A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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 \(x\ge0\).

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And if it does converge, is there an exact value?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here are the plots for \(2\le k\le5\).

    1 Attachment
  3. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\text{ if } n \le x < n+1 \text{ then } \lfloor x \rfloor =n \\ \\ \int\limits _1^2 \frac{x-1}{x^k} dx + \int\limits_2^3 \frac{x-2}{x^k}dx + \cdots + \int\limits_n^{n+1} \frac{x-n}{x^k}+ \cdots \\ \sum_{n=1}^\infty \int\limits _n^{n+1} \frac{x-n}{x^k} dx \\ \sum_{n=1}^\infty \int\limits_n^{n+1} (x^{1-k}-nx^{-k}) dx \\ \sum_{n=1}^{\infty} [\frac{x^{1-k+1}}{1-k+1}-n \frac{x^{-k+1}}{-k+1})|_n^{n+1} \\ \] brb this is what I have so far will check in a sec

  4. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    after pluggin in those limits it looks nasty

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sum_{n=1}^\infty \left(\frac{(n+1)^{2-k}}{2-k}-\frac{n(n+1)^{1-k}}{1-k}-\frac{n^{2-k}}{2-k}+\frac{n^{2-k}}{1-k}\right)\] It doesn't look so bad at first glance. It also looks like we need to have \(k>2\) for convergence.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Or maybe not? \[\lim_{k\to2}\frac{(n+1)^{2-k}-n^{2-k}}{2-k}=\ln\frac{n+1}{n}\]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^ since \(\{x\}<1\)

  9. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  10. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \[\frac{1}{k-2}+\sum_{n=1}^\infty \frac{n^{2-k}-n(n+1)^{1-k}}{1-k}\]

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    .

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  14. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I'm totally stuck for now need to sleep will check out problem tomorrow

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    surely you should be looking at \(k=1\)? it converges trivially for \(k=2\) by the integral bound you stated

  16. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    found this interesting result for \(s\ge 2\) \[\large \int\limits_{1}^{\infty} \dfrac{\{x\}}{x^{s+1}}\,dx = \dfrac{s}{s-1} - \dfrac{\zeta(s)}{s}\]

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    $$\begin{align*}\int_1^\infty\frac{x-\lfloor x\rfloor}{x^2}dx&=\sum_{n=1}^\infty\int_n^{n+1}\frac{x-n}{x^2}\ dx\\&=\sum_{n=1}^\infty\left[\log x+\frac{n}x\right]_n^{n+1}\\&=\sum_{n=1}^\infty\left(\log(n+1)-\log(n)+\frac{n}{n+1}-1\right)\\&=\sum_{n=1}^\infty\left(\log(n+1)-\log(n)-\frac1{n+1}\right)\end{align*}$$now we already know this has to converge but how can we massage the above to make it more clear

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(n\to\infty\)

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops, that should read \(-\sum_{n=2}^{N+1}\frac1n\)

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 $$\int_1^\infty\frac{\{x\}}{x^2}\ dx=1+\gamma\approx 1.57721$$

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    actually, oops, \(\log(N+1)-H_{N+1}\to-\gamma=-0.57721\) so \(\int_1^\infty\frac{\{x\}}{x^2}dx=1-\gamma\approx 0.42279\)

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(0.01\)

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll have to look more into harmonic analysis and that EM constant in the meantime.

  25. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.