I need to calculate the Fourier Coefficients $a_k$ of $\left | \cos(2 \pi f_c) \right |$ (a full-wave rectified cosine) so that $\left | \cos(2 \pi f_c) \right | = a_0 + \displaystyle\sum\limits_{k=1}^{\inf} a_k \cos(2 \pi k f_c t + \phi_k)$ and prove that for any odd value of $k$, $a_k=0$.
Applying (for $k\not=0$) the Fourier Cosine Series Coefficients Formula $a_k = \cfrac{1}{T} \displaystyle\int\limits_{T} f(x) \cos(k\,\pi\,x) \,\mathrm{d}x$, I get $a_k = 8\,f_c\displaystyle\int\limits_{0}^{\frac{1}{4f_c}} \cos(2 \pi f_c t)\,\cos(2\pi\,k\,f_c\,t) \, \mathrm{d}{x} = \frac{8\,f_c\,\cos(\frac{k\,\pi}{2})}{2\,f_c\,\pi-2\,f_c\,k^2\,\pi}$ (result of the integral confirmed with Mathematica).
Substituting $k=1$, we get an indeterminate form. If we calculate the limit at $k=1$, we get $1$. For any other odd $k$, $a_k=0$ as expected, with no indeterminate form.
When we use the Exponential Fourier Series Coefficients Formula, I end up with $m_k = \cfrac{2\,\sin(\frac{\pi}{2}\,(1-k))}{\pi\,(1-k)}$ and we also get an indeterminate form at $k=1$ with a limit of $1$.
I've also tried to convolve a cosine with a square wave (so that their multiplication would be the same as a full-wave rectified cosine) and I get the same formulas as above, with the same indeterminate form.
After a couple hours trying to solve this, I'm completely stumped. I have no idea how I can prove that $a_1=0$ when all attempts to calculate it result in an indeterminate form. What am I doing wrong? Thanks in advance for any help you can provide!