Fourier transform $f(x)=\sin(x/2)$ in $0 \leq x < 2\pi$

Question

Can you help me to find the coefficient of the Fourier series? I know that is ODD on $[0,4 \pi)$, but is it ODD also in $[0,2\pi )$? Thanks you a lot

Answer 1

To the Fourier transform:

In the title you are asking about the Fourier transform for $f$ in $0\leqslant x<2\pi$ but below you are asking about Fourier series coefficients. If it is about the Fourier transform one can calculate this directly by the definition (at least one version of the definition) $$\hat{f}(w)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)e^{-iwx}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\sin(\frac{x}{2})\mathbf{1}_{\{0\leqslant x<2\pi\}}e^{-iwx}dx=\frac{1}{\sqrt{2\pi}}\int_{0}^{2\pi}\sin(\frac{x}{2})e^{-iwx}dx$$

To Fourier series:

Assuming that the function is periodic (otherwise the series cannot determined only) with period $T=2\pi$ (and so $T=2\pi=2L\iff L=\pi$). Then the cfts are given by the usual Euler formulas $$a_{k}:=\frac{1}{L}\int_{c}^{c+2L}f(x)\cos(\frac{k\pi x}{L})dx=\frac{1}{\pi}\int_{0}^{2\pi}\sin(\frac{x}{2})\cos(kx)dx, \quad k\not=0$$ $$a_{0}:=\frac{1}{L}\int_{c}^{c+2L}f(x)dx=\frac{1}{\pi}\int_{0}^{2\pi}\sin(\frac{x}{2})dx$$ $$b_{k}:=\frac{1}{L}\int_{c}^{c+2\pi}f(x)\sin(\frac{k\pi x}{L})dx=\frac{1}{\pi}\int_{0}^{2\pi}\sin(\frac{x}{2})\sin(kx)dx,$$ choosing/setting $c=0$.

Then the expansion $$f(x)=\frac{a_0}{2}+\sum_{k\geqslant 1}(a_{k}\cos(\frac{k\pi x}{L})+b_{k}\sin(\frac{k\pi x}{L}))$$ $$\sin(\frac{x}{2})=\frac{a_0}{2}+\sum_{k\geqslant 1}(a_k \cos(kx)+b_{k}\sin(kx)),\quad (*)$$is valid for $0<x<2\pi$ (by Dirichlet condition for each continuity point in $]0,2\pi[$) and the series (the rhs of $(*)$) converges for $\frac{f_{+}(x_0)+f_{-}(x_0)}{2}$ in $x_{0}=0$, $x_{0}=2\pi$ (the discontinuity points). One has to be a little more careful with $x_0=0$ but I'll leave that up to you.

I think it answer for your comments

Can you help me to find the coefficient of the Fourier series?

I just want to understand how the coefficients are set.. I don't expect you to solve it for me!

In the other part where you talk about the parity of the function, I don't understand and I do not know if you are trying a odd-extension or even-extension but that is other story. Anyway you can calculate the integrals directly there is no need to simplify everything every time and you should complete all the details.

Answer 2

Although OP didnt request computations I want to note the amazing identity: $$\sum_{k=1}^{\infty}\frac1{1-4k^2}=-\frac12\tag 1$$ connected with the Fourier series of the periodic function $\sin(\frac x2)$, $0\leq x<2\pi$.

Since its even function (symmetric wrt. $y$-axis) $b_k$'s are zero.

$\frac{a_0}2=\frac{1}{2\pi}\int_0^{2\pi}\sin(\frac x2)dx=\frac2\pi$.

$a_k=\frac{1}{\pi}\int_0^{2\pi}\sin(\frac x2)\cos(kx)dx=\frac{4}{\pi(1-4k^2)}$ for$k\geq1$.

Hence, $f(x)=\frac2\pi+\sum_{k=1}^\infty \frac{4}{\pi(1-4k^2)}\cos(kx)$. Now we let $x=0$ to find $(1)$.

I think to find the Fourier transform of this function we can take the Fourier transform of its Foruier series term by term by using tables: $\cal{F} (1)=\delta(w)$, $\cal{F}(\cos(kx))=\frac{\delta(w-\frac k{2\pi})+\delta(w+\frac k{2\pi})}{2}$.