How to estimate the following integral?

Question

I would like to estimate the integral $I_{s,t}$ by a constant $C$ independent of $s, t \in \mathbb S^1$:

$$I_{s,t}=\int_{|r-s|\geq 2|t-s|} \frac{|t-s|}{|r-s|^2} \, dr,$$ where $r, s \, \mbox{and}\, t$ are in the unit circle $\mathbb S^1=\{\zeta\in \mathbb C: \, |\zeta|=1\}$. More precisely, I want to prove that, there exist a constant $C$ independent of $s, t \in \mathbb S^1$ such that $$I_{s,t}=\int_{|r-s|\geq 2|t-s|} \frac{|t-s|}{|r-s|^2} \, dr< C.$$

Thank you in advance

Answer 1

Hint:

If I understand correctly, we can write

$$r=e^{iw},s=e^{iu},t=e^{iv}$$

and the integral is

$$\int_{|r-s|\ge2|t-s|}\frac{|t-s|}{(\cos w-\cos u)^2+(\sin w-\sin u)^2}e^{iw}dw \\=|t-s|\int_{|r-s|\ge2|t-s|}\frac{e^{iw}dw}{2-2\cos w\cos u-2\sin w\sin u} \\=|t-s|\frac{e^{iu}}2\int_{|r-s|\ge2|t-s|}\frac{e^{i(w-u)}dw}{1-\cos(w-u)}.$$

With $z:=w-u$, the indefinite integral has an analytical form

$$\int\frac{\cos z+i\sin z}{1-\cos z}dz=\frac{\sin z+2\sin^2\dfrac z2\left(z-2i\log\sin\dfrac z2\right)}{\cos z-1}.$$

Finally, the integration domain is the portion of the unit circle which is outside a circle centered at $s$, with radius $2|t-s|$. Hence, $w$ varies in a single interval delimited by the intersections of the two circles.

This allows to write the integral in a (complicated) closed-form.

Answer 2

I don't think this is possible. The problem is, if $t\not=s,$ then the integrand is bounded in size, but it's bounded by something that is dependent on $t$ and $s$, which is itself not bounded. Let $D=\{r\in\mathbb{S}^2:|r-s|\ge 2|t-s| \}.$ From the domain $D$ over which we are integrating, we have that \begin{align*} |r-s|&\ge 2|t-s| \\ \frac{1}{|r-s|}&\le\frac{1}{2|t-s|} \\ \frac{1}{|r-s|^2}&\le\frac{1}{4|t-s|^2} \\ \frac{|t-s|}{|r-s|^2}&\le\frac{1}{4|t-s|}. \end{align*} The $ML$ estimate would then say that \begin{align*} \left|\int_D\frac{|t-s|}{|r-s|^2}\,dr\right|&\le \left|\int_D\frac{1}{4|t-s|}\,dr\right| \\ &=\frac{1}{4|t-s|}\left|\int_D\,dr\right| \\ &\le \frac{2\pi}{4|t-s|} \\ &=\frac{\pi}{2|t-s|}. \end{align*} This bound can get quite large if $t$ gets close to $s.$

I'm curious to try a few numerical calculations. Since $r,s,t\in\mathbb{S}^1,$ we can write \begin{align*} r&=e^{i\theta} \\ s&=e^{i\varphi} \\ t&=e^{i\xi}. \end{align*} Then $dr=i\,e^{i\theta}\,d\theta.$ Note a curious fact: this problem is symmetric under any rotation. Moreover, we have that $|r-s|\ge 2|t-s|$ if and only if $|\theta-\varphi|\ge 2|\xi-\varphi|,$ modulo $2\pi$. This is at least true when $t\approx s,$ which is what we're concerned about. So, let us set $\varphi-\xi=0,$ (from rotational symmetry) so that the limits on the $\theta$ integral will be $\xi$ to $2\pi.$ The integral becomes $$\int_D\frac{|t-s|}{|r-s|^2}\,dr=i\int_{\xi}^{2\pi}\frac{|e^{i\xi}-e^{i\varphi}|}{|e^{i\theta}-e^{i\varphi}|^2}\,e^{i\theta}\,d\theta. $$ This we can set up in Mathematica, and the results show, indeed, that the magnitude of this integral can be quite large, depending on how close $\varphi$ is to $\xi$.