How to estimate the following integral? 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
 A: 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 $\xi$. 
A: 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.
