Consider the following integral
$$
\int\_0^{\infty}\frac{e^{-x}-1}{x^{2+\frac{A}{\log b-5/6}}}\frac{1}{\log(b/x)-i\pi/2}\,dx
$$
where $A>0$ and $b>0$. I am interested in the small $b$ asymptotics of this integral. Any ideas on how to proceed?



How to get this integral's asymptotics?
The integral in question is 
\begin{equation\*}
I:=\int\_0^\infty f(x)\,dx, \tag{1} 
\end{equation\*}
where 
\begin{equation\*}
 f(x):=\frac{e^{-x}-1}{x^{2-\varepsilon}}\frac{1}{\ln(b/x)-i\pi/2},\quad
 \varepsilon:=-\frac{A}{\ln b-5/6}\downarrow0. 
\end{equation\*}
Let 
\begin{equation\*}
 C:=C\_\varepsilon:=A-\tfrac56\,\varepsilon, 
\end{equation\*}
so that 
\begin{equation\*}
 b=e^{-C/\varepsilon}. 
\end{equation\*}
We have 
\begin{equation\*}
 \int\_{b^{1/2}}^\infty f(x)\,dx
 \sim\int\_{b^{1/2}}^\infty \frac{e^{-x}-1}{x^{2-\varepsilon}}\frac{dx}{\ln(b/x)}
 =\frac{J\_1+J\_2}{b^{1-\varepsilon}}, 
\end{equation\*}
where
\begin{equation\*}
 J\_1:=\int\_{\varepsilon^{1/2}/b}^\infty \frac{1-e^{-by}}{y^{2-\varepsilon}}\frac{dy}{\ln y}
 \le\int\_{\varepsilon^{1/2}/b}^\infty \frac{dy}{y^{2-\varepsilon}}\frac{1}{\ln(\varepsilon^{1/2}/b)}
 \ll\varepsilon^{1/2}b^{1-\varepsilon}=o(b^{1-\varepsilon}) 
\end{equation\*}
and, with $u=\varepsilon\ln y$, 
\begin{equation\*}
 J\_2:=\int\_{b^{-1/2}}^{\varepsilon^{1/2}/b} \frac{1-e^{-by}}{y^{2-\varepsilon}}\frac{dy}{\ln y}
 \sim b \int\_{b^{-1/2}}^{\varepsilon^{1/2}/b} \frac{dy}{y^{1-\varepsilon}}\frac{1}{\ln y}
 = b \int\_{C/2}^{C+\varepsilon\ln(\varepsilon^{1/2})}\frac{du}u\,e^u
 \sim b(\text{Ei}(A)-\text{Ei}(A/2)). 
\end{equation\*}  

It follows that 
\begin{equation\*}
 \int\_{b^{1/2}}^\infty f(x)\,dx
 \to e^{-A}[\text{Ei}(A)-\text{Ei}(A/2)]. \tag{2}
\end{equation\*} 
Also, for 
\begin{equation\*}
 g(x):=-\frac1{x^{1-\varepsilon}}\frac{1}{\ln(b/x)-i\pi/2}, 
\end{equation\*}
we have 
\begin{equation\*}
 \int\_0^{b^{1/2}} |f(x)-g(x)|\,dx
 \ll\int\_0^{b^{1/2}} \frac{x^2}{x^{2-\varepsilon}}\frac{dx}{\pi/2}\to0. \tag{3}
\end{equation\*}
Then, using the substitution $t=\varepsilon\ln(b/x)$, we have 
\begin{equation\*}
 \int\_0^{b^{3/2}} g(x)\,dx\sim
 -\int\_0^{b^{3/2}}\frac{dx}{x^{1-\varepsilon}}\frac{1}{\ln(b/x)}
 =-e^{-C}\int\_{C/2}^\infty \frac{dt}t\,e^{-t}=-e^{-C}\text{E}\_1(C/2)\to
 -e^{-A}\text{E}\_1(A/2). \tag{4}
\end{equation\*}
Further, 
\begin{equation\*}
 \int\_{b^{3/2}}^{b^{1/2}} g(x)\,dx=
 \int\_{b^{3/2}}^b g(x)\,dx+\int\_b^{b^{1/2}} g(x)\,dx
 =\int\_{b^{1/2}}^1 g(bu)\,b\,du+\int\_{b^{1/2}}^1 g(b/u)\,\frac{b\,du}{u^2}
 =b\times\int\_{b^{1/2}}^1 h(u)\,du, 
\end{equation\*}
where 
\begin{equation\*}
 h(u):=g(bu)+g(b/u)/{u^2}
 =-2i e^{C(1/\varepsilon-1)}e^t\,\frac{2\pi\cosh(\varepsilon t)+4i t \sinh(\varepsilon t)}{\pi^2+4t^2} 
\end{equation\*}
and $t:=-\ln u$. 
(The latter identity is the key: note that $\Re g(x)$, as well as $\Re f(x)$, changes sign at $x=b$.)
So,
\begin{equation\*}
 \int\_{b^{3/2}}^{b^{1/2}} g(x)\,dx=
 -2i b e^{C(1/\varepsilon-1)}\int\_0^{C/(2\varepsilon)}dt\,\frac{2\pi\cosh(\varepsilon t)+4i t \sinh(\varepsilon t)}{\pi^2+4t^2}. \tag{5} 
\end{equation\*}
Next, 
\begin{equation\*}
 \int\_0^{C/(2\varepsilon)}dt\,\frac{2\pi\cosh(\varepsilon t)}{\pi^2+4t^2}
 \to\int\_0^\infty dt\,\frac{2\pi}{\pi^2+4t^2}=\frac\pi2 \tag{6}
\end{equation\*}
and 
\begin{equation\*}
 \int\_0^{C/(2\varepsilon)}dt\,\frac{4i t \sinh(\varepsilon t)}{\pi^2+4t^2}
 =\int\_0^{C/2}dz\,\frac{4i z \sinh z}{\varepsilon^2\pi^2+4z^2}
 \to\int\_0^{A/2}dz\,\frac{4i \sinh z}{4z}=i\,\text{shi}\frac A2, \tag{7}
\end{equation\*}
where $\text{shi}$ is the hyperbolic sine integral function. 
Collecting the pieces (1)--(7), we have 
\begin{equation\*}
 I\to e^{-A}(\text{Ei}(A)-\text{Ei}(A/2))-e^{-A}\text{E}\_1(A/2)-2i e^{-A} (\tfrac\pi2+i\,\text{shi}\tfrac A2)
 =e^{-A}(\text{Ei}(A)-i\pi), 
\end{equation\*}
which is the same expression as the one obtained by Carlo Beenakker. 


