Computation of a sum using Stirling's approximation and Watson's lemma $$Ω=\sum_{n=0}^{N-\frac{E}{\epsilon}}  \frac{Ν!}{\left(\frac{N-n-\frac{E}{\epsilon}}{2}\right)!\left(\frac{N-n+\frac{E}{\epsilon}}{2}\right)!n!}$$
I am supposed to calculate the above sum using first Stirling's approximation and then using Watson's lemma integrate for $x=n/N$. The point is to find the natural logarithm of Ω and the above are hints.
 A: Here's what I got:
For $ x= \frac nN$ and $ E'= \frac E{N\epsilon} $ we get:
$$Ω=\sum_{n=0}^{N-\frac{E}{\epsilon}} \frac{Ν!}{\left(\frac{N(1-x-E')}{2}\right)!\left(\frac{N(1-x+E')}{2}\right)!(xN)!}$$
And $\mathcal N=\frac{Ν!}{\left(\frac{N(1-x-E')}{2}\right)!\left(\frac{N(1-x+E')}{2}\right)!(xN)!}$
Now using Stirling's approximation as follows: $y!\approx \sqrt{2πy}(\frac ye)^y$ and if we set $1-x-E'=A$,$1-x+E'=B$ after some simplifications, we get:
$$\mathcal N=\frac{1}{N\sqrt{ABx}}(\frac{1}{2A^AB^Bx^x})^N$$
$N$ is big enough for the first fraction to be insignificant. Finally we get $\mathcal N= (f(x))^N$ where $f(x)$ is the second fraction of the previous product.
Now we turn our attention back to $Ω$. We can approximate the sum with an integral as follows:
$$Ω=\int_0^{N(1-E')}(g(x))^Ndx=\int_0^{N(1-E')}e^{Nln(g(x))}dx$$ $$(1)$$
We now use Watson's lemma: $\int_0^ne^{ng(x)}dx=\sqrt{\frac{2π}{n|g''(x_0)|}}e^{ng(x_0)}$ $(2)$
we want to find $x_0$ for which $g(x)=ln(f(x))$ has a maximum, which means that we need $\frac {dg}{dx}=0$ and $\frac {d^2g}{dx^2}<0$ .
Provided that I didn't make any algebraic mistakes, $x=\frac{3+\sqrt{5+4E'^2}}{2}$. And substituting that into $(1)$ using $(2)$ we get the desired result.
If anyone has a better suggestion, please let me know.
