Too long to fit in comments.

**Remark(?).**

$$\sum_{\sigma\in S_n}(-1)^{\ell(\sigma)}\sum_{i=1}^ni^{(any~real~number)}\sigma(i)=0.$$

That probably follows from Darij's comment. Here again: n>2 ( n=2 is really an exception).

The Python code below checks it. (One can use https://colab.research.google.com/
for free - you even do not need to install anything on your comp - use browser and code runs on google's servers: )

```
import numpy as np
import time
def inversion(permList): # http://code.activestate.com/recipes/579051-get-the-inversion-number-of-a-permutation/
"""
Description - This function returns the number of inversions in a
permutation.
Preconditions - The parameter permList is a list of unique positve numbers.
Postconditions - The number of inversions in permList has been returned.
Input - permList : list
Output - numInversions : int
"""
if len(permList)==1:
return 0
else:
numInversion=len(permList)-permList.index(max(permList))-1
permList.remove(max(permList))
return numInversion+inversion(permList)
# Get all permutations # https://www.geeksforgeeks.org/permutation-and-combination-in-python/
# using library function
from itertools import permutations # import lib
n = 7
lst = range(1,(n+1) ) # = [1, 2, 3, 4, 5, 6, 7,..., n]
perm = permutations(lst) # all n! permutations are here
start_time = time.time()
_sum = 0
_new_power = 4.44
for p in list(perm): # Loop over permutations
inv_count = inversion(list(p)) # Calculate inversion number for "p"
for i in range(1,(n+1)):
_sum += (-1)**inv_count * (i**_new_power) *p[i-1]
print(_sum)
print('n=',n, time.time() - start_time , 'seconds passed' )
```

For n= 9 it runs: 6.369966745376587 seconds passed on google's colab
(it should be faster than notebook, but not much)