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,

  • $\begingroup$ For small integer $n$ Mma gives $I_n(\infty,\epsilon)=C_n \epsilon^{1-n}$. To be expected with singularity at the origin. $\endgroup$
    – Andrew
    Commented Apr 12, 2019 at 16:49
  • $\begingroup$ That's interesting as without the absolute value there is no singularity at all. $\endgroup$
    – Ali
    Commented Apr 12, 2019 at 16:54

1 Answer 1


$\newcommand{\de}{\delta} \newcommand{\ga}{\gamma} \newcommand{\si}{\sigma} \newcommand{\ep}{\epsilon}$ 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 $n\in(0,1)$.

  • $\begingroup$ This does not work out when $n=1$ as you should not get the $\epsilon^{-1}$ asymptotic. $\endgroup$
    – Ali
    Commented Apr 12, 2019 at 19:12
  • $\begingroup$ @Ali : Oops! Previously, by mistake, I put $|\sin u|^n$ into the denominator, rather than the numerator. Now it's only significantly simpler, and I do get your $2\pi$ for $n=1$. $\endgroup$ Commented Apr 12, 2019 at 22:57
  • $\begingroup$ I don't think your solution is correct for the case when $n=\frac{1}{2}$, but it is correct for all the other cases. For the case $n=\frac{1}{2}$ there always is a strong singularity at one of the end points depending on the branch cut that will cancel out the $\epsilon^{\frac{1}{2}}$. $\endgroup$
    – Ali
    Commented Apr 19, 2019 at 15:40
  • 1
    $\begingroup$ @Ali : I have fixed the mistake. In fact, all values of $n\in(0,1)$ are like $n=1/2$. $\endgroup$ Commented Apr 19, 2019 at 21:33
  • $\begingroup$ yes this is what I meant as well. $\endgroup$
    – Ali
    Commented Apr 19, 2019 at 22:08

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