# Oscillatory integrals

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,

• For small integer $n$ Mma gives $I_n(\infty,\epsilon)=C_n \epsilon^{1-n}$. To be expected with singularity at the origin. – Andrew Apr 12 '19 at 16:49
• That's interesting as without the absolute value there is no singularity at all. – Ali Apr 12 '19 at 16:54

$$\begin{equation*} I_n(z,\ep)=2\ep^{1-n}(J_n^+(\de)+J_n^-(\de)), \tag{1} \end{equation*}$$ 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 $$0. Let $$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 $$n\in(0,1)$$.

• This does not work out when $n=1$ as you should not get the $\epsilon^{-1}$ asymptotic. – Ali Apr 12 '19 at 19:12
• @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 $n=1$. – Iosif Pinelis Apr 12 '19 at 22:57
• 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 $\epsilon^{\frac{1}{2}}$. – Ali Apr 19 '19 at 15:40
• @Ali : I have fixed the mistake. In fact, all values of $n\in(0,1)$ are like $n=1/2$. – Iosif Pinelis Apr 19 '19 at 21:33
• yes this is what I meant as well. – Ali Apr 19 '19 at 22:08