Consider the integrals


$$I\_n(\zeta,\epsilon)=\int\_{-\zeta}^\zeta \left|(t-i\epsilon)^{-n}-(t+i\epsilon)^{-n}\right|\,dt$$


I would like to know the asymptotic behavior of $I\_n(\zeta,\epsilon)$ for each fixed $\zeta>0$ as $\epsilon$ approaches zero, and hope that this will be independent of $\zeta$. For example, when $n=1$, it is easy to see that $I\_1(\zeta,\epsilon)$ tends to $2\pi$ as $\epsilon$ tends to $0$ independent of $\zeta$. Specifically, I would like to understand the cases where $n=\frac{1}{2},\frac{3}{2},\ldots$, but the case of integers would also be interesting.
Note that the question is trivial without the absolute value in the integrand but I would like to see how much does the presence of the absolute value changes the asymptotic behavior.
Thanks,



Oscillatory integrals
$\newcommand{\de}{\delta}
\newcommand{\ga}{\gamma}
\newcommand{\si}{\sigma}
\newcommand{\ep}{\epsilon}$
Take any real $n>0$ and any $z:=\zeta\in(0,\infty)$. Let $\ep\downarrow0$. 
For real $t\ne0$, let $u:=\arctan(\ep/t)$, so that $t=\ep\cot u$. 
Then for $t>0$
\begin{equation\*}
 \left|(t-i\ep)^{-n}-(t+i\ep)^{-n}\right|=2(t^2+\ep^2)^{-n/2}|\sin(nu)|
 =2\ep^{-n}|\sin(nu)|\,|\sin u|^n 
\end{equation\*}
and for $t<0$
\begin{equation\*}
 \left|(t-i\ep)^{-n}-(t+i\ep)^{-n}\right|
 =2\ep^{-n}|\sin(n(\pi+u))|\,|\sin u|^n. 
\end{equation\*}
So,  

\begin{equation\*}
 I\_n(z,\ep)=2\ep^{1-n}(J\_n^+(\de)+J\_n^-(\de)), \tag{1}
\end{equation\*}
where  

\begin{equation\*}
 \de:=\arctan\frac\ep z\downarrow0, \tag{1a}
\end{equation\*}
\begin{equation\*}
 J\_n^+(\de):=\int\_\de^{\pi/2}|\sin(nu)|(\sin u)^{n-2}\,du
 \to J\_n^+(0)\in(0,\infty), \tag{2}
\end{equation\*}
and 
\begin{equation\*}
 J\_n^-(\de):=\int\_{-\pi/2}^{-\de}|\sin(n(\pi+u))|\,|\sin u|^{n-2}\,du
=\int\_\de^{\pi/2}|\sin(n(\pi-v))|\,(\sin v)^{n-2}\,dv. 
\end{equation\*}


If $n\ge1$, then $J\_n^-(\de)\to J\_n^-(0)$ and hence 
\begin{equation\*}
 I\_n(z,\ep)\sim2\ep^{1-n}(J\_n^+(0)+J\_n^-(0))
 =2\ep^{1-n}\int\_0^{\pi/2}(|\sin(nu)|+|\sin(n(\pi-u))|)(\sin u)^{n-2}\,du. 
\end{equation\*}
So, 



> 
> $\displaystyle{I\_n(z,\ep)\sim2\ep^{1-n}\int\_0^\pi |\sin(nu)|(\sin u)^{n-2}\,du}$ for $n\ge1$. 
> 
> 
> 


In particular, for $n=1$ we get $I\_n(z,\ep)\to2\pi$. 


It remains to consider the case $0<n<1$. Let $a>0$ vary with $\ep$ in any way such that $a\downarrow0$ and $\ep=o(a)$, so that $\de=o(a)$. 
then 
\begin{equation\*}
 J\_n^-(\de)=\int\_\de^a+\int\_a^{\pi/2}, \tag{3}
\end{equation\*}
where 
\begin{equation\*}
 \int\_\de^a:=\int\_\de^a |\sin(n(\pi-v))|\,(\sin v)^{n-2}\,dv
 \sim\sin(n\pi)\int\_\de^a v^{n-2}\,dv
 \sim\sin(n\pi)\frac{\de^{n-1}}{1-n}, \tag{4}
\end{equation\*}
\begin{equation\*}
 \Big|\int\_a^{\pi/2}\Big|\le\int\_a^{\pi/2}v^{n-2}\,dv=o(\de^{n-1}). \tag{5}
\end{equation\*}
Collecting the pieces (1), (1a), (2), (3), (4), and (5), we see that 



> 
> $\displaystyle{I\_n(z,\ep)\to2z^{1-n}\frac{\sin(n\pi)}{1-n}}$ for $n\in(0,1)$. 
> 
> 
> 


