Question

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 \(x\ge0\).

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

after pluggin in those limits it looks nasty

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

.

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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 \(n\to\infty\)

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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 \(0.01\)

Answer 23

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.

Answer 24

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