Express g's Fourier coefficients using f's ones, if $g(x)=f(x+c)$. The Fourier coefficients are defined (in our course) as: $$\hat{f(n)}={1\over 2\pi}\int_{0}^{2\pi}{f(t)e^{-int}dt}$$
I am asked to express g's coefficients as a combination of f's ones, given $g(x)=f(x+c)$. The answer says: $$\hat{g(n)}=e^{2\pi inc}\hat{f(n)}$$ 
But I don't understand why. What I did is:
$$\hat{g(n)}={1\over 2\pi}\int_{0}^{2\pi}{f(t)e^{-in(t+c)}dt}=e^{-inc}\hat{f(n)}$$
Where am I wrong? I could really use your help.
 A: You forget to mention one vital thing: $f$ is periodic with periodicity $2\pi$.
The fourier coefficient is as you mention:
$$f_n=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}dx$$
An interesting property of this integral is that one may move the interval of integration: as $f(x+2\pi)=f(x)$ and also $e^{-in(x+2\pi)}=e^{-2n\pi i} \cdot e^{-inx}=e^{-inx}$, we may calculate $f_n$ as follows also:
$$f_n=\frac{1}{2\pi}\int_c^{c+2\pi}f(x)e^{-inx}dx$$ Thus:
\begin{equation}
\begin{split}
g_n &= \frac{1}{2\pi}\int_0^{2\pi}g(x)e^{-inx}dx
\\ &= \frac{1}{2\pi}\int_0^{2\pi}f(x+c)e^{-inx}dx
\\ &= \frac{1}{2\pi}\int_c^{c+2\pi}f(u)e^{-in(u-c)}d(u-c)
\\ &= e^{inc}\frac{1}{2\pi}\int_c^{c+2\pi}f(u)e^{-inu}du
\\ &= e^{inc}f_n
\end{split}
\end{equation}
A: I believe that your domain of interest is incorrect. As far as I can tell, you can only get this answer if you're considering the function on the domain $x\in [0,1]$. Starting from the definition of the nth Fourier coefficient on this domain:
$$f_n:=\int_{0}^{1} f(t) e^{-2\pi int}dt$$
Then, for $g(x)=f(x+c)$: 
$$\begin{aligned}g_n&=\int_0^{1} f(t+c) e^{-2\pi int}dt\\
&=\int_0^{1} f(u) e^{-2\pi in(u-c)}du\\
&=\int_0^{1} f(u) e^{-2\pi inu}e^{2\pi inc}du\\
&=e^{2\pi inc}\int_0^{1} f(u) e^{-2\pi inu}du\\
&=e^{2\pi inc}f_n\end{aligned}$$
where we used the substitution $u:=t+c$.
