Derive the EGF of Stirling number of the second kind with the OGF-EGF conversion formula

It is well known that, the EGF(Exponential Generating Function) of Stirling number of the second kind is $$\hat G_{k}(x)=\sum_{n=k}^{\infty} S(n, k) \frac{x^{n}}{n !}=\frac{1}{k !}\left(e^{x}-1\right)^{k}$$ And the OGF(Ordinary Generating Function) of Stirling number of the second kind is $$G_k(x) = \sum_{n\ge k} S(n,k) x^n = \frac{x^k}{(1-x)(1-2x)\cdots(1-kx)}$$ The OGF-EGF conversion formula is given here Generating function transformation, that is $$\hat F(z)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }F\left(ze^{-\imath \vartheta }\right)e^{e^{\imath \vartheta }}d\vartheta$$ where, $$\hat F(z)$$ means the EGF and $$F(z)$$ means the OGF.

However, I need to calculate $$\int_{-\pi }^{\pi } \frac{e^{e^{i t}} \left(-e^{-i t} z\right)^{-k} \left(e^{-i t} z\right)^k}{\left(1-\frac{e^{i t}}{z}\right){}_k} \, dt, \text{where } (x)_k \text{ the Pochhammer symbol.}$$ which is really difficult for me.

So my problem is to calculate the integral above.

1 Answer

Let $$\zeta = e^{i \vartheta}$$, then $$\widehat F(z) = \frac 1 {2 \pi i} \int_{|\zeta| = 1} \frac {e^\zeta F {\left( \frac z \zeta \right)}} \zeta d\zeta = \frac 1 {2 \pi i} \int_{|\zeta| = 1} \frac {e^\zeta} {\zeta \prod_{j = 1}^k \left( \frac \zeta z - j \right)} d\zeta, \quad |z| < \frac 1 k.$$ All the poles are inside the unit circle and are simple. The residues of the integrand are $$\operatorname*{Res}_{\zeta = 0} = \frac {(-1)^k} {k!}, \\ \operatorname*{Res}_{\zeta = l z} = \frac {e^{l z}} {l \prod_{j = 1}^{l - 1} (l - j) \prod_{j = l + 1}^k (l - j)} = \frac {e^{l z}} {l (l - 1)! (-1)^{k - l} (k - l)!} = \\ \frac {(-1)^{k - l}} {k!} \binom k l e^{l z}, \quad 1 \leq l \leq k.$$ The sum of the residues is the binomial expansion of $$(e^z - 1)^k/k!$$.