This is a problem from my exam:
A periodic function is defined by: $$f(x)= \begin{cases} -1 & \textrm{ for }x\in[-\pi,\pi)\setminus\{0\} \\ 0 & \textrm{ for }x=0 \\ \end{cases}\space , \space\space\space f(x+2π)=f(x)$$ Determine the Fourier expansion for f(x).
Can I use the improper integral to solve the problem? If not, could you tell me how to solve it? Thanks in advance.
$f$ is even and $2\pi$ periodic so you only need to evaluate the cosine coefficients : $$a_0(f) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)\,\mathrm{d}t=\frac{1}{2\pi}\int_{-\pi}^{0} -\,\mathrm{d}t+\frac{1}{2\pi}\int_{0}^{\pi} -\,\mathrm{d}t=-1$$ Indeed $f$ is discontinuous in only one point so $f$ is Riemann integrable.
Also $\forall n>0$ : $$a_n(f) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos\left(nt\right)\,\mathrm{d}t=\frac{1}{\pi}\int_{-\pi}^{0} -\cos\left(nt\right)\,\mathrm{d}t+\frac{1}{\pi}\int_{0}^{\pi} -\cos\left(nt\right)\,\mathrm{d}t\\=-\dfrac{\sin\left({\pi}n\right)}{n}-\dfrac{\sin\left({\pi}n\right)}{n}=-\dfrac{2\sin\left({\pi}n\right)}{n}=0$$
So the Fourier expansion of $f$ in $x$ is : $$-1$$