Asymptotics of $\int_0^{2\pi} {\rm d}t \, e^{ikt} \left(1+e^{-it}\right)^{1/2+n\left(1+e^{it}\right)}$ Sort of short question, but I'm just interested in the asymptotics of the following integral $$\int_0^{2\pi} {\rm d}t \, e^{ikt} \left(1+e^{-it}\right)^{1/2+n\left(1+e^{it}\right)}$$
that is for $n\rightarrow \infty$ and $k\in\{1,2,...,n\}$. Any idea how to estimate for large $n$?
Background: From the Stirling-Expansion of the rising factorial $$(x+1)(x+2)\cdots(x+n)=\sum_{k=0}^n {n+1 \brack {k+1}} x^k \tag{1}$$ the coefficient of $x^k$ is actually the sum of all products of $(n-k)$ distinct integers in $\{1,2,3,...,n\}$. So I'm not sure how to proof it, but I figured I can write the coefficient of $x^k$ by $$\frac{1}{2\pi i} \oint_C x^{-k} \, \frac{\Gamma(x+n+1)}{\Gamma(x+1)} \, \frac{{\rm d}x}{x}$$ where I expressed the LHS of (1) by the ratio of Gamma functions. $C$ is the circle around $0$ with some arbitrary radius. While I was only interested in the large $n$ behaviour, I thought choosing $x=ne^{it}$ with $t\in(-\pi,\pi)$ is beneficial. Then by the Stirling approximation for the Gamma-function I eventually arrived at $$\frac{n^{n-k} e^{-n}}{2\pi} \int_{-\pi}^\pi e^{it(n-k)} \left(1+e^{-it}\right)^{1/2+n\left(1+e^{it}\right)} \, {\rm d}t \, .$$
 A: Obviously $z=e^{it}$ is the most suitable substitution.
It gives
$$\mathcal I_{k,n}:=\int_0^{2\pi} e^{ikt} \left(1+e^{-it}\right)^{1/2+n\left(1+e^{it}\right)}dt=-i\oint_{|z|=1}z^{k-1} \underbrace{\left(1+\frac1z\right)^{1/2+n+nz}}_{h(z)}dz$$
Let's study $h(z)$'s analytic behaviour.
$$\begin{align}
\ln h(z)
&=\left(\frac12+n+ nz\right)\ln\left(1+\frac1z\right) \\
&=-\left(\frac12+n+nz\right)\sum^\infty_{j=1}\frac{(-z)^{-j}}{j} \qquad\text{for }|z|>1\\
&=-\sum^\infty_{j=1}\frac{(1/2+n)(-z)^{-j}}{j}+\sum^\infty_{j=1}\frac{n(-z)^{-j+1}}{j}\\
&=-\sum^\infty_{j=1}\frac{(1/2+n)(-z)^{-j}}{j}+\sum^\infty_{j=0}\frac{n(-z)^{-j}}{j+1}\\
&=n+\sum^\infty_{j=1}\left(-\frac{1+2n}{2j}+\frac{n}{j+1}\right)(-z)^{-j} \\
&=n-\sum^\infty_{j=1}\frac{j+2n+1}{2j^2+2j}(-z)^{-j} \\
h(z)&=e^n\cdot\exp\left(-\sum^\infty_{j=1}\frac{j+2n+1}{2j^2+2j}(-z)^{-j}\right)
\end{align}
$$
Therefore, $h(z)$ is analytic on $|z|>1$. Hence, by residue theorem
$$\begin{align}
-i\oint_{|z|=1}z^{k-1}h(z)dz
&=-i\cdot-2\pi i\operatorname{Res}_{\infty}z^{k-1}h(z) \\
&=-2\pi \operatorname*{Res}_{z=0}\frac{-1}{z^2}z^{1-k}h\left(\frac1z\right) \\
&=2\pi e^n\cdot [z^k]\exp\left(-\sum^\infty_{j=1}\frac{j+2n+1}{2j^2+2j}(-z)^{j}\right) \\
\mathcal I_{k,n}
&=2\pi e^n\cdot [z^k]\exp\left(-\sum^k_{j=1}\frac{j+2n+1}{2j^2+2j}(-z)^{j}\right)
\end{align}
$$
Notice that up until this point we did not make any approximations. I have no idea how to further simplify it. However, for small $k$, the calculations are managable by hands. For instance,
$$\mathcal I_{1,n}=\pi e^n(n+1)\sim \pi ne^n$$
$$\mathcal I_{2,n}=\frac{\pi e^n}{12}(3n^2+2n-3)\sim \frac{\pi}4 n^2 e^n$$
$$\mathcal I_{3,n}=\frac{\pi e^n}{24}(n^3-n^2-3n+3)\sim\frac{\pi}{24} n^3 e^n$$
In general, it can be proved that
$$\color{red}{\mathcal I_{k,n}\sim\frac{\pi}{2^{k-1}k!}n^k e^n}$$
Derivation:
Let $\displaystyle{c_j=(-1)^{j+1}\frac{j+2n+1}{2j^2+2j}}$.
Then,
$$\begin{align}
\mu_k :=[z^k]\exp\left(-\sum^k_{j=1}\frac{j+2n+1}{2j^2+2j}(-z)^{j}\right)
&=[z^k]\exp\left(\sum^k_{j=1}c_j z^{j}\right) \\
&=[z^k]\prod^k_{j=1}\exp(c_j z^j) \\
&=[z^k]\prod^k_{j=1}\sum^\infty_{m=0}\frac{c_j^m}{m!} z^{mj} \\
\end{align}
$$
To collect the coefficient of $z^k$, we need to find all the non-negative solutions $(m_1,m_2,\cdots,m_k)$ to the diophantine equation
$$k=1m_1+2m_2+3m_3+\cdots+km_k$$
Once all the solutions (call the solution set, which is a set of $k$-tuples, $M$) are found, we get 
$$\mu_k=\sum_M \frac{c_1^{m_1}}{m_1!}\frac{c_2^{m_2}}{m_2!}\cdots\frac{c_k^{m_k}}{m_k!}$$
Obviously, $\mu_k$ is a polynomial of $n$, and we are only interested in the coefficient of the leading monomial term.
As $c_{(\cdot)}=O(n)$, a solution $k$-tuple $(m_1,\cdots,m_k)$ would contribute  a $$O(n^{m_1})O(n^{m_2})\cdots O(n^{m_k})=O(n^{m_1+\cdots+m_k})$$ term to $\mu_k$.
Clearly, the index maximizes when and only when $m_1=k$ and $m_2=\cdots=m_k=0$. Consequently, the coefficient of the leading monomial term is
$$[n^k]\mu_k=[n^k]\frac{c_1^k}{k!}=[n^k]\left(\frac{2+2n}{4}\right)^k\cdot\frac1{k!}=\frac1{2^k k!}$$
As a result, for fixed $k$, as $n\to\infty$
$$\mathcal I_{k,n}=2\pi e^n\mu_k\sim 2\pi e^n\cdot\frac{n^k}{2^k k!}=\frac{\pi}{2^{k-1}k!}n^k e^n\qquad \blacksquare$$
