I've been reading Imayoshi & Taniguchi's text "An Introduction to Teichmuller Spaces" and while going through the proof of Lemma 4.20 realized that I'm not comfortable integrating functions over the complex plane (i.e. over an area form, not a contour integral.)
In the proof they state that for $1<p<2,$ the function $$\frac{\zeta}{z(z-\zeta)}\in L^p(\mathbb{C})$$ where $z$ is our variable and $\zeta=a+ib$ is a fixed complex number. I want to verify this claim, my attempt is as follows: $$\int \int_{\mathbb{C}}\left\lvert \frac{\zeta}{z(z-\zeta)} \right\rvert^p dxdy =\lvert\zeta\rvert^p \int\int_{\mathbb{C}} \frac{dxdy}{\lvert x+iy\rvert^p \lvert x+iy-(a+ib)\rvert^p}$$ $$=\lvert\zeta\rvert^p\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{dxdy}{(x^2+y^2)^{p/2} ((x-a)^2+(y-b)^2)^{p/2}}.$$
My puestions are: 1) Is what I've written above correct so far? 2) Any ideas on how to proceed or a better way to justify that this function is in fact in $L^p(\mathbb{C})$?
Really, if all you're doing is integrating the modulus of a function, then you just need to check an integral over $\mathbb R^2$ written in a funny way. First prove that $1/z \in L^q_{loc}$ for $q<2$. This follows from the fact that $1/|x|\in L^q_{loc}(\mathbb R^n)$ for $q<n$, which can be seen by using radial coordinates around $0$. This implies that $\frac1{z(z-\zeta)} \in L^q_{loc}$. To finish, notice that outside of a ball $B$ so large that it contains both $0$ and $\zeta$, you should be able to prove you have sufficient decay at infinity for the function to be integrable.