Fourier Series/Cosine of x over Minus Pi to Zero, Minus Cosine of x over Zero to Pi

Theorem
Let $f \left({x}\right)$ be the real function defined on $\left({-\pi \,.\,.\, \pi}\right)$ as:


 * $f \left({x}\right) = \begin{cases}

\cos x & : -\pi < x < 0 \\ -\cos x & : 0 < x < \pi \end{cases}$

Then its Fourier series can be expressed as:


 * $\displaystyle f \left({x}\right) \sim -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$

Proof
It is apparent by inspection that $f \left({x}\right)$ is an odd function over $\left({-\pi \,.\,.\, \pi}\right)$.

It follows from Fourier Series for Odd Function over Symmetric Range:


 * $\displaystyle f \left({x}\right) \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$

where for all $n \in \Z_{> 0}$:
 * $b_n = \displaystyle \frac 2 \pi \int_0^\pi f \left({x}\right) \sin n x \rd x$

for all $n \in \Z_{>0}$.

Thus by definition of $f$:


 * $\displaystyle b_n = -\frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$

When $n \ne 1$, we have:

When $n = 1$, we have:

Hence:

When $n$ is odd, we have $n = 2 r + 1$ for $r \ge 1$, and so:

When $n$ is even, we have $n = 2 r$ for $r \ge 1$, and so: