## 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 $$x\ge0$$.

1. anonymous

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

2. anonymous

Here are the plots for $$2\le k\le5$$.

3. 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

4. freckles

after pluggin in those limits it looks nasty

5. 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.

6. anonymous

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

7. 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.

8. anonymous

^ since $$\{x\}<1$$

9. 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

10. 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

11. 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 $\frac{1}{k-2}+\sum_{n=1}^\infty \frac{n^{2-k}-n(n+1)^{1-k}}{1-k}$

12. ganeshie8

.

13. 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.

14. freckles

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

15. anonymous

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

16. ganeshie8

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

17. 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

18. 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 $$n\to\infty$$

19. anonymous

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

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

21. anonymous

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

22. 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 $$0.01$$

23. 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.

24. anonymous

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