I'm learning about Fourier analysis and need help with the following problem:
Suppose $f \in L^1[-\pi, \pi]$ and $\alpha_n, \beta_n$ are the Fourier coefficients of $f$. Show that if $f$ is odd and $f(x + \pi) = f(x)$ for $x \in \mathbb{R}$, then $\beta_{2k - 1} = 0, \forall{k} \in \mathbb{N}$.
My work and thoughts:
Since $f$ is odd, then the Fourier coefficient $\alpha_n =0, \forall n \in \mathbb{N^*}$. Therefore, The Fourier series of $f$ can be written as
$$f(x) \sim \sum_{k = 1}^{+\infty}\beta_k\sin{kx}.$$
Thus
$$\beta_{2k - 1} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(2k - 1)xdx.$$
Since $f(x + \pi) = f(x)$ for $x \in \mathbb{R}$, we have
$$\beta_{2k - 1} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x + \pi)\sin(2k - 1)xdx.$$
I think I'm close but I don't know how to manipulate the above integral to get the desired result.
On the one hand you have:
$$\beta_{2k-1}=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin\{(2k-1)\cdot x\}dx$$
on the other:
$$\beta_{2k-1}=\frac{1}{\pi}\int_{-\pi}^\pi f(u)\sin\{(2k-1)\cdot u\}du$$
and after setting $u=x+\pi$ you have: $f(u)=f(x)$ from periodicity, $du=dx$ from integration rules and $\sin\{(2k-1)u\}=\sin\{(2k-1)\cdot(x+\pi)\}=-\sin\{(2k-1)x\}$, for all appropriate $k\in\mathbb{N}$.
Therefore, the second expression for $\beta_{2k-1}$ becomes:
$$\beta_{2k-1}=-\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin\{(2k-1)\cdot x\}dx$$
Consequently,
$$\beta_{2k-1}=-\beta_{2k-1}\Rightarrow$$
$$2\beta_{2k-1}=0\Rightarrow$$
$$\beta_{2k-1}=0$$
[Note: Please correct your symbols for the Fourier coefficients. Either all $a_n$, $b_n$ or all $\alpha_n$, $\beta_n$]