I have trouble solving a). How do I approach this problem?
Let $-\pi\leq a<b\leq\pi$. Consider the function
$$f(x)=\left\{\begin{array}{ll}1, & x \in] a, b[ \\ 0, & x \in]-\pi, \pi[\backslash[a, b]\end{array}\right.$$
It is shown that the (complex) Fourier coefficients are given by: $$c_0=\frac{b-a}{2\pi}$$ $$ c_{n}=\frac{i}{2 \pi} \frac{e^{-i n b}-e^{-i n a}}{n}, n\neq0 $$
a) Use the relevant theory about fourier series and the Fourier coefficients above to show$$ \sum_{n=1}^{\infty}\left(\frac{\sin (n b)}{n}\right)^{2}=\frac{b \pi-b^{2}}{2} $$ for $b\in[0,\pi]$
Immediately when I see an expression squared like that, I think of Parseval's theorem. However, I do not see what's going on with the limits when calculating the norm? Where did the $a$ disappear?
This is what I started doing, until I realized I shouldn't have any $a$. $$ \begin{aligned} &\|f\|^{2}=\left.\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x)\right|^{2} d x=\frac{1}{2 \pi} \int_{a}^{b} 1 d x=\frac{b-a}{2 \pi}\\ &\sum_{n=-\infty}^{\infty}\left|c_{n}\right|^{2}=\left|c_{0}\right|^{2}+\sum_{n=1}^{\infty}\left(\left|c_{n}\right|^{2}+\left|c_{-n}\right|^{2}\right)\\ &=\left|c_{0}\right|^{2}+\sum_{n=1}^{\infty}\left|\frac{i}{2 \pi} \frac{e^{-i n b}-e^{-i n a}}{n}\right|^{2}+ \left| \frac{i}{2 \pi} \frac{e^{i n b}-e^{i n a}}{n}\right|^{2} \end{aligned} $$ $$ \begin{array}{l} =\left|c_{0}\right|^{2}+\frac{1}{2 \pi} \sum_{n=1}^{\infty} \frac{1}{n^{2}}\left[2\left(e^{-i n b}-e^{-i n a}\right)\left(e^{i n b}-e^{i n a}\right)\right] \\ =\left|c_{0}\right|^{2}+\frac{1}{2 \pi} \sum_{n=1}^{\infty} \frac{1}{n^{2}}\left[2\left(1-e^{i n(a-b)}-e^{-i n(a-b)}+1\right)\right] \\ =\left|c_{0}\right|^{2}+\frac{1}{2 \pi} \sum_{n=1}^{\infty} \frac{2}{n^{2}}[2-2 i \sin (n(a-b))] \end{array} $$
So am I on the right track, what should I do instead?
- another question. Is it also understood correct, that if $a=-\pi$ and $b=\pi$, $f$ would be continuous on all of $\mathbb{R}$ and thus the Fourier series would converge uniformly on $\mathbb{R}$ in that case?
