Consider a real analytic, positive-valued function $f:(1,\infty)\rightarrow\mathbb{R}$ satisfying \begin{equation} f(x)\sim(x-1)^{-\alpha}\qquad(x\to1) \end{equation} for some $\alpha\in(0,1)$. For all $y>0$, consider the integral \begin{equation} I(y)=\int_1^{\infty}\frac{f(x)}{x-1-iy}\,\mathrm{d}x, \end{equation} supposing that $f(x)$ vanishes at $x\to\infty$ fast enough to ensure $I(y)$ to converge. As $y\to0$, numerical evidence suggests me that $I(y)$ diverges like this: \begin{equation} I(y)\sim y^{-\alpha}\qquad(y\to0), \end{equation} i.e. \begin{equation} \lim_{y\to0}y^\alpha I(y)\text{ is finite, } \end{equation} but I am currently unable to prove it, nor to find sufficient additional conditions for this to hold.
My attempt: for every $\epsilon>0$, we can split the integral as such: \begin{equation} y^\alpha I(y)=\int_1^{1+\epsilon}\frac{y^\alpha f(x)}{x-1-iy}\,\mathrm{d}x+\int_{1+\epsilon}^{\infty}\frac{y^\alpha f(x)}{x-1-iy}\,\mathrm{d}x. \end{equation} When taking the limit, the second term in the right-hand side should converge to zero; as for the first term, heuristically speaking, when taking $\epsilon$ "sufficiently small" we should be able to "approximate" it by substituting $f(x)$ with $(x-1)^{-\alpha}$, and thus its asymptotic behaviour should be close to the one of the integral \begin{equation} \int_1^{1+\epsilon}\frac{y^\alpha (x-1)^{-\alpha}}{x-1-iy}\,\mathrm{d}x, \end{equation} which can be computed exactly and does converge to a finite value when $y\to0$. However, this is far from being an actual proof.
OK, that was easier than I thought. I will write down here my procedure for completeness.
Without loss of generality, set $f(x)=g(x)/(x-1)^\alpha$ for some real analytic function $g$ in $(1,\infty)$ such that $\lim_{\lambda\to1^+}g(x)\equiv g(1)$ is finite. Then \begin{equation} y^\alpha I(y)=\int_1^\infty\frac{g(x)}{(x-1)^\alpha}\frac{y^\alpha}{x-1+iy}\,\mathrm{d}x. \end{equation} Now, performing the change of variable $x'=(x-1)/y$, we obtain \begin{equation} y^\alpha I(y)=\int_0^\infty\frac{g(xy+1)}{x^\alpha}\frac{1}{x+i}\,\mathrm{d}x. \end{equation} On the other hand, we have \begin{equation} \int_0^\infty\frac{x^{-\alpha}}{x+i}\,\mathrm{d}x=\pi(-i)^\alpha\csc\pi\alpha, \end{equation} thus, by dominated convergence, \begin{equation} \lim_{y\to0}y^\alpha I(y)=g(1)\pi(-i)^\alpha\csc\pi\alpha. \end{equation}