# A formula to find $\lim_{s \to 1} \frac{1}{\zeta(s)} (\sum_{x_i=1}^\infty \dots \sum_{x_1 =1}^\infty)\frac{1}{ (\sum_{k=1}^i (x_k)^i)^s}$?

## 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?

• Reason for downvote? – More Anonymous Apr 3 at 22:42
• What does $x_n$ mean here? – stuart stevenson Apr 3 at 23:04
• $x_n$ is a bound variable ... All $x_i$ are summed from $1$ to $\infty$ – More Anonymous Apr 3 at 23:07

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 $$(\frac{\Gamma(1/m)}{m})^m = (\int_0^\infty e^{-t^m}dt)^m$$