# 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 $$r(k) = \Theta\left(\frac{1}{k}\right)$$.

Update the Third: Atal (along with Rushil) has found an even more effective version of the limit. Suppose we define $$3(x) = \int_0^1 f(t)^x \,dt.$$ Then from the above fact that $$f(t)^x \leq g(t)^x$$ and $$\int_0^1 g(t)^x \,dt = \frac{1}{x+1}$$, we can see that $$x3(x) \leq \frac{x}{x+1}$$; similarly, we can show that $$g(t)^2 \leq f(t)$$ (and so $$g(t)^{2x} \leq f(t)^x$$), which is equivalent to showing $$\left((n+1)(1-nt)\right)^2 \leq \frac{1}{t}-n \qquad \text{ for } t \in \left[ \frac{1}{n+1}, \frac{1}{n} \right], \qquad \text{ for any } n \in \mathbb{N}.$$ This essentially comes down to showing that $$- \left((n+1)n\right)^2\left(x-\frac{1}{n}\right)\left(x-\frac{1}{n+1}\right)\left(x-\frac{1}{n(n+1)}\right) \geq 0 \qquad \text{ on } x \in \left[ \frac{1}{n+1}, \frac{1}{n} \right],$$ which is clear. By some wacky monotonicity considerations it should be possible to show that $$\alpha = \lim_{x\to\infty} x3(x)$$ exists, and the above bounds tell us that $$\frac{x}{2x+1} \leq x 3(x) \leq \frac{x}{x+1}, \qquad \text{ so } \frac{1}{2} \leq \alpha \leq 1.$$ But this doesn't tell us what $$\alpha$$ is. However, a more succinct approach does: Note that $$3(x) = \int_0^1 \left\{ \frac{1}{t} \right\}^x \,dt = \int_1^\infty \frac{\{u\}^x}{u^2} \,du,$$ by applying $$u = \frac{1}{t}$$. Then we see that $$\int_1^\infty \frac{\{u\}^x}{u^2} \,du = \sum_{n=1}^\infty \int_n^{n+1} \frac{\{u\}^x}{u^2} \,du = \sum_{n=1}^\infty \int_n^{n+1} \frac{(u-n)^x}{u^2} \,du \geq \sum_{n=1}^\infty \frac{1}{(n+1)^2} \int_n^{n+1} (u-n)^x \,du = \frac{1}{x+1} \sum_{n=1}^\infty \frac{1}{(n+1)^2} = \frac{\zeta(2)-1}{x+1}.$$ (This method is due to Rushil Raghavan. Thanks Rushil! He also showed that $$\alpha \leq \frac{\pi^2}{6}$$ by using $$\frac{1}{u^2} \leq \frac{1}{n^2}$$ in the same way).
For the upper bound, we use the fact that $$\frac{(u-n)^x}{u^2} \leq \frac{(u-n)^{x-1}}{(n+1)^2} \qquad \text{ for } n \leq u \leq n+1,$$ which is clear since for $$n \leq u \leq n+1$$, we have $$0 \leq (u-(n+1))(u-n(n+1)) = u^2 - u(n+1)^2 + n(n+1)^2.$$ Then $$\sum_{n=1}^\infty \int_n^{n+1} \frac{(u-n)^x}{u^2} \,du \leq \sum_{n=1}^\infty \frac{1}{(n+1)^2} \int_n^{n+1} (u-n)^{x-1} \,du = \frac{\zeta(2) - 1}{x}.$$ So $$\alpha = \zeta(2) - 1$$. Well, this should also tell us that the giant sum above is $$\sum_{n=0}^{k-2} (-1)^n \left( \frac{B_{n+1} H_n}{n+1} - \zeta'(-n) \right) k \binom{k-1}{n} = \gamma + \frac{\pi^2}{6k} + O\left( \frac{1}{k^2} \right).$$ Can we see a pattern emerging?