- So I make the Fourier transform of$ f(t)= e^{iat} $on $[- \pi, \pi]$ for some real $a$ and i get:
$$a_n=\frac{2a \sin(a \pi)(-1)^n}{\pi(a^2-n^2)}$$ $$b_n=\frac{2i(n\sin(a \pi) (-1)^n)}{\pi(a^2 - n^2)}$$
But now if i want to evaluate the sum at $t=\pi$, at the part of the series with $b_n \sin (nx)$ i get 0 and therefore there is no imaginary part in the whole series. Why is this the case?
- $f \in C^{\infty}$ but the series don't seen to be normally convergent (which should be the case if $f$ was a real function). Is that because we are now working with complex numbers?
Re 2: It is not true that $f \in C^\infty$ (for general values of $a$). Note that we are looking at the $2\pi$-periodic extension of the given $f$. For most values of $a$, $f(\pi) \neq f(-\pi)$ which implies that the periodic extension of $f$ is not continuous at odd multiples of $\pi$.
However, $f$ is smooth enough to guarantee that the Fourier series of $f$ evaluated at $t=\pi$ converges the average of $\lim_{t\to \pi^-} f(t)$ and $\lim_{t\to \pi^+} f(t)$ i.e. to $$ \frac{e^{ia\pi} + e^{-ia\pi}}{2} = \cos a\pi $$ which is real.