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Question

Using a conjectured formula of mine I believe the following relation to be true:

$$ (\int_0^\infty e^{-x^\lambda} dx)^\lambda = \lim_{s \to 1} \frac{1}{\zeta(s)} (\sum_{x_\lambda=1}^\infty \dots\sum_{x_2 =1}^\infty \sum_{x_1 =1}^\infty)\frac{1}{ (\sum_{k=1}^\lambda (x_k)^\lambda)^s} $$

where $\lambda$ is any positive integer $\geq 1$ and $\zeta(s)$ is the zeta function.

Can someone prove/disprove(or find a counter-example) this formula?

Background

It's derivation using the conjecture is quite similar to: What is the limit of this Dirichlet series?

More Anonymous
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1 Answers1

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For $n \in \Bbb{Z}^m$ let $d_m(n) = \sum_{j=1}^m |n_j|^m$ then $$\Gamma(s)\sum_{n \in \Bbb{Z}^m - (0)} d_m(n)^{-s} = \int_0^\infty \sum_{n \in \Bbb{Z}^m - (0)} t^{s-1}e^{-d_m(n) t}dt = \int_0^\infty ( \theta_m(t)^m-1) dt$$ Where $$\theta_m(t) = \sum_{k \in \Bbb{Z}}e^{-t |k|^m}$$

From $$\int_0^\infty t^{s-1} (\theta_m(t)-1) dt = \Gamma(s) 2\zeta(sm)$$ We know as $t \to 0^+$ $$ \theta_m(t)-1 = Res(\Gamma(s) 2\zeta(sm),s=1/m)(1+O(t^\epsilon))= t^{-1/m} \frac{2\Gamma(1/m)}{m} +t^{\epsilon-1/m}$$ Therefore $$\theta_m(t)^m = t^{-1} (\frac{2\Gamma(1/m)}{m})^m +O(t^{\epsilon-1})$$ So that $$\Gamma(s) \sum_{n \in \Bbb{Z}^m - (0)} d_m(n)^{-s}$$ has a simple pole at $s = 1$ of residue $(\frac{2\Gamma(1/m)}{m})^m$

And $ \sum_{n \in \Bbb{Z}_{\ge 1}^m } d_m(n)^{-s}$ has a simple pole at $s=1$ of residue $(\frac{\Gamma(1/m)}{m})^m = (\int_0^\infty e^{-t^m}dt)^m$

reuns
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