Let $f:\mathbb{R}\to\mathbb{R}$,$f$ is $2\pi$ periodic such that for $x\in (-\pi,\pi]$: $f(x) = \cos \pi x$. We define $g(x) = f(x + 2010)$. Find $\sum_{n=-\infty}^\infty \hat{g}(n)$. Where $\hat{g}(n)$ is the n-th Fourier coefficient.
My Work:
$$ \hat{g}(n) = \frac{1}{2\pi} \int_0^{2\pi} \cos (\pi x + 2010\pi)e^{-inx} \ dx = \frac{1}{2\pi} \left( \int_0^{2\pi}\cos(\pi x)\cos(2010\pi)e^{-inx} \ dx + \int_0^{2\pi}\sin(\pi x)\sin(2010\pi)e^{-inx} \ dx \right) \\ = \frac{1}{2\pi} \int_0^{2\pi}\cos(\pi x)e^{-inx} \ dx = \frac{1}{\pi} \int_0^\pi \cos(\pi x)e^{-inx} \ dx $$
From here, it doesn't get so pretty. I think another way should be taken. I thought about utilizing Parseval's theorem but wasn't sure how to apply it here.
I'd be glad for help.
Thanks.
EDIT:
A thought came to my mind right now; since $g(x)$ is continuous:
$$g(t) = \sum_{n=-\infty}^\infty \hat{g}(n) e^{int} = \cos (\pi t + 2010\pi) \implies \sum_{n=-\infty}^\infty \hat{g}(n) = \frac{\cos(\pi t + 2010\pi)}{e^{int}} = \frac{\cos(\pi t)}{e^{int}} = \frac{e^{i\pi t} + e^{-i \pi t}}{2 e^{int}} = \frac{e^{\pi / n} + e^{-\pi / n}}{2}$$
You take the denominator $\Bbb e ^{\Bbb i nt}$ out of your sum and this is forbidden! Not to mention that $$\frac { e ^{\Bbb i \pi t} } {e ^{\Bbb i nt}} \ne \Bbb e ^{\frac \pi n} !!!$$
Once you write $\sum \limits _{n = -\infty} ^\infty \hat g (n) \Bbb e ^{\Bbb i nt} = \cos (\pi t + 2010 \pi)$ just apply in $t=0$ and get $\sum \limits _{n = -\infty} ^\infty \hat g (n) = \cos (2010 \pi) = 1$.