Show that $\widehat{f}(n)$ is zero for odd $n$

Question

The following problem is from Stein´s Introduction to Fourier analysis:

Suppose that $f(\theta + \pi)=f(\theta)$ for all $\theta \in \mathbb{R}$ Show that $\widehat{f}(n)$ is zero for odd $n$.

My attempt:

$$\widehat{f}(n)= \frac{1}{2 \pi} \int_{0}^{2 \pi}f(\theta)e^{-in \theta}d\theta= \frac{1}{2 \pi} \int_{\pi}^{3 \pi}f(\theta+ \pi)e^{-in (\theta+ \pi)}d\theta = \frac{1}{2 \pi} \int_{-\pi}^{ \pi}f(\theta)e^{-in (\theta+ \pi)}d\theta $$ $$= \frac{1}{2 \pi}[ \int_{-\pi}^{\pi}f(\theta)cos(n (\theta + \pi)) d\theta +i \int_{-\pi}^{ \pi}f(\theta)sin(n(\theta + \pi))d\theta ] $$ $$= -\frac{1}{2 \pi}[ \int_{-\pi}^{\pi}f(\theta)cos(n (\theta )) d\theta +i \int_{-\pi}^{ \pi}f(\theta)sin(n(\theta ))d\theta ]=0 $$

only for odd n

well the thing is that I am not sure about my last step, can you help me to fix the proof or tell me if I am right, thanks a lot :)

Answer 1

You don't need to switch to $\cos$ and $\sin$. In the first line:

$$\begin{align} \widehat{f}(n) &= \frac{1}{2 \pi} \int_{0}^{2 \pi}f(\theta)e^{-in \theta}d\theta \\ &= \frac{1}{2 \pi} \int_{\pi}^{3 \pi}f(\theta+ \pi)e^{-in (\theta+ \pi)}d\theta \\ &= \frac{1}{2 \pi} \int_{-\pi}^{ \pi}f(\theta)e^{-in \theta}e^{n\pi}d\theta \\ & = -\widehat{f}(n) \tag{since $e^{n\pi}=-1$ for odd $n$} \end{align}$$

So $\widehat{f}(n)=0$

Answer 2

$\hat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi} f(\theta)e^{-in\theta}=\frac{1}{2\pi}\int_0^{\pi} f(\theta)e^{-in\theta}+\frac{1}{2\pi}\int_{\pi}^{2\pi} f(\theta)e^{-in\theta}=\frac{1}{2\pi}\int_0^{\pi} f(\theta)e^{-in\theta}+\frac{1}{2\pi}\int_0^{\pi} f(\theta)e^{-i(n+\pi)\theta}=0$

Change the variable of the second term.(Happens in the third equality)