Let $-1/2<a<1/2$ I am asked to use the fourier series of the function $$f(x)=\begin{cases}1,\text{ if } |x|\leq a \\ 0,\text{ if }a<|x|<1/2\end{cases}$$ to calculate $$\sum_{n\geq1}\frac{\text{sin}^{2}(2\pi an)}{n^2}$$
I have noticed that the function $f(x)$ is even. The fourier series of $f(x)$ is $$2a +2\sum_{n\geq 1}\frac{\text{sin}(2\pi an)}{\pi n}\text{cos}(2\pi nx)$$
Since $$\int_{-1/2}^{1/2}f(x)dx=2a,\text{ }\int_{-1/2}^{1/2}f(x)\text{cos}(2\pi nx)dx=\frac{\text{sin}(2\pi an)}{\pi n}$$
And I am stuck here, the fourier series does not seem to help since $f(x)$ is even, what am I doing wrong?
On the one hand $sin^2(2\pi na)=\frac{1-cos(4\pi na)}{2}$.
On the other hand $sin(2\pi na).cos(2\pi nx)=\frac{1}{2}(cos(2\pi n(a-x))-cos(2\pi n(a+x)))$ so taking $x=a$ yields $sin(2\pi na).cos(2\pi na)=\frac{1}{2}(1-cos(4\pi na))=sin^2(2\pi na)$.
These both come from trigonometric formulae. Once you are here only need to make a substitution in your series which is convergent to $f$ and use Dini since $f$ is not continuous at $x=a$ (you will state the series is equal to $\frac{1}{2}$ in stead of $1$) and a couple more calculations will land you on the desired result.