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anonymous
 one year ago
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 \(x\ge0\).
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And if it does converge, is there an exact value?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here are the plots for \(2\le k\le5\).

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ if } n \le x < n+1 \text{ then } \lfloor x \rfloor =n \\ \\ \int\limits _1^2 \frac{x1}{x^k} dx + \int\limits_2^3 \frac{x2}{x^k}dx + \cdots + \int\limits_n^{n+1} \frac{xn}{x^k}+ \cdots \\ \sum_{n=1}^\infty \int\limits _n^{n+1} \frac{xn}{x^k} dx \\ \sum_{n=1}^\infty \int\limits_n^{n+1} (x^{1k}nx^{k}) dx \\ \sum_{n=1}^{\infty} [\frac{x^{1k+1}}{1k+1}n \frac{x^{k+1}}{k+1})_n^{n+1} \\ \] brb this is what I have so far will check in a sec

freckles
 one year ago
Best ResponseYou've already chosen the best response.2after pluggin in those limits it looks nasty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^\infty \left(\frac{(n+1)^{2k}}{2k}\frac{n(n+1)^{1k}}{1k}\frac{n^{2k}}{2k}+\frac{n^{2k}}{1k}\right)\] It doesn't look so bad at first glance. It also looks like we need to have \(k>2\) for convergence.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or maybe not? \[\lim_{k\to2}\frac{(n+1)^{2k}n^{2k}}{2k}=\ln\frac{n+1}{n}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0At any rate, we have \[\int_1^\infty \frac{\{x\}}{x^k}\,dx\le\int_1^\infty \frac{dx}{x^k}\]so convergence is guaranteed.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2just 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We can simplify the sum a little bit:\[\sum_{n=1}^\infty \frac{(n+1)^{2k}n^{2k}}{2k}=\frac{2^{2k}1+3^{2k}2^{2k}+\cdots}{2k}=\frac{1}{k2}\]so the integral is equivalent to \[\frac{1}{k2}+\sum_{n=1}^\infty \frac{n^{2k}n(n+1)^{1k}}{1k}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still not quite sure if we can have \(k=2\). Looking at the partial sums of the "first" series: \[\begin{align*}\sum_{n=1}^\infty \frac{(n+1)^{2k}n^{2k}}{2k}&=\lim_{N\to\infty}\sum_{n=1}^N\frac{(n+1)^{2k}n^{2k}}{2k}\\\\ &=\lim_{N\to\infty}\frac{N^{2k}1}{2k} \end{align*}\] As \(k\to2\), I'm getting \(\dfrac{N^{2k}1}{2k}\to\ln N\). Not quite sure what to make of this.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I'm totally stuck for now need to sleep will check out problem tomorrow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0surely you should be looking at \(k=1\)? it converges trivially for \(k=2\) by the integral bound you stated

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0found this interesting result for \(s\ge 2\) \[\large \int\limits_{1}^{\infty} \dfrac{\{x\}}{x^{s+1}}\,dx = \dfrac{s}{s1}  \dfrac{\zeta(s)}{s}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\begin{align*}\int_1^\infty\frac{x\lfloor x\rfloor}{x^2}dx&=\sum_{n=1}^\infty\int_n^{n+1}\frac{xn}{x^2}\ dx\\&=\sum_{n=1}^\infty\left[\log x+\frac{n}x\right]_n^{n+1}\\&=\sum_{n=1}^\infty\left(\log(n+1)\log(n)+\frac{n}{n+1}1\right)\\&=\sum_{n=1}^\infty\left(\log(n+1)\log(n)\frac1{n+1}\right)\end{align*}$$now we already know this has to converge but how can we massage the above to make it more clear

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0basically 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 \(n\to\infty\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, that should read \(\sum_{n=2}^{N+1}\frac1n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so denoting the \(n\)th harmonic number \(H_n=\sum_{k=1}^n\frac1n\) we get: $$\log(N+1)\left(\sum_{n=1}^{N+1}\frac1n1\right)=\log(N+1)H_{N+1}+1$$and using our knowledge of the EulerMascheroni constant we find: $$\log(N+1)H_{N+1}\to\gamma\approx 0.57721$$ so $$\int_1^\infty\frac{\{x\}}{x^2}\ dx=1+\gamma\approx 1.57721$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually, oops, \(\log(N+1)H_{N+1}\to\gamma=0.57721\) so \(\int_1^\infty\frac{\{x\}}{x^2}dx=1\gamma\approx 0.42279\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0note 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 \(0.01\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll have to look more into harmonic analysis and that EM constant in the meantime.
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