A function in the intersectino of Hardy spaces for $p < 1$ I'm working on an exercise from Peter Duren's Theory of $H^p$ Spaces. The question is:

Show that $(1-z)^{-1}$ is in $H^p$ for every $p < 1$, but not in $H^1$.

I have been able to show the result for $H^1$ but for $p<1$, I'm stuck. The definition of $H^p$ space is rather difficult to work with (at least for me).
I would like to show that $$\sup\limits_{0 < r < 1}\left(\frac{1}{2\pi}\int_0^{2\pi}\frac{dt}{|1-re^{it}|^p}\right)^{\frac{1}{p}}=:M_p ~<~ \infty.$$

Note that $|1-re^{it}|^2=1-2r\cos t+r^2$, so, in the integral, after the substitutions $s=t-\pi$ and $t=\tan s/2$, we are reduced to find an uniform bound in $r$ of 
$$I_r:=\int_{-\infty}^{+\infty}\frac 1{\left(1+2r\frac{1-t^2}{1+t^2}+r^2\right)^{p/2}}\frac{dt}{1+t^2}.$$
Note that 
\begin{align}
1+2r\frac{1-t^2}{1+t^2}+r^2&=\left(r+\frac{1-t^2}{1+t^2}\right)^2+1-\left(\frac{1-t^2}{1+t^2}\right)^2\\
&\geqslant \frac{(1+t^2)^2-(1-t^2)^2}{(1+t^2)^2}\\
&=\frac{4t^2}{(1+t^2)^2}\\
&\overset{\infty}{\sim}4t^{-2}.
\end{align}
