I am solving an exercise about the Fourier series and I am having trouble with the last request.
Let $f$ be a function with period 1, $f(x):=x$, $ \forall x\in [0,1)$.
- Show that its Fourier series is $\displaystyle Sf(x)=\frac{1}{2}+\sum_{n\in Z\setminus\{0\}}\frac{ie^{2\pi i n x}}{2\pi n}$.
- Show that $$ Sf(x)= \left\{ \begin{array}{ll} x & \mbox{, if }x\in R \setminus Z\\ \frac{1}{2} & \mbox{, otherwise} \end{array} \right. . $$
- Evaluate $\displaystyle \sum_{j=0}^{+\infty}\frac{\sin(\pi(2j+1)/2)}{2j+1}$. ( Hint: Use $x=\frac{1}{4}$).
I was able to solve the first two requests.
- $T=1$ and I know the following:
for $k \ge 1$
$a_k= 2 \int_{0}^{1} f(x) \cos(2\pi kx) dx$ and $b_k= 2 \int_{0}^{1} f(x) \sin(2\pi kx) dx \Rightarrow Sf(x)=\sum_{k=1}^{+\infty} a_k \cos(2\pi kx)+\sum_{k=1}^{+\infty} b_k \sin(2\pi kx) + \hat{f}(0)$
I have $$\hat{f}(0)=\int_{0}^{1} f(x)dx =\int_{0}^{1} x dx=\frac{1}{2}$$ $$a_k=2 \int_{0}^{1}x \cos (2\pi x k) dx = 2[\frac{2\pi k sin(2\pi k) + \cos (2\pi k)}{4 \pi^2 k^2} - \frac{1}{4\pi^2 k^2}] = 2[\frac{0 + 1}{4 \pi^2 k^2} - \frac{1}{4\pi^2 k^2}] =0 $$ $$ b_k =2 \int_{0}^{1}x \sin (2\pi x k) dx = 2 [\frac{sin (2 \pi k) - 2\pi k \cos(2\pi k)}{4\pi^2 k^2}] =2 [\frac{0 - 2\pi k \cdot 1}{4\pi^2 k^2}] = - \frac{1}{ \pi k}$$ $$Sf(x)= +\sum_{k=1}^{+\infty} [(- \frac{1}{ \pi k}) \sin(2\pi kx)] +\frac{1}{2}= \frac{1}{2} +\sum_{k=1}^{+\infty} [(- \frac{1}{ \pi k}) \frac{e^{i 2 \pi xk}-e^{-i 2 \pi xk}}{2i}] $$ $$ = \frac{1}{2} +\sum_{k=1}^{+\infty} [(- \frac{e^{i 2 \pi xk}}{ 2\pi ki })] +\sum_{k=1}^{+\infty} [( \frac{e^{-i 2 \pi xk}}{ 2\pi ki })] = \frac{1}{2} +\sum_{k=1}^{+\infty} [(\frac{ie^{i 2 \pi xk}}{ 2\pi k })] +\sum_{k=-\infty}^{-1} [( \frac{ie^{i 2 \pi x(-k)}}{ 2\pi (-k) })] $$ $$ = \frac{1}{2} + \sum_{k \in Z \setminus \{ 0 \} } [(\frac{ie^{i 2 \pi xk}}{ 2\pi k })]$$ 2.$Sf(x)=f(x)$ if $f$ is continuous , so $Sf(x)= x \hspace{0.5 cm} \forall x \in R \setminus Z$
In general $Sf(x)= \frac{f(x^{+})- f(x^{-}) }{2} = \frac{1}{2} $
so $\forall x \in Z$ $Sf(x)= \frac{x^{+}- x^{-} }{2} = \frac{1}{2} $ 3. Using the hint, I can write: $$\frac{1}{4} = \frac{1}{2} + \sum_{k \in Z \setminus \{ 0 \} } (\frac{ie^{i \pi k / 2}}{ 2\pi k })$$ Then I don't know how to proceed. I thought about changing the index or using $\sinh$ but it seems useless.
Thank you!
From 1 and 2 it follows that $$ \frac 1 4=\frac{1}{2}+\sum_{n\in \mathbb Z \setminus\{0\}}\frac{ie^{2\pi i n \frac 1 4}}{2\pi n}. $$ Using the fact that $\sin t = \frac {e^{it}-e^{-it}}{2i}$ we obtain $$ \frac{1}{2}+\sum_{n\in \mathbb Z \setminus\{0\}}\frac{ie^{2\pi i n \frac 1 4}}{2\pi n} = \frac{1}{2}-\frac 1 \pi \sum_{n=1}^\infty\frac{\sin(\frac \pi 2 n)}{n}. $$ If $n=2j$ for $j \in \mathbb N$ then $\sin(\frac \pi 2 n) =0$ and therefore $$ \frac{1}{2}-\frac 1 \pi \sum_{n=1}^\infty\frac{\sin(\frac \pi 2 n)}{n} = \frac{1}{2}-\frac 1 \pi \sum_{j=1}^\infty\frac{\sin(\frac \pi 2(2j+1))}{2j+1} = \frac{1}{2}-\frac 1 \pi \left ( \sum_{j=0}^\infty\frac{\sin(\frac \pi 2(2j+1))}{2j+1} -1 \right). $$ The last expression equals $\frac 1 4$ from which the value of the desired series can be derived.