I tried to solve a question of true or false. The affirmation is "The function $g(x)= \tan(\pi x)$ to $\displaystyle x \in [\frac{-1}{4};\frac{1}{4}]$ and extended for period $\dfrac{1}{2}$ have only sin in the Fourier Series."
I tried to solve $a_0$, and I find $a_0=0$.
But when I tried to solve $a_n$ I stayed stuck, I don't know what to do. Let's my tried:
$$a_n = \frac{1}{L}\int^{\frac{1}{4}}_{-\frac{1}{4}}{g(x) \cdot \cos(\frac{n\pi x}{L}) dx}$$
$$a_n= 4 \cdot \int^{\frac{1}{4}}_{-\frac{1}{4}}{\tan(\pi x) \cdot \cos(\frac{n\pi x}{L}) dx}$$
$$a_n= 4 \cdot \int^{\frac{1}{4}}_{-\frac{1}{4}}{\frac{\sin(\pi x)}{\cos(\pi x)} \cdot \cos(\frac{n\pi x}{L}) dx}$$
I tried too many identities and too many ways to solve this, I tried Wolfram Alpha, I don't get it, I tried to solve with series, but it seems too wrong to me. Going on:
$$a_n= 4 \cdot \int^{\frac{1}{4}}_{-\frac{1}{4}}{\frac{\sum^{\infty}_{n=0}\frac{(-1)^{n} x^{2n+1}}{\mathrm{(2n+1)!}}}{\sum^{\infty}_{n=0}{\frac{(-1)^{n} x^{2n}}{\mathrm{(2n)!}}}}\cdot \sum^{\infty}_{n=0}{\frac{(-1)^{n}.({\frac{\pi.n.x}{L}})^{2n}}{\mathrm{(2n)!}}}} dx$$
I realize now, that I can't unify sums product. Hence, all that I tried is wrong and I don't know what to do. Sorry for bad english.