Let $C$ be the unit circle centered at the origin and $a \in \mathbb{R}$.
$$\int_0^{2\pi}\frac{dt}{1 + a^2 – 2a\cos(t)} = \int_C \frac{i\;dz}{(z-a)(az-1)}$$
Use Cauchy's integral formula to deduce if $0 \leq a < 1$ then,
$$\int_0^{2\pi}\frac{dt}{1 + a^2 – 2a\cos(t)} = \frac{2\pi}{1 – a^2}$$
I was unsure how to go about the first part. I could just try to compute both integrals and show they are equal but that doesn't seem to be what is wanted. Is there is a trick that I am missing?
As for the second part, I keep getting $0$. I use the first part, and see there are singularities at $z = a$ and $z = 1/a$. If $a < 1$ then the singularities lie within $C$ and Cauchy's integral formula can be used.
I think I can split the integral into two, where I take $1 / (z – a)$ to be my function and evaluate the integral around a circle centered at $1/a$, and vice versa take $1/(az – 1)$ to be my function and evaluate the integral around a circle centered at $a$.
Many thanks.
Best Answer
For the first part: Assume $z=e^{it}$ as a substitution for the integral on the right side. Then
$\int_C \dfrac{i dz}{(z-1)(az-1)} = - \int_0^{2\pi} \dfrac{e^{it}dt}{a(1+e^{2it})-(1+a^2)e^{it}} = - \int_0^{2\pi} \dfrac{e^{it}dt}{a(1+\cos{2t}+i\sin{2t})-(1+a^2)(\cos{t}+i\sin{t})} = - \int_0^{2\pi} \dfrac{e^{it}dt}{(\cos{t}+i\sin{t})(2a\cos{t} -(1+a^2))} = \int_0^{2\pi} \dfrac{dt}{1+a^2 -2a\cos{t}}$.
For the second part,when 0≤a<1, the only singularity lies on the real axis within the unit circle at $a$ and it is a simple pole. So the residue can be computed as $\lim_{z\rightarrow a}(z-a)\dfrac{i}{(z-a)(az-1)}=\dfrac{i}{a^2−1}$.
Hence the integral is $2πi× Residue= \dfrac{2\pi}{1-a^2}$