$\int_0^{2\pi} \frac{rie^{it}}{re^{it}-a}dt$ without Cauchy integral theorem, residues, series.

Question

I have to solve $$\int_0^{2\pi} \frac{rie^{it}}{re^{it}-a}dt.$$ There are many solutions to this but they all use Cauchy's integral formula, or residue theorem which I am not allowed to use. I also don't like the series expansion solutions. Is there a way to directly calculate it? If $r=a=1$ then I can multiply by $ \frac{e^{ \frac{-it}{2} }}{e^{ \frac{-it}{2} }}$ and then by $\frac{e^{it/2}+e^{-it/2}}{e^{it/2}+e^{-it/2}}$ and then the bottom becomes just the $\cos$. But if these values are not nice I get stuck.

Answer 1

Split into two cases: $|a|>1$ and $|a|<1$. In each case one can check that the integral $I(a)$ satisfies $I'(a) = \int_{0}^{2\pi} \frac{ d}{dt} p(t) dt$ for an explicit periodic function $p(t) = (r e^{it} -a)^{-1}$. Thus $I'(a)= p(2\pi)-p(0) =0$ in both cases. This shows that the integral is constant in each of the two cases. Now determine the values of those two constants. In one case you can set $a=0$ to evaluate the constant $I(0)$. In the other case let $|a|\to \infty$ to see $I(\infty)=0$.

Answer 2

$$I=\int_0^{2\pi} \frac{rie^{it}-ia+ia}{re^{it}-a}dt.$$
$$=i(2\pi)+\int_0^{2\pi} \frac{ia}{re^{it}-a}dt.$$
$$=2\pi i+\int_0^{2\pi} \frac{iae^{-it}}{r-ae^{-it}}dt.$$
$$=2\pi i+(\ln ( r-ae^{-it}))_0^{2\pi}$$
$$=\color{red}{2\pi i}$$ for $r>|a|$