I ran into an expression calculating the expected value of $\exp(i t \sigma)$ where $\sigma$ is the total number of cycles in a uniformly chosen $S\_n$ element. The expression is
$$E\_n (\exp(i t \sigma)) = \Gamma(n + \exp(it)) / (\Gamma(\exp(it)) n!)$$
where $E\_n$ denotes the expectation under the uniform distribution on $S\_n$. The paper then claims that using Binet's form of Stirling approximation one can get
$$E\_n (\exp(it \sigma)) = n^{\exp(it) -1}/\Gamma(\exp(it)) (1 + o(1))$$


Then here comes the derivation I cannot understand:
using the last expression, they claim one gets the following central limit theorem
$$\lim\_{n \to \infty} E\_n(\exp(it (\sigma - \log n)/\sqrt{\log n})) = \exp( -1/2 t^2)$$
for any real $t$.


I would highly appreciate anyone who can tell me why this is true. It appears to be related to some property of the Gamma function over the complex number.
The relevant paper is Shepp and Lloyd: Ordered cycle lengths in a random permutation.



Generalized binomial coefficients and Gaussian density
Note that $exp(it) = 1 + it - t^2/2 + O(t^3)$ uniformly in $t \in \mathbb{R}$. Thus $n^{exp(it)-1} = exp(it \cdot \log n - \log n \cdot t^{2}/2 + O(t^3 \cdot \log n))$ and also by Taylor's theorem $1/\Gamma(exp(it)) = 1 + O(t)$ when $t$ is small (but in fact also for all real $t \in \mathbb{R}$ by periodicity). Thus $$n^{exp(it)-1}/\Gamma(exp(it)) = exp(it \cdot \log n - \log n \cdot t^2/2 +O(t^3 \cdot \log n))$$
Multiplying both sides by $exp(- it \cdot \log n)$ and substituting $t := t \cdot (\log n)^{-1/2}$ we obtain as $n \rightarrow \infty$ the desired limit, $exp(-t^2/2)$.


