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