Consider a function $h$ holomorphic for $\Re(s)>1-\delta$ (for some $\delta \in (0,1/2)$ fixed), and let $f(s):=\frac{1}{(s-1)^{1-\alpha}}h(s)$ (with $\alpha \in (0,1)$ fixed). Let the power series development of the function $h(s)/s$ in some neighbourhood of $s=1$ be $\sum_{j \geq 0} \alpha_j(s-1)^j$ (where $\alpha_j$ are fixed complex numbers). I want to find an asymptotic development of the function $g: (1, \infty) \longrightarrow \mathbb{C}$ given by $g(x):= \int_\gamma \frac{f(s)x^s}{s^2} ds$ where $\gamma$ is the piecewise differentiable path consisting of the directed segment $\gamma_1$ joining $1-\delta$ to $1$ (in this direction along the 'bottom' of the real axis) and the directed segment $\gamma_2$ from $1$ to $1-\delta$ (in this direction along the 'top' of the real axis, so basically $x^s$ is a multivalued function here, with different branches along $\gamma_1$ and $\gamma_2$, chosen accordingly). The asymptotic development should be of the form $$\frac{x}{\log^\alpha x} \left( \sum_{j=0}^k \frac{\beta_j}{\log^j x} + O\left( \frac{1}{\log^{k+1} x} \right) \right) \hspace{10mm} \cdots \hspace{5mm} (1)$$ and such an estimate should be obtainable for every $k \geq 1$.
My Attempts and Approaches: I have tried looking at integrals of the form $\int_{1-\delta}^{1} e^{ut} (t-1)^{j+\alpha-1} t^{-1} dt$ (where $u>0$), and expanded $(1-t)^{j+\alpha-1}$ binomially. Then I used the general formula for the integral of $v^ne^v$ for integers $n \ge 0$ along with the asymptotic estimate of the exponential integral:
$$\int_{-\infty}^x e^t t^{-1} dt = \Theta \left( \frac{e^x}{x} \left( \sum_{r \geq 0} \frac{r!}{x^r} \right) \right)$$
However, despite doing all of this, I am unable to get the asymptotic estimate in the desired form $(1)$. I would really appreciate some hints or suggestions regarding the same. Thanks a lot.
P.S.: It is also really desirable to find the relation between the sequences $(\alpha_j)_{j \in \mathbb{N}}$ and $(\beta_j)_{j \in \mathbb{N}}$, but of course we'd first need to get to the above form $(1)$.