While working on another problem, I came up with the following expression. This involved many manual definite integral evaluations and not at all elegant. So I am not going into the details of the derivation.
$$ \zeta(3) = \frac{\pi^2}{96} \lim_{\text{N} \to \infty} \left[1444 + 15232\text{log}(2)+3456\text{log}(3)+\left\{\sum\limits_{\text{k}=5}^{\text{N}-1}[88+256\text{log}(2)+32\text{k}^2 + 384\text{k}^2\text{log}(\text{k})]\right\} + 32\text{N}^2 -64(\text{N}-3)\text{N}^2\text{log}(\text{N}) - 64(\text{N}-2)(\text{N}+1)^2\text{log}(\text{N}+1)-(2\text{N}+7)(2\text{N}-1)^3\text{log}(2\text{N}-1)+(2\text{N}-5)(2\text{N}+3)^3\text{log}(2\text{N}+3)\right] $$
Notes:
- This seems like a known result, as Wolfram Cloud (guessing same as Mathematica underneath) gives the exact expression $\zeta(3)$ when you type in the right hand side of the expression above.
Limit[(Pi^2/96)(1444+15232Log[2]+3456Log[3]+Sum[88+32k^2+256Log[2]+384k^2Log[k],{k,5,N-1}]+32N^2-64(N-3)N^2Log[N]-64(N-2)(N+1)^2Log[N+1]-(2N+7)(2N-1)^3Log[2N-1]+(2N-5)(2N+3)^3Log[2N+3]),N->Infinity]
- This result is not surprising either, given that the Fourier Transform of $\text{x}^2\text{log}(\text{x})$ has some $\frac{1}{\text{y}^3}$ term and the extra terms in my expression are probably helping with nullifying the effect of $\frac{1}{\text{y}^3}$ at $\text{y} = 0$
Questions:
- Can someone help with simplifying my clumsy looking expression to reveal the simplest relationship between $\zeta(3)$ and Logarithm function? I am not asking for a closed form (i.e. I am fine with a simpler looking infinite sum involving logarithm). I tried simplifying $\text{log}(\text{N}+1)$ as $\text{log}(\text{N})$ as $\text{N} \to \infty$ and similarly for $\text{log}(2\text{N}+1)$ as $\text{log}(2\text{N})$, but it looks like such simplifications are not mathematicaly correct (and yield non-converging answers).
- Why am I not seeing any literature (upon Google search) for such a relationship? Note: I am not a mathematician or a formal student of mathematics and therefore I have not studied this problem in a rigorous way. Maybe it is an obvious and uninteresting result for this community. I do see several non-closed forms for expressing $\zeta(3)$ in the form of infinite sums (that mostly focus on fast convergence, not really revealing any new relationship) in Wikipedia and linked references. In that sense, regardless of convergence rate, I was expecting to see such a relationship just for completeness. Note that I am not talking about PolyLogarithm functions, which readily yield to expressing $\zeta(3)$ in the form of $\text{Li}_{3}(1)$ etc, but in terms of ordinary logarithm function.
$\color\red{\text{Edit1}}$:
I did minor simplification as below
$$ \zeta(3) = \frac{\pi^2}{96} \lim_{\text{N} \to \infty} \left[1400 + 15104\text{log}(2)+3456\text{log}(3)+\left\{\sum\limits_{\text{k}=5}^{\text{N}-1}[88+256\text{log}(2)+32\text{k}^2 + 384\text{k}^2\text{log}(\text{k})]\right\} + 32\text{N}^3 + 16\text{N}^2 -(\frac{376}{3}+256\text{log}(2))\text{N} -64\text{N}(\text{N}-1)(2\text{N}-1)\text{log}(\text{N})\right] $$
This can be checked again in Wolfram Cloud with the code below.
Limit[(Pi^2/96)(1400+15104Log[2]+3456Log[3]+Sum[88+32k^2+256Log[2]+384k^2Log[k],{k,5,N-1}]+32N^3+16N^2-(376/3+256Log[2])N-64N(N-1)(2N-1)Log[N]),N->Infinity]
$\color\red{\text{Edit2}}$:
I did another minor simplification as below
$$ \zeta(3) = \frac{\pi^2}{96} \lim_{\text{N} \to \infty} \left[\left\{\sum\limits_{\text{k}=1}^{\text{N}}[32\text{k}^2 + 384\text{k}^2\text{log}(\text{k})]\right\} + 32\text{N}^3 - 16\text{N}^2 -\frac{112}{3}\text{N} -64\text{N}(\text{N}+1)(2\text{N}+1)\text{log}(\text{N})\right] $$
This can be checked again in Wolfram Cloud with the code below.
Limit[(Pi^2/96)(Sum[32k^2+384k^2Log[k],{k,1,N}]+32N^3-16N^2-(112/3)N-64N(N+1)(2N+1)Log[N]),N->Infinity]
$\color\red{\text{Edit3}}$:
I did another simplification as below
$$ \zeta(3) = \pi^2 \lim_{\text{N} \to \infty} \left[\left\{\sum\limits_{\text{k}=1}^{\text{N}}[ 4\text{k}^2\text{log}(\text{k})]\right\} + \frac{4}{9}\text{N}^3 -\frac{1}{3}\text{N} -\frac{2}{3}\text{N}(\text{N}+1)(2\text{N}+1)\text{log}(\text{N})\right] $$
This can be checked again in Wolfram Cloud with the code below.
Limit[(Pi^2)(Sum[4k^2Log[k],{k,1,N}]+(4/9)N^3-N/3-(2N/3)(N+1)(2N+1)Log[N]),N->Infinity]
$\color\red{\text{Edit4}}$:
This can also be written as
$$ \zeta(3) = 4\pi^2 \lim_{\text{N} \to \infty} \left[\left\{\sum\limits_{\text{k}=1}^{\text{N}}[ \text{k}^2\text{log}(\text{k})]\right\} - \left\{\text{log}(\text{N})\sum\limits_{\text{k}=1}^{\text{N}} \text{k}^2\right\}+ \frac{\text{N}^3}{9} -\frac{\text{N}}{12}\right] $$
This can be checked again in Wolfram Cloud with the code below.
Limit[(4Pi^2)(Sum[k^2Log[k],{k,1,N}] - Log[N]Sum[k^2,{k,1,N}]+(N^3)/9-(N/12)),N->Infinity]
Edit: A general result that is already known to others can be found in this : thread on MO
I am posting an answer that attempts to answer my question 1 partially. I already made some simplifications by reaching Edit4 (within the body of my question) after I posted my original question. Thanks for many useful suggestions in the comments section. Here, I will try to list a couple of additional relations in the same fashion. I am posting an answwer only because there seems to be some pattern emerging. I need to spend more time on this to establish a more generalized result.
$$ \zeta(1) = -2 \pi^{0} \lim\limits_{\text{N} \to \infty} \left [ \left\{\sum\limits_{\text{k}=1}^{\text{N}}\text{k}^{0}\text{log}(\text{k})\right\} - \left\{\text{log}(\text{N})\sum\limits_{\text{k}=1}^{\text{N}}\text{k}^{0}\right\}+\text{N}-\text{log}(\text{N})\right] $$
$$ \zeta(3) = 4 \pi^{2} \lim\limits_{\text{N} \to \infty} \left [ \left\{\sum\limits_{\text{k}=1}^{\text{N}}\text{k}^{2}\text{log}(\text{k})\right\} - \left\{\text{log}(\text{N})\sum\limits_{\text{k}=1}^{\text{N}}\text{k}^2\right\}+\frac{\text{N}^{3}}{9} -\frac{\text{N}}{12}\right] $$
$$ \zeta(5) = -\frac{4}{3} \pi^{4} \lim\limits_{\text{N} \to \infty} \left [ \left\{\sum\limits_{\text{k}=1}^{\text{N}}\text{k}^{4}\text{log}(\text{k})\right\} - \left\{\text{log}(\text{N})\sum\limits_{\text{k}=1}^{\text{N}}\text{k}^4\right\}+\frac{\text{N}^{5}}{25}-\frac{\text{N}^{3}}{12} +\frac{\text{13N}}{360}\right] $$