Integral of a function.

Question

If $$I(x)=(1-|x|^2)^s \int_{|y|>1} \frac{dy}{(|y|^2-1)^s|y-x|}$$ where $x\in (-1, 1)$ and $s\in (0,1).$ Is it true that $I(x)$ is a constant depending on $s$ only?

Answer 1

$$\int_{|y|>1}\frac{dy}{(|y|^2-1)^s|y-x|}=\int_{1}^{+\infty}\frac{dy}{(y^2-1)^s(y-x)}+\int_{1}^{+\infty}\frac{dy}{(y^2-1)^s(y+x)}\,dy $$ equals $$ \int_{1}^{+\infty}\frac{2y\,dy}{(y^2-1)^s(y^2-x^2)}=\int_{1}^{+\infty}\frac{dt}{(t-1)^s(t-x^2)}=\int_{0}^{+\infty}\frac{du}{u^s(u+(1-x^2))} $$ or, by setting $u=(1-x^2)v$, $$ (1-x^2)\int_{0}^{+\infty}\frac{dv}{v^s(v+1)(1-x^2)^{s+1}}=\frac{C_s}{(1-x^2)^s}. $$ It is also interesting to point out that by Euler's Beta function we have $C_s=\frac{\pi}{\sin(\pi s)}$.