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

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freckles
  • freckles
after pluggin in those limits it looks nasty
anonymous
  • anonymous
\[\sum_{n=1}^\infty \left(\frac{(n+1)^{2-k}}{2-k}-\frac{n(n+1)^{1-k}}{1-k}-\frac{n^{2-k}}{2-k}+\frac{n^{2-k}}{1-k}\right)\] It doesn't look so bad at first glance. It also looks like we need to have \(k>2\) for convergence.
anonymous
  • anonymous
Or maybe not? \[\lim_{k\to2}\frac{(n+1)^{2-k}-n^{2-k}}{2-k}=\ln\frac{n+1}{n}\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
^ since \(\{x\}<1\)
freckles
  • freckles
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
freckles
  • freckles
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
anonymous
  • anonymous
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 \[\frac{1}{k-2}+\sum_{n=1}^\infty \frac{n^{2-k}-n(n+1)^{1-k}}{1-k}\]
ganeshie8
  • ganeshie8
.
anonymous
  • anonymous
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.
freckles
  • freckles
I'm totally stuck for now need to sleep will check out problem tomorrow
anonymous
  • anonymous
surely you should be looking at \(k=1\)? it converges trivially for \(k=2\) by the integral bound you stated
ganeshie8
  • ganeshie8
found this interesting result for \(s\ge 2\) \[\large \int\limits_{1}^{\infty} \dfrac{\{x\}}{x^{s+1}}\,dx = \dfrac{s}{s-1} - \dfrac{\zeta(s)}{s}\]
anonymous
  • anonymous
$$\begin{align*}\int_1^\infty\frac{x-\lfloor x\rfloor}{x^2}dx&=\sum_{n=1}^\infty\int_n^{n+1}\frac{x-n}{x^2}\ dx\\&=\sum_{n=1}^\infty\left[\log x+\frac{n}x\right]_n^{n+1}\\&=\sum_{n=1}^\infty\left(\log(n+1)-\log(n)+\frac{n}{n+1}-1\right)\\&=\sum_{n=1}^\infty\left(\log(n+1)-\log(n)-\frac1{n+1}\right)\end{align*}$$now we already know this has to converge but how can we massage the above to make it more clear
anonymous
  • anonymous
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 \(n\to\infty\)
anonymous
  • anonymous
oops, that should read \(-\sum_{n=2}^{N+1}\frac1n\)
anonymous
  • anonymous
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 $$\int_1^\infty\frac{\{x\}}{x^2}\ dx=1+\gamma\approx 1.57721$$
anonymous
  • anonymous
actually, oops, \(\log(N+1)-H_{N+1}\to-\gamma=-0.57721\) so \(\int_1^\infty\frac{\{x\}}{x^2}dx=1-\gamma\approx 0.42279\)
anonymous
  • anonymous
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 \(0.01\)
anonymous
  • anonymous
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.
anonymous
  • anonymous
I'll have to look more into harmonic analysis and that EM constant in the meantime.

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