Atal's Integrals

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Define a function \(f\) on \([0,1]\) as follows: $$f(x) =\begin{cases}\frac{1}{x}-n & \text{if } n \in \mathbb{N} \text{ and } \frac{1}{n+1} \lt x \leq \frac{1}{n}, \\ 0 & \text{if } x = 0. \end{cases}$$ Let \(k\) be fixed, and let \(\{x\}\) denote the fractional part of \(x\). Then \(f(x) = \{ \frac{1}{x} \}\) and we have the following asymptotic as \(N \to \infty\): $$\sum_{n=1}^N \left\{ \frac{N}{n} \right\}^k \sim N \int_0^1 (f(x))^k \,dx,$$ This page contains the values of the integral \(\int_0^1 (f(x))^k \,dx\) as \(k\) varies.
Here \(\gamma\) is the Euler-Mascheroni constant, \(A\) is the Glaisher-Kinkelin constant, and \(\zeta\) is the Riemann zeta function.

For \(k=1\): $$1-\gamma$$ For \(k=2\): $$-1-\gamma +\log (2)+\log (\pi )$$ For \(k=3\): $$\frac{1}{2} (-12 \log (A)-1-2 \gamma +\log (8)+3 \log (\pi ))$$ For \(k=4\): $$-12 \log (A)+\frac{3 \zeta (3)}{\pi ^2}-\frac{1}{3}-\gamma +\log (4)+2 \log (\pi )$$ For \(k=5\): $$-20 \log (A)+20 \zeta '(-3)+\frac{15 \zeta (3)}{2 \pi ^2}+\frac{1}{18}-\gamma +\frac{5}{2} \log (2 \pi )$$ For \(k=6\): $$-30 \log (A)+60 \zeta '(-3)+\frac{15 \zeta (3)}{\pi ^2}-\frac{45 \zeta (5)}{2 \pi ^4}+\frac{43}{60}-\gamma +\log (8)+3 \log (\pi )$$ For \(k=7\): $$-42 \log (A)+42 \zeta '(-5)+140 \zeta '(-3)+\frac{105 \left(\pi ^2 \zeta (3)-3 \zeta (5)\right)}{4 \pi ^4}+\frac{191}{120}-\gamma +\frac{7}{2} \log (2 \pi )$$ For \(k=8\): $$-56 \log (A)+168 \zeta '(-5)+280 \zeta '(-3)+\frac{21 \left(2 \pi ^4 \zeta (3)-10 \pi ^2 \zeta (5)+15 \zeta (7)\right)}{\pi ^6}+\frac{823}{315}-\gamma +\log (16)+4 \log (\pi )$$ For \(k=9\): $$-72 \log (A)+72 \zeta '(-7)+504 \zeta '(-5)+504 \zeta '(-3)+\frac{63 \left(2 \pi ^4 \zeta (3)-15 \pi ^2 \zeta (5)+45 \zeta (7)\right)}{2 \pi ^6}+\frac{7951}{2100}-\gamma +\frac{9}{2} \log (2 \pi )$$ For \(k=10\): $$-90 \log (A)+360 \zeta '(-7)+1260 \zeta '(-5)+840 \zeta '(-3)+\frac{45 \left(4 \pi ^6 \zeta (3)-42 \pi ^4 \zeta (5)+210 \pi ^2 \zeta (7)-315 \zeta (9)\right)}{2 \pi ^8}+\frac{13091}{2520}-\gamma +\log (32)+5 \log (\pi )$$ For \(k=11\): $$-110 \log (A)+110 \zeta '(-9)+1320 \zeta '(-7)+2772 \zeta '(-5)+1320 \zeta '(-3)+\frac{495 \left(\pi ^6 \zeta (3)-14 \pi ^4 \zeta (5)+105 \pi ^2 \zeta (7)-315 \zeta (9)\right)}{4 \pi ^8}+\frac{14803}{2160}-\gamma +\frac{11}{2} \log (2 \pi )$$
Challenge: Find a general expression for this integral in terms of \(k\).

Progress: Atal has conjectured the following formula for the integral: $$\int_0^1 (f(x))^k \,dx = - \gamma + \sum_{n=0}^{k-2} \left[ (-1)^{n+1} (k-n) \binom{k}{n} \zeta'(-n) \right] - c_k,$$ where \(\{c_k\}_{k=1}^\infty\) is a sequence of rational numbers (the first few are \(-1,1,1,\frac{4}{3},\frac{29}{18},\frac{107}{60},\frac{229}{120},\frac{647}{315},\frac{4649}{2100},\frac{5809}{2520},\frac{4997}{2160}\)).

New Challenge: Explain the sequence \(\{c_k\}_{k=1}^\infty\).

Another New Challenge: What about non-integer \(k\)?

Update: Atal has found the following general formula for integers \(k > 1\): $$\int_0^1 (f(x))^k \,dx = - \gamma + \sum_{n=0}^{k-2} \left[ (-1)^n \left( \frac{B_{n+1} H_n}{n+1} - \zeta'(-n) \right) k \binom{k-1}{n} \right] - \frac{1}{k-1},$$ where \(B_n\) is the \(n\)th Bernoulli number and \(H_n\) is the \(n\)th harmonic number.
From this, it also follows, interestingly, that $$\lim_{k \to \infty} \left[ \sum_{n=0}^{k-2} (-1)^n \left( \frac{B_{n+1} H_n}{n+1} - \zeta'(-n) \right) k \binom{k-1}{n} \right] = \gamma.$$

Another Update: Atal has found an effective version of the above limit. First, let $$g(x) = \begin{cases} (n+1)(1-nx) & \text{if } n \in \mathbb{N} \text{ and } \frac{1}{n+1} \lt x \leq \frac{1}{n}, \\ 0 & \text{if } x = 0.\end{cases}$$ Then, by noticing that \(f(x) \leq g(x)\) on \([0,1]\), we can observe that $$\sum_{n=0}^{k-2} (-1)^n \left( \frac{B_{n+1} H_n}{n+1} - \zeta'(-n) \right) k \binom{k-1}{n} = \gamma + r(k),$$ where the remainder term \(r(k)\) satisfies \(r(k) = \Theta\left(\frac{1}{k}\right)\).