Definite Integral $\int_0^1 \left \{\frac{1}{x^\frac{1}{6}} \right\}\, dx$

Question

The curly brackets mean 'FractionalPart' which, I believe, is defined as {${x}$}$=x-\lfloor x \rfloor$ where $x \in \mathbb{R}$.

My best approximation so far is: .182657 , however, I suspect there is a closed form expression for this definite integral. My approximation was attained by using a lower bound of .00000001 (a bit to the right of 0) with a graphing program.

Both Answers below are perfect responses to my question but MSE only lets me Accept one of them. The second answer is a general solution to any integrand $\displaystyle \int_{0}^{1}\left { \frac{1}{\sqrt[m]{x}} \right } dx$ — Bob Kadylo, Apr 11 '16 at 06:09
Great question! You're so good in making original questions! — GohP.iHan, Apr 11 '16 at 12:03
It turns out that $m$ does not have to be an integer. The formula is correct for any real $m > 1$. — marty cohen, Apr 11 '16 at 18:12
@BobKadylo Even better the techniques I referred you in the above link handles more general cases! — Mhenni Benghorbal, Apr 12 '16 at 01:08

score 9 · Accepted Answer · answered Apr 11 '16 at 03:01

The integral is equal to

$$\int_0^1 dx \, x^{-1/6} - \int_0^1 dx \lfloor x^{-1/6} \rfloor $$

Let's focus on the second integral. Note that when $x \in \left [(n+1)^{-6},n^{-6} \right ]$, $\lfloor x^{-1/6} \rfloor = n$. Thus,

$$\begin{align} \int_0^1 dx \lfloor x^{-1/6} \rfloor &= \sum_{n=1}^{\infty} n \left [\frac1{n^6} - \frac1{(n+1)^6} \right ] \\ &= \sum_{n=1}^{\infty} \left [ \frac1{n^5} - \frac1{(n+1)^5} \right ] + \sum_{n=1}^{\infty} \frac1{(n+1)^6} \\ &= 1+\zeta(6)-1\\ &= \frac{\pi^6}{945} \end{align}$$

Thus, the integral is equal to

$$\int_0^1 dx \lbrace x^{-1/6} \rbrace = \frac65 - \frac{\pi^6}{945}$$

which checks out numerically.

Yours is simpler than mine, though less general. However, it could certainly be made as general. It's nice that our answers agree. — marty cohen, Apr 11 '16 at 03:44

score 4 · Answer 2 · answered Apr 11 '16 at 03:43

Let $g_m =\int_0^1 \{x^{-1/m}\} dx $.

I get $g(m) =\dfrac{m}{m-1}-\zeta(m) $.

Here's how.

I want to partition $[0, 1]$ into intervals over which $\{x^{-1/m}\} $ is between two consecutive integers, and then sum the integral over these intervals.

So, for each positive integer $n$, I want $n = x^{-1/m} $ or $x =\frac1{n^m} $.

Let

$\begin{array}\\ g_{m, n} &=\int_{\frac1{(n+1)^m}}^{\frac1{n^m}} \{x^{-1/m}\} dx\\ &=\int_{\frac1{(n+1)^m}}^{\frac1{n^m}} (x^{-1/m}-n) dx\\ &=\int_{\frac1{(n+1)^m}}^{\frac1{n^m}} x^{-1/m}dx -n(\frac1{n^m}-\frac1{(n+1)^m})\\ &=\dfrac{x^{1-1/m}}{1-1/m}\big|_{\frac1{(n+1)^m}}^{\frac1{n^m}} -(\frac{n}{n^{m}}-\frac{n}{(n+1)^{m}})\\ &=\dfrac{m}{m-1}x^{(m-1)/m}\big|_{\frac1{(n+1)^m}}^{\frac1{n^m}} -(\frac{1}{n^{m-1}}-\frac{n+1-1}{(n+1)^{m}})\\ &=\dfrac{m}{m-1}(\frac1{n^{m-1}}-\frac1{(n+1)^{m-1}}) -(\frac{1}{n^{m-1}}-\frac{1}{(n+1)^{m-1}}+\frac{1}{(n+1)^{m}})\\ &=(\dfrac{m}{m-1}-1)(\frac1{n^{m-1}}-\frac1{(n+1)^{m-1}}) -\frac{1}{(n+1)^{m}}\\ &=\dfrac{1}{m-1}(\frac1{n^{m-1}}-\frac1{(n+1)^{m-1}}) -\frac{1}{(n+1)^{m}}\\ \end{array} $

Therefore

$\begin{array}\\ g_m &=\sum_{n=1}^{\infty} g_{m, n}\\ &=\sum_{n=1}^{\infty} (\dfrac{1}{m-1}(\frac1{n^{m-1}}-\frac1{(n+1)^{m-1}}) -\frac{1}{(n+1)^{m}})\\ &=\dfrac{1}{m-1}\sum_{n=1}^{\infty} (\frac1{n^{m-1}}-\frac1{(n+1)^{m-1}}) -\sum_{n=1}^{\infty}\frac{1}{(n+1)^{m}})\\ &=\dfrac{1}{m-1}-(\zeta(m)-1)\\ &=\dfrac{m}{m-1}-\zeta(m)\\ \end{array} $

For $m=6$, Wolfy says this is $0.182656938015... $, so this has a good chance of being correct,

I think this is a perfect Generalization !! I wish I could accept two answers. — Bob Kadylo, Apr 11 '16 at 06:14
You can upvote it, if you haven't already. Also, a further generalization might be to have a rational exponent, rather than just $1/m$. This might be entertaining. — marty cohen, Apr 11 '16 at 06:28
I did upvote your solution as soon as I saw it. My head hurts just thinking about the rational exponent extension. — Bob Kadylo, Apr 11 '16 at 07:40
Actually, my proof works just fine for any real $m > 1$. $m$ does not have to be an integer. — marty cohen, Apr 11 '16 at 18:11

Definite Integral $\int_0^1 \left \{\frac{1}{x^\frac{1}{6}} \right\}\, dx$

2 Answers2