Does \(\displaystyle\int_1^\infty \dfrac{\{x\}}{x^k}\,dx\) converge for integers \(k\color{red}>2\)? \(\{x\}\) denotes the fractional part of \(x\), i.e. \[\{x\}=x-\lfloor x\rfloor\]if \(x\ge0\).

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Does \(\displaystyle\int_1^\infty \dfrac{\{x\}}{x^k}\,dx\) converge for integers \(k\color{red}>2\)? \(\{x\}\) denotes the fractional part of \(x\), i.e. \[\{x\}=x-\lfloor x\rfloor\]if \(x\ge0\).

Mathematics
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And if it does converge, is there an exact value?
Here are the plots for \(2\le k\le5\).
1 Attachment
\[\text{ if } n \le x < n+1 \text{ then } \lfloor x \rfloor =n \\ \\ \int\limits _1^2 \frac{x-1}{x^k} dx + \int\limits_2^3 \frac{x-2}{x^k}dx + \cdots + \int\limits_n^{n+1} \frac{x-n}{x^k}+ \cdots \\ \sum_{n=1}^\infty \int\limits _n^{n+1} \frac{x-n}{x^k} dx \\ \sum_{n=1}^\infty \int\limits_n^{n+1} (x^{1-k}-nx^{-k}) dx \\ \sum_{n=1}^{\infty} [\frac{x^{1-k+1}}{1-k+1}-n \frac{x^{-k+1}}{-k+1})|_n^{n+1} \\ \] brb this is what I have so far will check in a sec

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after pluggin in those limits it looks nasty
\[\sum_{n=1}^\infty \left(\frac{(n+1)^{2-k}}{2-k}-\frac{n(n+1)^{1-k}}{1-k}-\frac{n^{2-k}}{2-k}+\frac{n^{2-k}}{1-k}\right)\] It doesn't look so bad at first glance. It also looks like we need to have \(k>2\) for convergence.
Or maybe not? \[\lim_{k\to2}\frac{(n+1)^{2-k}-n^{2-k}}{2-k}=\ln\frac{n+1}{n}\]
At any rate, we have \[\int_1^\infty \frac{\{x\}}{x^k}\,dx\le\int_1^\infty \frac{dx}{x^k}\]so convergence is guaranteed.
^ since \(\{x\}<1\)
just for fun I wanted to see what our sum was doing for some value k (it wouldn't let me go passed k=10) like for k=5 I have the sum=0.06272525 and for k=10 I have sum=0.0136657 I just wanted to see if I could find some kind of pattern from the sums there and give myself an idea to work forward in a backwards sort of way but these numbers are icky to me :p
well instead of aiming to find a sum I will just see if I can see if it converges or find a way to show it does converge
We can simplify the sum a little bit:\[\sum_{n=1}^\infty \frac{(n+1)^{2-k}-n^{2-k}}{2-k}=\frac{2^{2-k}-1+3^{2-k}-2^{2-k}+\cdots}{2-k}=\frac{1}{k-2}\]so the integral is equivalent to \[\frac{1}{k-2}+\sum_{n=1}^\infty \frac{n^{2-k}-n(n+1)^{1-k}}{1-k}\]
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Still not quite sure if we can have \(k=2\). Looking at the partial sums of the "first" series: \[\begin{align*}\sum_{n=1}^\infty \frac{(n+1)^{2-k}-n^{2-k}}{2-k}&=\lim_{N\to\infty}\sum_{n=1}^N\frac{(n+1)^{2-k}-n^{2-k}}{2-k}\\\\ &=\lim_{N\to\infty}\frac{N^{2-k}-1}{2-k} \end{align*}\] As \(k\to2\), I'm getting \(\dfrac{N^{2-k}-1}{2-k}\to\ln N\). Not quite sure what to make of this.
I'm totally stuck for now need to sleep will check out problem tomorrow
surely you should be looking at \(k=1\)? it converges trivially for \(k=2\) by the integral bound you stated
found this interesting result for \(s\ge 2\) \[\large \int\limits_{1}^{\infty} \dfrac{\{x\}}{x^{s+1}}\,dx = \dfrac{s}{s-1} - \dfrac{\zeta(s)}{s}\]
$$\begin{align*}\int_1^\infty\frac{x-\lfloor x\rfloor}{x^2}dx&=\sum_{n=1}^\infty\int_n^{n+1}\frac{x-n}{x^2}\ dx\\&=\sum_{n=1}^\infty\left[\log x+\frac{n}x\right]_n^{n+1}\\&=\sum_{n=1}^\infty\left(\log(n+1)-\log(n)+\frac{n}{n+1}-1\right)\\&=\sum_{n=1}^\infty\left(\log(n+1)-\log(n)-\frac1{n+1}\right)\end{align*}$$now we already know this has to converge but how can we massage the above to make it more clear
basically we can study that infinite sum in terms of partial sums, especially since the first part telescopes: $$\sum_{n=1}^N\left(\log(n+1)-\log(n)-\frac1{n+1}\right)=\log(N+1)-\sum_{n=2}^N\frac1n$$ as \(n\to\infty\)
oops, that should read \(-\sum_{n=2}^{N+1}\frac1n\)
so denoting the \(n\)-th harmonic number \(H_n=\sum_{k=1}^n\frac1n\) we get: $$\log(N+1)-\left(\sum_{n=1}^{N+1}\frac1n-1\right)=\log(N+1)-H_{N+1}+1$$and using our knowledge of the Euler-Mascheroni constant we find: $$\log(N+1)-H_{N+1}\to\gamma\approx 0.57721$$ so $$\int_1^\infty\frac{\{x\}}{x^2}\ dx=1+\gamma\approx 1.57721$$
actually, oops, \(\log(N+1)-H_{N+1}\to-\gamma=-0.57721\) so \(\int_1^\infty\frac{\{x\}}{x^2}dx=1-\gamma\approx 0.42279\)
note this series converges pretty slow as evidenced by the asymptotic formula for the harmonic numbers $$H_n\sim \ln n+\gamma+\frac1{2n}+O(1/n^2)$$ i.e. it takes about \(50\) terms to get precision to the second decimal place \(0.01\)
Ah, got it. I've been using Mathematica to try to get a numerical approximation and I kept getting an error, thinking it implied the integral didn't converge. After looking into it, the error indeed has to do with the fact that convergence is too slow.
I'll have to look more into harmonic analysis and that EM constant in the meantime.

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