Consider the following limit:
$$\lim_{m\to\infty} \log_k(m)^2-\int_0^{\log_k(m)} \log_k(m-k^x)dx = \frac{\zeta(2)}{ \ln(k)^2} \label{eq:1}\tag{1} $$
The Question:
Does this sit in a generalization that gives us access to other zeta values?
Exposition:
I think that this has to be true but I cannot seem to describe the space I want to integrate over. I am asking for a little help with generalization into $s$ dimensions for $s\neq 2$ which is already shown above. Answering "How did I get the equation above?" should help to elucidate my strategy here.
Consider the following equation $\color{red}{2^{|x|}+2^{|y|}=m}$ which is red in the graph above the square given by $$\color{blue}{\begin{cases} x=\pm \log_2(m-1) \\ y=\pm \log_2(m-1)\end{cases}}$$ which is blue in the graph above. Let's compute the area of the shaded region in the top right. These curves meet at $\big(0,\log_2(m-1)\big)$ and $\big(0,\log_2(m-1)\big)$.
This region's area is given by
$$ \color{green}{ \int_0^{\log_2(m-1)}{ \log_2(m-1) - \log_2(m-2^x) dx} \label{eq:2}\tag{2}} $$
The constant out front it easy to integrate and (apparently by wolfram alpha) we have
$$\int_0^{\log_2(m-1)}{\log_2(m-2^x) dx}=\frac{\operatorname{Li}_2(\frac{1}{m})-\operatorname{Li}_2{(\frac{m-1}{m})}+\log_2(m)\log_2(m-1)}{\ln(2)^2}$$
where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty {\frac{x^n}{n^2}}$
By taking $m\to \infty$ equation (\ref{eq:2}) becomes
$$\lim_{m\to \infty} \int_0^{\log_2(m-1)}{ \log_2(m-1) - \log_2(m-2^x) dx}=\frac{\operatorname{Li}_2(1)-\operatorname{Li}_2{0}}{\ln(2)^2}= \frac{\pi^2/6-0}{\ln(2)^2}=\frac{\zeta(2)}{\ln(2)^2}$$
One can change some $2$s to $k$s and mess with the indexing slightly (this washes out as we let $m\to \infty$) to get to equation (\ref{eq:1}) above.
So how should we expect to bring about $\zeta(3)$? Well, I have some intuition (and could be wrong) that this should come about by asking ourselves what the space is between the cube with side lengths $\log_2(m-2)$ and the region bounded by $x=0, y=0, z=0$ and $2^x+2^y+2^z=m$. How do I describe the space here as integrals? And for arbitrary dimension $s$ can it be demonstrated that this limits to $\zeta(s)/ \ln(k)^s$?
So this is my attempt at this:
Let $m\in \mathbb{R}$ and $s\in \mathbb{N}$. Let $R_m$ be the region described bounded by $x_1 = x_2 = \dots x_s=\log_2(m-s)$ and $m=\sum_{2^{x_i}}$. Then the limit of the volume of the region I am interested in computing is (hopefully, I don't trust myself on this:) $$\lim_{m\to\infty} \log_2(m-s)^s - \int_{R_m} \log_2 \bigg( m-\sum_{j=1}^{s-1}2^{x_i} \bigg)$$ Is this the correct description? And is there away to compute this?
Let's look specifically at $s=3$. The volume you're looking for is $$\iiint\limits_{\substack{0\leqslant x,y,z\leqslant\log_2(m-2)\\2^x+2^y+2^z\geqslant m}}dx\,dy\,dz=\frac{1}{(\ln 2)^3}\iiint\limits_{\substack{1\leqslant x_1,x_2,x_3\leqslant m-2\\x_1+x_2+x_3\geqslant m}}\frac{dx_1\,dx_2\,dx_3}{x_1 x_2 x_3}$$ and, after substituting $x_k=1+(m-3)y_k$, it is equal to $f\big(1/(m-3)\big)$, where $$f(a)=\frac{1}{(\ln 2)^3}\iiint\limits_{\substack{0\leqslant y_1,y_2,y_3\leqslant 1\\y_1+y_2+y_3\geqslant 1}}\frac{dy_1\,dy_2\,dy_3}{(a+y_1)(a+y_2)(a+y_3)};$$ now $m\to\infty$ corresponds to $a\to 0$. Can you see what happens?
(We could go even simpler with $x_k=my_k$. In the 2D case, the corresponding integral converges.)