Im following the book "Analytic Combinatorics" by Flajolet and Sedgwick. I'm having trouble understanding the last part of the proof of the theorem regarding the standard function scale. In particular im confused on how to bound the right part of the integral following a Hankel contour. Namely I do not see how to show the following
\begin{equation} \int_{\mathcal{H}} (-t)^{-\alpha} (1 + \frac{t}{n})^{-n-1} dt = \int_{\mathcal{H}}(-t)^{-\alpha}e^{-t} dt \, \big(1 + O(1/n)\big) \end{equation}
The theorem in question is Theorem VI. 1. I add at the end the relevant pages of the book in case anyone wants to look at the original source.
Here is how the statement of the theorem and the argument goes as explained by my own understanding
Theorem:[Standard Function Scale]. Let $\alpha$ be an arbitrary complex number in $\mathbb{C} \setminus \mathbb{Z}_{\leq 0}$. The coefficient of $z^n$ in \begin{equation} f(z) = (1-z)^{-\alpha} \end{equation}
admits for large $n$ a complete asymptotic expansion in descending powers of $n$
\begin{equation} [z^n]f(z) \sim \frac{n^{\alpha -1}}{\Gamma{(\alpha)}}(1 + \sum_{k=1}^\infty \frac{e_k}{n^k}) \end{equation}
where $e_k$ is a polinomial on $\alpha$ of degree $2k$. In the sequel, we shall restrict ourselves in expansions of order of $\frac{1}{n}$. In particular we often use the result \begin{equation} [z^n]f(z) \sim \frac{n^{\alpha -1}}{\Gamma{(\alpha)}}(1 + O(\frac{1}{n})) \end{equation}
Proof:
Our objective is to derive an asymptotic expression for the coefficients of the function
\begin{equation} f(z) = (1 - z)^{\alpha} \end{equation}
where $\alpha \in \mathbb{C} \setminus \mathbb{Z}_{\geq 0}$. Note that $f(z)$ is singular along the strip $\{|z| \geq 1, \arg(z) = 0\}$. We start with the Cauchy Formula for our coefficients $f_n$.
\begin{equation} f_n = \frac{1}{2\pi i} \int_C \frac{f(z)}{z^{n+1}} dz \label{eq: CF standard scale} \end{equation}
Where $C$ is for example a circumference of radius $1/2$ around the origin. We deform $C$ to a contour $C'$ given by an outer open circle of radius $R> 1$ and an inner semicircle centered at $1$ of radius $\frac{1}{n}$ and two rectilinear paths that are paral.lel to the real axis joining the circles. We note that the in the contribution of the outer circle, the integrand decreases as $R^{-n-1}$, hence its contribution goes to $0$ as $R\to \infty$. Hence we can deform $C'$ into a path $\gamma = \gamma_1 \cup \gamma_2 \cup \gamma_3$ that comes from infinity at distance $1/n$ from the real axis, winds clockwise around $1$ and goes back to infinity. This is properly defined below.
Thus, we need to look at the contribution of the integrals over the following integration curves
\begin{align} \gamma_1 = \{z : z = w - \frac{i}{n}, w\geq 1\} \\ \gamma_2 = \{z: |z-1| = 1/n, \arg(z) \in [-\pi/2 , \pi/2] \}\\ \gamma_3 = \{z: z = w + \frac{i}{n}, w \geq 1 \} \end{align}
Performing the change of variable $z = 1 + \frac{t}{n}$ we get from (\ref{eq: CF standard scale})
\begin{equation} f_n = \frac{n^{\alpha -1}}{2\pi i} \int_{\mathcal{H}} (-t)^{-\alpha} (1 + \frac{t}{n})^{-n-1} dt \end{equation}
Where the countor of integration is now $\mathcal{H} = n(\gamma -1)$, so a Hankel countour that winds around $0$ and at distance $1$ from the real line. We note that the integrand $(-t)^{-\alpha} (1 + \frac{t}{n})^{-n-1}$ converges point-wise to $-t^{-\alpha}e^{-t}$ as $n \to \infty$. Thus if we can take the limit on the of the integrand and moreover ensure that the convergence is uniform, that is it is not dependant on $t$ we would be very close to the result! This can be achieved by expanding the expression $(1 + \frac{t}{n}) = e^{-(n+1)\log(1 + \frac{t}{n})}$ and bounding the the contributions of the integrand for along $[T, \infty)$ for some big enough $T$.
By expanding $e^{u}$ and $-n\log(1 + \frac{t}{n})$ one gets,
\begin{align} -n\log(1 + \frac{t}{n}) = -t + \frac{t^2}{2n} - \frac{t^3}{3n^2} - \dots \\ e^{u} = 1 + u + \frac{u^2}{2} + \frac{u^3}{3!} + \dots \end{align}
Thus, for fixed $t$ one has
\begin{equation} (1 + \frac{t}{n})^{-n -1 }= e^{- (n+1) log( 1 + \frac{t}{n})} = e^{-t} [1 + \frac{t^2 - 2t}{2n} + \frac{3t^4 - 20 t^3 +24 t^2}{24 n^2} + \dots ] \label{eq: as e} \end{equation}
In fact we have the expression
\begin{equation} (1 + \frac{t}{n})^{-n} = e^{-t} ( \sum_k \frac{p_{2k}(t)}{n^k}) \end{equation}
where $p_{2k}(t)$ is a polinomial on $t$ of oder $2k$.
As we hinted before, this convergence becomes uniform for any bounded $t-$domain, and these considerations lead us to make the following claim
Claim:
\begin{equation} \int_{\mathcal{H}} (-t)^{-\alpha} (1 + \frac{t}{n})^{-n-1} dt = \int_{\mathcal{H}}(-t)^{-\alpha}e^{-t} dt \, \big(1 + O(1/n)\big) \end{equation}
We can see this by first splitting the integration contour in $Re(t) > log^2(n)$ and $Re(t) \leq log^2(n)$.
Next we check that the part $Re(t) > log^2(n)$ is negligible on the scale we are working on our problem. Note that for any $K$
\begin{equation} \int_{\log^2(n)}^\infty e^{-t}t^{K} dt = O(e^{-log(n)^2}) \end{equation}
THIS IS WHERE MY QUESTION ARISES
Thus,
\begin{equation} \begin{split} \int_{\log(n^2)}^\infty t^{-\alpha} e^{-t} (\sum_{k=0}^K \frac{p_{2k}(t)}{n^k})) dt = O(e^{-\log(n)^2}) \end{split} \end{equation}
for any $K$, which I guess leads to
\begin{equation} \int_{\log^2(n)} ^\infty (-t)^{-\alpha} (1 + \frac{t}{n})^{-n} dt = O(e^{-\log^2(n)}) \end{equation}
but i dont really know how to justify. On the other hand one has
\begin{equation} \int_{\log(n^2)}^{(0)} t^{-\alpha} (1 + \frac{t}{n})^{-n} dt \end{equation}
where now we can expand as in (\ref{eq: as e}) with uniform error terms since $t = \log(n)^2/n$ is small. Thus justifying term by temr integration. We then get
\begin{equation} \int_{\mathcal{H}} (-t)^{-\alpha} (1 + \frac{t}{n})^{-n-1} dt = \sum_k \int_{\log(n^2)} ^{(0)} (-t)^{-\alpha} \frac{p_{2k}(t)}{n^{k}} + O(e^{-\log^2(n)}) \end{equation}
Note that the term $O(e^{-\log^2(n)})$ becomes irrelevant in front of any polynom on $n$. Finally, each term of the form $\frac{t^r}{n^s}$ induces by Hankel's formula a term of the form
\begin{equation} \frac{n^{-s}}{\Gamma(\alpha - r)} = \frac{n^{-s}}{\Gamma(\alpha)} (\alpha -1) \dots (\alpha - r) \end{equation}
and the statement of the theorem eventually follows.
