integral involving floor function $\int_0^1 \left \lfloor{ (a^{n}x) ^{ \frac{n}{2} }}\right \rfloor dx $ 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).
