Estimate on a simple-looking integral arising from harmonic analysis/harmonic extensions

Question

Let $z\in \mathbb{D}, t\in S^1, \beta\in \mathbb{R}$. I was dealing with the following integral arising from some other calculation regarding harmonic extension on $\mathbb{D}$:

$$I(z)=\int_{S^1}|t-z|^{\beta}|dt|= \int_0^{2\pi}\bigl({1+r^2-2r\cos(\theta-\phi)}\bigr)^{\beta/2}d\theta,$$ $t=e^{i\theta}$, $z= re^{i\phi}$, $|dt| $ denote the arc length measure on $S^1$.

My question is: 1) can we, at best, evaluate this integral?

Or if not, 2) can we get $I(z)\leq K(1-|z|)^{1+\beta}$, $\beta \ne -1$, and $I(z)\leq -K\ln(1-|z|)$ if $\beta = -1$?

Answer 1

It looks to me (with Maple's help) like $$I(z) = 2 \pi \sum_{k=0}^\infty \frac{\Gamma(k-\beta/2)^2}{(k!)^2 \Gamma(-\beta/2)^2} r^{2k} = 2 \pi\; {}_2F_1([-\beta/2,-\beta/2],[1],r^2)$$ for $|r|<1$.

Answer 2

You can have such an estimate when $\beta<-1$, at least. Consider the slightly rewritten form $$I(r)=\int_{-\pi}^\pi\bigl((1-r)^2+2r(1-\cos\theta)\bigr)^{\beta/2}\,d\theta.$$ The integral is bounded except when $r\to1$, so there is no need to consider small $r$. On the given interval we can find some $a>0$ with $1-\cos\theta\ge a\theta^2$. Thus $$\begin{aligned}\tfrac12I(r)&<\int_0^{1-r}(1-r)^\beta\,d\theta+\int_{1-r}^\pi(2ar)^{\beta/2}\theta^\beta\,d\theta \\&=(1-r)^{\beta+1}+\frac{(2ar)^{\beta/2}}{\beta+1}\bigl(\pi^{\beta+1}-(1-r)^{\beta+1}\bigr), \end{aligned}$$ which implies an estimate on the desired form. I think you can mimic this for $\beta=-1$ as well, whereas it fails for $\beta>-1$. When $\beta>0$, we already know that the estimate is wrong (see my comment above).

Update: The estimate is wrong for all $\beta>-1$, since the integral converges to a positive value as $r\to1$ in that case – while the right hand side of the estimate goes to zero. When $-1<\beta<0$ you need the dominated convergence theorem to see this. (For fixed $\theta$, the integrand is maximal when $r=\cos\theta$. Substituting that in the integral results in a finite integral.)