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Background & Motivation

I'm trying to verify/disprove the conjectured formula of the weighted integral of $f(x)$: The Definite Integral Problem (with a twist)?

$$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}}_{\text{removable singularity}} \int_0^\infty f(x) \, dx $$

Where f(x) is a smooth continuous function whose integral from 0 to infinty is absolutely convergent and $\zeta(s)$ is the zeta function.

My idea is to choose $f(x) = e^{-x}$ and $g(x) = e^{-x^2}$. Notice: $$(\int_{0^+}^\infty g(x) dx)^2 = \lim_{k' \to \infty} \lim_{n' \to \infty}\ \sum_{i,j=1}^\infty e^{-(i^2 +j ^2) (\frac{k'}{n'})^2} (\frac{k'}{n'})^2$$

Compare $(\int_{0^+}^\infty g(x) dx)^2$ to the weighted integral of $f(x)$ and choose $k$ and $n$ such that $ k \to k'^2$ and $n \to n' ^2$ and also make $a_r$ such that both the series match.

$$(\int_{0^+}^\infty g(x) dx)^2 = \lim_{k' \to \infty} \lim_{n' \to \infty}\ \sum_{i,j=1}^\infty e^{-(i^2 +j ^2) (\frac{k'}{n'})^2} (\frac{k'}{n'})^2 $$ $$= \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}}_{\text{removable singularity}} \int_0^\infty f(x) \, dx $$

For example $a_1 = 0$, $a_2 = 1$, $a_3=0$, $a_4 = 0$, $a_5 =2$ and so on ... In general $$a'_r =\begin{cases} 0 & r\neq i^2 + j^2 \\ 1 & r = i^2 + j^2 , i = j\\ 2 & r = i^2 + j^2 , i \neq j \\ \end{cases}$$ where $i$ and $j$ are arbitrary positive integers $\neq 0$. and $a_r = \sum_{i\geq j} a'_{r(i,j)}$ For example $r=50$: $$a_{50} = a'_{50(5,5)} + a'_{50(7,1)} = 2+1 = 3$$

Questions

Can you tell the limit of:

$$\lim_{s \to 1} {\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}} = ?$$

where as mentioned previously

$$a'_r =\begin{cases} 0 & r\neq i^2 + j^2 \\ 1 & r = i^2 + j^2 , i = j\\ 2 & r = i^2 + j^2 , i \neq j \\ \end{cases}$$ where $i$ and $j$ are arbitrary positive integers $\neq 0$. Also if $a_{r(i,j)} = a_{r(i,j)}$ in t and $a_r = \sum_{i \geq j} a'_{r(i,j)}$ For example $r=50$: $$a_{50} = a'_{50(5,5)} + a'_{50(7,1)} = 2+1 = 3$$

Is there any flaw in the idea of the verification?

More Anonymous
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    @Clayton thanks ... I've edited the question hopefully there aren't any loopholes left – More Anonymous Apr 03 '19 at 15:03
  • Looks good; thanks for the clarification! – Clayton Apr 03 '19 at 15:17
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    Your Dirichlet Series is the one shown here, specifically in $(8)$ of the answer, to be $\zeta(s)\beta(s)-\zeta(2s)$, so your limit should be $\beta(1) = \frac{\pi}{4}$. Also, $d_4$ in your definition should be $0$, not $1$, since $i,j>0$. – Paul LeVan Apr 03 '19 at 16:18
  • Wow! that means the formula is valid!! Because $ \int g(x) dx = \frac{\pi^{1/2}}{2}$ – More Anonymous Apr 03 '19 at 16:23
  • @PaulLeVan Also, your right about $d_4$ ... I'll edit that .. – More Anonymous Apr 03 '19 at 16:25
  • The factorization $\zeta_{\Bbb{Q}(i)}(s)=\sum_{(n,m) \ne (0,0)} (n^2+m^2)^{-s} =4 \zeta(s)L(s,\chi_4)$ is obtained from https://en.wikipedia.org/wiki/Splitting_of_prime_ideals_in_Galois_extensions#Example_%E2%80%94_the_Gaussian_integers For any $F$ then $\lim_{s \to 1} F(s)/\zeta(s) = \lim_{s \to 1}(s-1)F(s)$ – reuns Apr 03 '19 at 22:11
  • @reuns any idea if the first equation (the conjectured formula) is already in the number theory literature? – More Anonymous Apr 03 '19 at 22:16
  • I already told you those $\lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) $ are not interesting – reuns Apr 03 '19 at 22:34
  • "already" ... must have forgotten ... But I think one can do "cool" things with the formula for example: https://math.stackexchange.com/questions/3173934/a-formula-to-find-lim-s-to-1-frac1-zetas-sum-x-1-1-infty-dots – More Anonymous Apr 03 '19 at 22:41

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