Consider the function $f(x) = \begin{cases} \frac{\pi}{2}+x & & x \in (-\pi, 0] \\ \frac{\pi}{2}-x & & x \in (0, \pi]\\ \end{cases}$ extended 2$\pi$ periodically to $\mathbb{R}$. Calculate $a_0, a_n, b_n$
I understand how to work out a fourier series but I am unsure what to set for $f(x)$ due to the way its set out.
Would I have $a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\pi}{2}+x dx$ due to splitting it into odd and even parts?
Divide it two parts and calculate $$a_{ 0 }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi } f\left( x \right) dx=\frac { 1 }{ \pi } \int _{ -\pi }^{ 0 }{ \left( \frac { \pi }{ 2 } +x \right) dx } +\frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ \left( \frac { \pi }{ 2 } -x \right) dx } =\\ ={ \left( \frac { \pi }{ 2 } x+\frac { { x }^{ 2 } }{ 2 } \right) }_{ -\pi }^{ 0 }{ +\left( \frac { \pi }{ 2 } x-\frac { { x }^{ 2 } }{ 2 } \right) }_{ -\pi }^{ 0 }=0\\ { a }_{ n }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi } f\left( x \right) \cos { \left( nx \right) } dx=\frac { 1 }{ \pi } \left( \int _{ -\pi }^{ 0 }{ \left( \frac { \pi }{ 2 } +x \right) \cos { \left( nx \right) } dx } +\int _{ 0 }^{ \pi }{ \left( \frac { \pi }{ 2 } -x \right) \cos { \left( nx \right) } dx } \right) \\ { b }_{ n }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi } f\left( x \right) \sin { \left( nx \right) } dx=\frac { 1 }{ \pi } \left( \int _{ -\pi }^{ 0 }{ \left( \frac { \pi }{ 2 } +x \right) \sin { \left( nx \right) } dx } +\int _{ 0 }^{ \pi }{ \left( \frac { \pi }{ 2 } -x \right) \sin { \left( nx \right) } dx } \right) $$ Can you take from here?