Proving $frac1{2pi} int_0^{2pi} frac{R^2-r^2}{R^2-2Rrcostheta+r^2} dtheta =1$ by integrating $frac{R+z}{z(R-z)}$ without residue theorem.

I would like to thank everyone. Your answers were very useful and helped me to approach the problem in different ways. Unfortunately, none of them was the answer I was expecting to get to.

The final solution I finally came to was this (considering the themes I have studied):

for any $0<r<R$, let $gamma = re^{it}$, then:

$$int_gamma f dz = int_gamma frac1z + frac2{R-z} dz$$ $$Rightarrow int_gamma f dz = int_0^{2pi} frac{ire^{it}}{re^{it}}dt + int_0^{2pi} frac{2ire^{it}}{R-re^{it}}dt$$ $$ = 2pi i + int_0^{2pi} frac{2ire^{it}}{R-re^{it}} dt$$

Since $r<R$, the right function is holomorphic inside $gamma$, which means $2pi i + int_0^{2pi} frac{2ire^{it}}{R-re^{it}} dt = 2pi i$.

So we can make:

$$ 2pi i = int_0^{pi} frac{(R+re^{it})ire^{it}}{(R-re^{it})re^{it}}dt$$ $$ Rightarrow 2pi i = i int_0^{2pi} frac{R+re^{it}}{R-re^{it}} dt \ = i int_0^{2pi} frac{R+rcos t + irsin t}{R-rcos t - irsin t}left( frac{R-rcos t + irsin t}{R-rcos t + irsin t} right) dt \ = iint_0^{2pi} frac{R^2 - r^2+2iRrsin t}{R^2 -2rRcos t + r^2} dt $$

By taking the real and imaginary parts in both $2pi i$ and the integral we get:

$$ 2pi = int_0^{2pi} frac{R^2 - r^2}{R^2 -2rRcos t + r^2} dt \ Rightarrow 1 = frac{1}{2pi} int_0^{2pi} frac{R^2 - r^2}{R^2 -2rRcos t + r^2} dt$$

$$I = frac{1}{pi}int_0^pi frac{R^2-r^2}{R^2+r^2 - 2Rrcos(theta)}:dtheta$$ Use the tangent half-angle substitution to get $$I = frac{2}{pi}int_0^infty frac{(R-r) (R+r)}{t^2 (R+r)^2+(R-r)^2},dt$$ $$t=frac{ (R-r)}{r+R}x implies I=frac{2}{pi}int_0^infty frac{dx}{x^2+1}=frac{2}{pi}frac pi 2=1$$

For each $0<r<1$ $$ P_{r}(x)=frac{1-r^2}{1-2rcos(x)+r^2}. $$ can be expressed as $$ P_r(x)=sum_{ninmathbb{Z}}r^{|n|}text{e}^{ixn}= text{Re}Big( frac{1+z}{1-z}Big)=frac{1-|z|^2}{|1-z|^2}, $$ where $z=rtext{e}^{ix}$ and $|x|leqpi$. Since the trigonometric series is uniformly convergent, the order of summation and integration can be change to get

$$frac{1}{2pi}int^{pi}_{-pi} P_t(x),dx=frac{1}{2pi}sum_{ninmathbb{Z}}r^{|n|}int^{pi}_{-pi}e^{ixn},dx = 1$$

since $frac{1}{2pi}int^{pi}_{-pi}e^{inx},dx=0$ for all $ninmathbb{Z}setminus{0}$.

We could avoid complex analysis altogether

$$I = frac{1}{pi}int_0^pi frac{R^2-r^2}{R^2+r^2 - 2Rrcostheta}:dtheta$$

$$ = frac{1}{pi}int_0^pi frac{R^2-r^2}{R^2+r^2left(cos^2frac{theta}{2}+sin^2frac{theta}{2}right) - 2Rrleft(cos^2frac{theta}{2}-sin^2frac{theta}{2}right)}:dtheta$$

$$frac{1}{pi}int_0^pi frac{(R-r)(R+r)}{(R-r)^2cos^2frac{theta}{2}+(R+r)^2sin^2frac{theta}{2}}:dtheta = frac{2}{pi}int_0^pi frac{left(frac{R-r}{R+r}right)cdotfrac{1}{2}sec^2theta:dtheta}{left(frac{R-r}{R+r}right)^2+tan^2frac{theta}{2}}$$

$$= frac{2}{pi}tan^{-1}left[left(frac{R-r}{R+r}right)tanfrac{theta}{2}right]Biggr|_0^{pi^-} = 1$$

with the assumption that $|R|neq|r|$