My function is $f(x)= \dfrac{1}{1-2e^{ix}} + \dfrac{1}{1-2e^{-ix}} $, which has been periodically extended by $2\pi$.
I found $C_0$ to be $\pi$. I'm having trouble expressing $C_n$.
All I have is
$$C_n = \dfrac{1}{2\pi} \int\limits_{-\pi}^{\pi} \! \dfrac{4-2cosx}{5-4cosx} e^{-inx} \, \mathrm{d}x $$
where I added the fraction to get
$\dfrac{2-e^{ix}+2 - e^{-ix}}{4 + e^0 - 2e^{ix} - 2 e^{-ix}} $
I've messed around with it for a long time but I can't figure out where to go. Any help? Thank you
Use the expansion $$ \frac1{1-2\exp(ix)} = -\frac 12 \exp(-ix) \frac1{1-\frac 12 \exp(-ix)} = -\frac 12 \exp(-ix) \sum_{n=0}^\infty \frac 1{2^n}\exp(-inx) $$
so $$ \frac1{1-2\exp(-ix)} = -\frac 12 \exp(ix) \sum_{n=0}^\infty \frac 1{2^n}\exp(inx) $$
and you can easily check that the convergence is uniform: $$ \sum_{n=0}^\infty \sup_{x\in\Bbb R}\left|\frac 1{2^n}\exp(inx) \right| = \sum_{n=0}^\infty \frac 1{2^n}<\infty $$ hence you can invert integral and summation in the definition of $c_p$:
$$ c_p = \frac 1{2\pi} \int_{-\pi}^\pi \exp(-ipx) f(x) dx %%%%%%%%%%%%%%%%%%% \\= \frac 1{2\pi} \int_{-\pi}^\pi \exp(-ipx) \left[-\frac 12 \sum_{n=0}^\infty \frac 1{2^n}\exp(-i(n+1)x) -\frac 12 \sum_{n=0}^\infty \frac 1{2^n}\exp(i(n+1)x) \right] dx %%%%%%%%%%% \\= -\frac 12 \sum_{n=0}^\infty \frac 1{2\pi} \int_{-\pi}^\pi \exp(-ipx) \left[\frac 1{2^n}\exp(-i(n+1)x) \right] dx +\\ -\frac 12 \sum_{n=0}^\infty \frac 1{2\pi} \int_{-\pi}^\pi \exp(-ipx) \left[ \frac 1{2^n}\exp(i(n+1)x) \right]\exp(-ipx) dx %%%%%%%%%%%%%%%%%% \\= \frac1{2^{p-1}} $$if $p\neq 0$ and $0$ otherwise.