • my mistake, I meant $\Gamma(2)x^{-2}$ (around $s =2 $ : $\Gamma(s) \zeta(s/2) \sim \frac{\Gamma(2)}{s-2}$) – reuns Jul 19 '16 at 15:41
  • From Mathematica it seems that the residue of $\Gamma(s) \zeta(s/2) x^{-s}$ at $s=2$ is $2 x^{-2}$, but seems $\Gamma(2)=1$, so I think we might be off by a factor of 2. – sigma Jul 19 '16 at 15:45
  • I just told you I wrote $\Gamma(1/2)$ instead of $\Gamma(2)$ – reuns Jul 19 '16 at 15:48
  • I see. Sorry; thank you for your help. – sigma Jul 19 '16 at 15:51
  • the only one problem is to show that $\int_{\sigma+i\infty}^{-\infty+i\infty}+\int_{-\infty+i\infty}^{-\infty-i\infty}+\int_{-\infty-i\infty}^{\sigma-i\infty} \Gamma(s) \zeta(s/2) x^s dx = 0$ and I think it is true, that's why I wrote "the residue theorem should tell us this is a series representation for $g(x)$" because this has to be checked (the growth rate of $\Gamma(s) \zeta(s/2) $ being given by the functional equation for $\zeta(s)$ and the Stirling approximation) – reuns Jul 19 '16 at 15:59

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