For convenience, denote $R_{2\pi}:=\{f:\Bbb R\to \Bbb C$ of $2\pi$ periodic and integrable on $[-\pi,\pi]\}$, and $f_a(x):=f(a+x)$
It is clear that $\forall f\in R_{2\pi}$, $f_a$ is also in $R_{2\pi}$.
I am proving that $\hat{f_a}=e^{ina}\hat{f}$. I almost got it, but with a little bit of uncertainty.
$\hat{f_a}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f_a(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+a)e^{-inx}dx$
$=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+a)e^{-in(x+a)}e^{ina}d(x+a)$
$=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+a)e^{-in(x+a)}e^{ina}d(x+a)$
$=e^{ina}\frac{1}{2\pi}\int_{-\pi+a}^{\pi+a}f(u)e^{-inu}du $
?$=e^{ina}\hat{f}(n)$
My problem is: by definition, $\hat{f}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(u)e^{-inu}du$, can I have the last step? if yes, why$ \frac{1}{2\pi}\int_{-\pi+a}^{\pi+a}f(u)e^{-inu}du=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(u)e^{-inu}du$ ? Any suggestion would be appreciated.