i need to find the result of this integral
$$\int_0^1 \left \lfloor{ (a^{n}x) ^{ \frac{n}{2} }}\right \rfloor dx $$
with $$a \in \mathbb{N}$$
i tried to transform it to a finite sum and i found this:
$$ \frac{1}{a^{n}} \sum_0^{ a^{n}-1 } k((k+1)^{ \frac{2}{n}}-(k)^{ \frac{2}{n} }) $$
but i couldn't calculate it and i don't know if it is right or false
Let's put $$ \eqalign{ & y = \left( {a^{\,n} x} \right)^{\,n/2} \cr & x = {{y^{\,2/n} } \over {a^{\,n} }} \cr & dx = {2 \over {na^{\,n} }}y^{\,2/n - 1} dy \cr & u = y(1) = a^{\,n^{\,2} /2} \cr} $$
then $$ \eqalign{ & I(a,n) = \int_{x = 0}^1 {\left\lfloor {\left( {a^{\,n} x} \right)^{\,n/2} } \right\rfloor dx} = {2 \over {na^{\,n} }}\int_{y = 0}^u {\left\lfloor y \right\rfloor y^{\,2/n - 1} dy} = \cr & = {2 \over {na^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {\int_{y = k}^{k + 1} {k\,y^{\,2/n - 1} dy} } \right)} + \int_{y = \left\lfloor u \right\rfloor }^u {\left\lfloor u \right\rfloor \,y^{\,2/n - 1} dy} } \right) = \cr & = {2 \over {na^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {{{nk} \over 2}\left( {\left( {k + 1} \right)^{\,2/n} - k^{\,2/n} } \right)} \right)} + \left\lfloor u \right\rfloor {n \over 2}\left( {u^{\,2/n} - \left\lfloor u \right\rfloor ^{\,2/n} } \right)} \right) = \cr & = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k\left( {\left( {k + 1} \right)^{\,2/n} - k^{\,2/n} } \right)} \right)} + \left\lfloor u \right\rfloor \left( {u^{\,2/n} - \left\lfloor u \right\rfloor ^{\,2/n} } \right)} \right) \cr} $$
The last term is $$ \eqalign{ & \left\lfloor u \right\rfloor \left( {u^{\,2/n} - \left\lfloor u \right\rfloor ^{\,2/n} } \right) = \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor \left( {a^{\,n} - \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor ^{\,2/n} } \right) = \cr & = R(a,n) \cr} $$ and is null for $a,n$ positive integers and $n$ even, but it is not if $n$ is odd. Let's call it $R(a,n)$.
The sum can be further simplified to give $$ \eqalign{ & I(a,n) = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k\left( {\left( {k + 1} \right)^{\,2/n} - k^{\,2/n} } \right)} \right)} + R(a,n)} \right) = \cr & = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {\left( {k + 1 - 1} \right)\left( {k + 1} \right)^{\,2/n} - k^{\,2/n + 1} } \right)} + R(a,n)} \right) = \cr & = {1 \over {a^{\,n} }}\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n + 1} } - \sum\limits_{\left( {0\, \le } \right)\,1\, \le \,\,k\, \le \,\left\lfloor u \right\rfloor - 1} {k^{\,2/n + 1} } - \sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } + R(a,n)} \right) = \cr & = {1 \over {a^{\,n} }}\left( {\left\lfloor u \right\rfloor ^{\,2/n + 1} + R(a,n) - \sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } } \right) = \cr & = \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - {1 \over {a^{\,n} }}\sum\limits_{0\, \le \,k\, \le \,\left\lfloor u \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } \cr} $$
So we can write $$ \bbox[lightyellow] { I(a,n) = \int_{x = 0}^1 {\left\lfloor {\left( {a^{\,n} x} \right)^{\,n/2} } \right\rfloor dx} = \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - {1 \over {a^{\,n} }}\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } } \tag{1}$$
Now, we have the sum of powers with constant exponent and variable basis.
This is related to Generalized Harmonic Numbers ( and to Bernoulli Polynomials, Digamma Function).
However, being the exponent fractional, the above are expressable through the Hurwitz zeta function, i.e.
$$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,\left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor - 1} {\left( {k + 1} \right)^{\,2/n} } = \cr & = \sum\limits_{0\, \le \,k\,} {{1 \over {\left( {k + 1} \right)^{\, - \,2/n} }}} - \sum\limits_{0\, \le \,k\,} {{1 \over {\left( {k + \left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor + 1} \right)^{\, - \,2/n} }}} = \cr & = \zeta \left( { - 2/n,\;1} \right) - \zeta \left( { - 2/n,\;\left\lfloor {a^{\,n^{\,2} /2} } \right\rfloor + 1} \right) \cr} } \tag{2}$$
Example
$$ \eqalign{ & I(2,3) = 8.57135699... \cr & I(2,4) = 84.84616346... \cr} $$ checked with the original integral, eq. (1), and eq.(1) with substitution of (2).