3
$\begingroup$

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,

$\endgroup$
  • $\begingroup$ 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 Apr 12 '19 at 16:49
  • $\begingroup$ That's interesting as without the absolute value there is no singularity at all. $\endgroup$ – Ali Apr 12 '19 at 16:54
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ This does not work out when $n=1$ as you should not get the $\epsilon^{-1}$ asymptotic. $\endgroup$ – Ali Apr 12 '19 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 $n=1$. $\endgroup$ – Iosif Pinelis Apr 12 '19 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 Apr 19 '19 at 15:40
  • 1
    $\begingroup$ @Ali : I have fixed the mistake. In fact, all values of $n\in(0,1)$ are like $n=1/2$. $\endgroup$ – Iosif Pinelis Apr 19 '19 at 21:33
  • $\begingroup$ yes this is what I meant as well. $\endgroup$ – Ali Apr 19 '19 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.