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

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 $\beta = -1$?

-
Function names like $\cos$ and $\ln$ are interpreted as strings of letters representing variables and thus get italicized if you write them like that. To get the proper font and spacing for them, you can use the predefined commands like \cos and \ln to get $\cos$ and $\ln$, respectively, or, if you need a function for which there's no predefined command, \operatorname{name} to get $\operatorname{name}$. – joriki Oct 21 '12 at 18:18
Note that you can get rid of $\phi$ by transforming to $\psi=\theta-\phi$. – joriki Oct 21 '12 at 18:20
I found the bad typesetting so distracting that my brain refused to consider the mathematical content. Fixed. – Harald Hanche-Olsen Oct 21 '12 at 18:25
The estimate is certainly wrong when $\beta>0$, since the integral approaches a positive limit when $r\to1$ in that case. From your final question, I gather that you are primarily interested in the case $\beta<0$; but it might be worth a mention anyhow. – Harald Hanche-Olsen Oct 21 '12 at 22:23

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.)

-
Dear Prof. Olsen: Thanks for your response. Could you explain why the second term in your estimate, i.e. $\frac{(2ar)^{\beta/2}. {\pi}^{\beta+1}}{\beta+1}$ should imply an estimate of the form $K.(1-r)^{1+\beta}$, if $r$ is close to $1$,i.e. $(1-r)$ is small? – Mathmath Oct 23 '12 at 14:59
That term is bounded, while $(1-r)^{1+\beta}\to\infty$ when $r\to1$. Recall that this is for the case $\beta<-1$ only. – Harald Hanche-Olsen Oct 23 '12 at 18:16
My mistake, thanks. Your estimate is also true for $\beta=-1$, I checked. – Mathmath Oct 24 '12 at 3:07

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 $|r|<1$.

-
Robert Israel: does the maple give any kind of estimate ? And what are the symbols $F_1([\beta/2,\beta/2],[1],r^2)$ ? Could you explain ? – Mathmath Oct 21 '12 at 20:20
${}_2F_1$ is the hypergeometric function defined by that power series. See en.wikipedia.org/wiki/Hypergeometric_function – Robert Israel Oct 22 '12 at 15:09