Axially Symmetric Lens Integration

Let $$I(r_0,r;a,b)=int_a^{b}frac{r_0-r cosphi}{r_0^2+r^2-2 r_0 r cosphi}dphi$$

The integrand plots like

Note that

$$ I(r_0,r;0,2pi)=I(r_0,r;0,pi)+I(r_0,r;pi,2pi) $$ Upon the change of varibale $phitophi-pi$ in the second integral above, and then explicitly simplifying, one gets the recursion

$$ begin{align} I(r_0,r;0,2pi)&=r_0int_0^{2pi}frac{r_0^2-r^2 cosphi}{r_0^4+r^4-2 r_0^2 r^2 cosphi}dphi &=r_0~I(r_0^2,r^2;0,2pi) end{align} $$

Therefore $$ begin{align} I(r_0,r;0,2pi)&=lim_{ktoinfty}r_0^{k -1}I(r_0^k,r^k;0,2pi)hspace{1cm} kin1,2,4,8ldots end{align} $$

For $r<r_0$, $$ begin{align} I(r_0,r;0,2pi)&=lim_{ktoinfty}frac{r_0^{k-1} r_0^k}{r_0^{2k}}int_0^{2pi}frac{1-(r/r_0)^kcosphi}{1+(r/r_0)^{2k}-2(r/r_0)^kcosphi}dphi &=frac{2pi}{r_0} end{align} $$

For $r_0<r$, $$ begin{align} I(r_0,r;0,2pi)&=lim_{ktoinfty}frac{r_0^{k-1} r^k}{r^{2k}}int_0^{2pi}frac{(r_0/r)^k-cosphi}{1+(r_0/r)^{2k}-2(r_0/r)^kcosphi}dphi &=lim_{ktoinfty}(r_0/r)^{k-1}ldots &=0 end{align} $$

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