Extend an asymptotic relation with integer parameter to real parameters? A text I'm studying derives the following, for integer s>0 and $n \to \infty$:
$$S(s,n):=\sum_{k=0}^{2n}(-1)^{n+k}\binom{2n}{k}^s \sim
\frac{1}{\sqrt{s}}\big(2\cos\big(\frac{\pi}{2s}\big)\big)^{2ns+s-1}\,2^{2-s}
(\pi\,n)^{(1-s)/2} $$
The derivation involves an s-fold multiple integral and a step occurs where the Hessian, a determinant, is estimated. It is therefore definitely depending on $s$ being a positive integer. 
I tested this numerically for many $real$ $s>s_0(n)$ and observe that it also seems to hold. 
For instance, here is a small table with n=30 and n=300. $S_e$ = 'exact,' the brute-force summation with lots of precision.  $S_a$ =the asymptotic approximation.
$$
\begin{array}{c|lcr}
s & S_e(\cdot,30) &S_a(\cdot,30) & S_e(\cdot,300)  &S_a(\cdot,300) \\
\hline
149/100 & 0.1267 & 0.1271 & 2.465\,\text{x}\,10^{-6} & 3.721\,\text{x}\,10^{-6} \\
150/100 & 0.3696 & 0.3706 & 0.2084 & 0.2084 \\
18/10 & 1.0260\,\text{x}\,10^{11} &1.0296\,\text{x}\,10^{11} &  
4.521\,\text{x}\,10^{116} &4.523\,\text{x}\,10^{116} \\
23/10 & 1.028\,\text{x}\,10^{25} &1.033\,\text{x}\,10^{25} & 1.7817\,\text{x}\,10^{261} &1.7826\,\text{x}\,10^{261}  \\
\end{array}
$$ 
If $s$ drops much below 1.5, the the exact sum appears to go negative and the asymptotic relationship no longer holds.
I seek a proof that the asymptotic relationship holds for all real $s\ge 1.5$

This problem is worked out in de Bruijn, Asymptotic Methods in Analysis, Ch. 4.7 for natural $s$ and Ch. 6.4-6.7 for real $s$.
