I'm stuck trying to solve this problem: Let $f \in C(\mathbb{R}/\mathbb{Z};\mathbb{C})$ and define the trigonometric Fourier coefficients $a_n,b_n$ for $n=0,1,2,3,...$ by $$a_n=2\int_0^1f(x)\cos(2\pi nx)dx, b_n=2\int_0^1f(x)\sin(2\pi nx)dx.$$ Show that the Fourier series of $f, \sum_{n=-\infty}^\infty\hat{f}(n)e_n$, can be rearranged to equal the series $$\frac{1}{2}a_0+\sum^\infty_{n=1}(a_n\cos(2\pi nx)+b_n\sin(2\pi nx)).$$
So far I have $$\sum^\infty_{n=-\infty}\hat{f}(n),e_n$$ $$=\sum^\infty_{n=-\infty}\int_0^1f(x)e^{-2\pi inx}dxe_n$$ $$= 2\sum^\infty_{n=1} [\int_0^1f(x)\cos(2\pi nx)dx-i\int_0^1f(x)\sin(2\pi nx)dx](\cos(2\pi nx)+i\sin(2\pi nx)$$ $$= 2\sum^\infty_{n=1}\cos(2\pi nx)\int_0^1f(x)\cos(2\pi nx)dx+\sin(2\pi nx)\int_0^1f(x)\sin(2\pi nx)dx+i\sin(2\pi nx)\int_0^1f(x)\cos(2\pi nx)dx-i\cos(2\pi nx)\int_0^1f(x)\sin(2\pi nx)dx$$ $$= \sum^\infty_{n=1}[a_n\cos(2\pi nx)+b_n\sin(2\pi nx)]+2i\sin(2\pi nx)\int_0^1f(x)\cos(2\pi nx)dx-2i\cos(2\pi nx)\int_0^1f(x)\sin(2\pi nx)dx$$
But I can't get the last piece to simplify.