Fourier Series for $|\cos(x)|$

I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.

I understand its an even function, so all the $b_n$s are $0$

$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$

$$a_n = \frac 2 \pi \int _0^\pi |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$

since for all $j,k$ not equal the integral is zero.

so only $a_1$ remains. is this correct?

How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\ {}$?

Although $ \int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx. $$

We can evaluate it as follows, as can be seen in the plot below

$$a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx = \frac 4 \pi \int_0^{\pi/2} \cos(x)\,dx=\frac 4 \pi.$$ $$\tag{1}$$

Plot of $\cos x$ (doted line) and $|\cos x|$ (solid line) in the interval $[-\pi,\pi]$.

The coefficients $b_n=0$ as you concluded. As for the $a_n$ coefficients only the odd ones are equal to $0$ (see below). The functions $\cos(x)$ and $\cos(nx)$ are orthogonal in the interval $[-\pi,\pi]$, but $|\cos(x)|$ and $\cos(nx)$ are not. Since

\begin{equation*} \left\vert \cos (x)\right\vert =\left\{ \begin{array}{c} \cos (x) \\ -\cos (x) \end{array} \begin{array}{c} \text{if} \\ \text{if} \end{array} \begin{array}{c} 0\leq x\leq \pi /2 \\ \pi /2\leq x\leq \pi, \end{array} \right. \tag{2} \end{equation*}

we have that

\begin{eqnarray*} a_{n} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }\left\vert \cos (x)\right\vert\cos (nx)\,dx=\frac{2}{\pi }\int_{0}^{\pi }\left\vert \cos (x)\right\vert \cos (nx)\,dx \\ &=&\frac{2}{\pi }\int_{0}^{\pi /2}\left\vert \cos (x)\right\vert \cos (nx)\,dx+\frac{2}{\pi }\int_{\pi /2}^{\pi }\left\vert \cos (x)\right\vert \cos (nx)\,dx \\ &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos (x)\cos (nx)\,dx-\frac{2}{\pi } \int_{\pi /2}^{\pi }\cos (x)\cos (nx)\,dx. \\ a_{1} &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos ^{2}(x)\,dx-\frac{2}{\pi }\int_{\pi /2}^{\pi }\cos ^{2}(x)\,dx=0. \end{eqnarray*}

Using the following trigonometric identity, with $a=x,b=nx$, \begin{equation*} \cos (a)\cos (b)=\frac{\cos (a+b)+\cos (a-b)}{2},\tag{3} \end{equation*}

we find

\begin{eqnarray*} a_{2m} &=&\frac{4}{\pi \left( 1-4m^{2}\right) }\cos (\frac{2m\pi }{2})=\frac{ 4}{\pi \left( 1-4m^{2}\right) }(-1)^{m} \\ a_{2m+1} &=&\frac{4}{\pi ( 1-4(2m+1)^{2}) }\cos (\frac{(2m+1)\pi }{2})=0,\qquad m=1,2,3,\ldots.\tag{4} \end{eqnarray*}

The expansion of $\left\vert \cos (x)\right\vert $ into a trigonometric Fourier series in the interval $[-\pi ,\pi ]$ is thus

\begin{equation*} \left\vert \cos x\right\vert =\frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left( a_{n}\cos (nx)+b_{n}\sin (nx)\right) =\frac{2}{\pi }+\frac{4}{\pi } \sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx)\tag{5} \end{equation*}

$$|\sin(x)|\ \text{(blue) and the partial sum }\frac{2}{\pi }+\frac{4}{\pi } \sum_{m=1}^{5 }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx) \ \text{(red) in }[-\pi,\pi]$$

Setting $x=0$ in $(5)$, we obtain

\begin{equation*} 1=\frac{2}{\pi }+\frac{4}{\pi }\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}=\frac{2}{\pi }-\frac{4}{\pi }\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}.\tag{6} \end{equation*}

Hence

\begin{equation*} \sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}=\frac{1}{2}-\frac{\pi }{4}.\tag{7} \end{equation*}

It is not correct. $$ \int_0^\pi|\cos x|\cos(n\,x)\,dx=\int_0^{\pi/2}\cos x\cos(n\,x)\,dx-\int_{\pi/2}^\pi\cos x\cos(n\,x)\,dx. $$ Compute the integrals and you will see that the result is not $0$.

You must breakup the integral into three intervals: $\left[-\pi \cdots -\frac{\pi}{2} \right]$, $\left[-\frac{\pi}{2} \cdots \frac{\pi}{2} \right]$, and $\left[\frac{\pi}{2} \cdots \pi \right]$

Which represent the regions where the sign of $\cos x$ changes. \begin{equation*} \left\vert \cos x\right\vert = \begin{cases} -\cos x & -\pi \le x \le -\frac{\pi}{2} \\ \cos x & \frac{\pi}{2} \le x \le \frac{\pi}{2} \\ -\cos x & \frac{\pi}{2} \le x \le \pi \end{cases} \end{equation*}

When I plugged the integral over the three regions into Maple I got:

\begin{align*} a_n &=\frac{1}{2 \pi} \int\limits_{t=-\pi}^{\pi} \left\vert \cos(t) \right\vert \cos(nt) \\ &= \frac{1}{2 \pi} \int\limits_{t=-\pi}^{-\frac{\pi}{2}}(-\cos(t)) \cos(nt) \\ &+ \frac{1}{2 \pi} \int\limits_{t=-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(t) \cos(nt) \\ &+ \frac{1}{2 \pi} \int\limits_{t=\frac{\pi}{2}}^{\pi}(-\cos(t)) \cos(nt) \\ &=-4{\frac {\cos \left( \frac{1}{2} \pi n \right) }{ \left( -1+{n}^{2} \right) \pi }} \end{align*}

Since, $\left\vert \cos t \right\vert$ is even you could break the integral in two and find $a_n$ as in Julián's answer. \begin{align*} a_n &=\frac{2}{\pi} \int\limits_{t=0}^{\pi} \left\vert \cos(t) \right\vert \cos(nt) \\ &= \frac{2}{\pi} \int\limits_{t=0}^{\frac{\pi}{2}}\cos(t) \cos(nt) \\ &+ \frac{2}{\pi} \int\limits_{t=\frac{\pi}{2}}^{\pi}(-\cos(t)) \cos(nt) \\ &=-4{\frac {\cos \left( \frac{1}{2} \pi n \right) }{ \left( -1+{n}^{2} \right) \pi }} \end{align*} for $1 < n$ and $a_0 = \frac{2}{\pi}$

My answer is simply an amplification of Julián's answer, but, I hope it helps.