My textbook states this question as follows :
Let $f$ be a $2\pi$-periodic Riemann integrable function defined on ${\bf R}$ .Then show that $f$ can be written as $$f(\theta)\sim\widehat{f}(0)+\sum_{n\geq1}\bigg\{\bigg(\widehat{f}(n)+\widehat{f}(-n)\bigg)\cos(n\theta)+i\bigg(\widehat{f}(n)-\widehat{f}(-n)\sin(n\theta)\bigg)\bigg\}~,$$ provide that we given $f(\theta)\sim\displaystyle\sum_{n=-\infty}^{\infty}\widehat{f}(n)e^{in\theta}$ and $\widehat{f}$ is the Fourier transform of $f$ .
$\rule{18cm}{2pt}$
However, I think the conclusion may be not true , in general , when we don't make an assumption $\displaystyle\sum_{n\in{\bf Z}}|\widehat{f}(n)|<+\infty$ since this question may involve the issue of absolute convergence of a function. Or is there another way to show this conclusion which does not use the property of absolute convergence ?
The following is my attempt via the additional hypothesis $\displaystyle\sum_{n\in{\bf Z}}|\widehat{f}(n)|<+\infty$ :
\begin{align} &\displaystyle\sum_{n=-\infty}^{\infty}\widehat{f}(n)e^{in\theta}\\ =&\widehat{f}(0)+\sum_{n\in{\bf Z}-\{0\}}\widehat{f} {(n)}\bigg(\cos(n\theta)+i\sin(n\theta)\bigg)\\ =&\widehat{f}(0)+\sum_{n\in{\bf Z}_{<0}}\widehat{f} {(n)}\bigg(\cos(n\theta)+i\sin(n\theta)\bigg)+\sum_{n\in{\bf Z}_{>0}}\widehat{f} {(n)}\bigg(\cos(n\theta)+i\sin(n\theta)\bigg)\\ =&\color{red}{\widehat{f}(0)+\sum_{n\in{\bf Z}_{>0}}\widehat{f} {(-n)}\bigg(\cos((-n)\theta)+i\sin((-n)\theta)\bigg)+\sum_{n\in{\bf Z}_{>0}}\widehat{f} {(n)}\bigg(\cos(n\theta)+i\sin(n\theta)\bigg)}\\ =&\color{blue}{\widehat{f}(0)+\sum_{n\in{\bf Z}_{>0}}\bigg\{\widehat{f}(-n)\bigg(\cos((-n)\theta)+i\sin((-n)\theta)\bigg)+\widehat{f}(n)\bigg(\cos(n\theta)+i\sin(n\theta))\bigg)\bigg\}}\\ =&\widehat{f}(0)+\sum_{n\in{\bf Z}_{>0}}\bigg\{\widehat{f}(-n)\bigg(\cos(n\theta)-i\sin(n\theta)\bigg)+\widehat{f}(n)\bigg(\cos(n\theta)+i\sin(n\theta))\bigg)\bigg\}\\ =&\widehat{f}(0)+\sum_{n\geq1}\bigg\{\bigg(\widehat{f}(n)+\widehat{f}(-n)\bigg)\cos(n\theta)+i\bigg(\widehat{f}(n)-\widehat{f}(-n)\sin(n\theta)\bigg)\bigg\} \end{align}
I think the equality is true form red to blue if we impose this following condition : $$\sum_{n=-\infty}^{\infty}|\widehat{f}(n)e^{in\theta}|=\sum_{n\in{\bf Z}}|\widehat{f}(n)||\cos(n\theta)+i\sin(n\theta)|=\sum_{n\in{\bf Z}}|\widehat{f}(n)|<+\infty$$
Any advice or comment will be appreciated . Thanks for considering my request .
Actually we define \begin{align*} \sum_{n=-\infty}^{\infty}\widehat{f}(n)e^{in\theta}=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N}\widehat{f}(n)e^{in\theta}, \end{align*} where \begin{align*} &\sum_{n=-N}^{N}\widehat{f}(n)e^{in\theta}\\ &=\widehat{f}(0)+\sum_{n=1}^{N}\widehat{f}(n)e^{in\theta}+\sum_{n=1}^{N}\widehat{f}(-n)e^{-in\theta}\\ &=\widehat{f}(0)+\sum_{n=1}^{N}\left(\widehat{f}(n)+\widehat{f}(-n)\right)\cos(n\theta)+i\sum_{n=1}^{N}\left(\widehat{f}(n)-\widehat{f}(n)\right)\sin(n\theta). \end{align*} So \begin{align*} \widehat{f}(0)+\sum_{n=1}^{N}\left(\widehat{f}(n)+\widehat{f}(-n)\right)\cos(n\theta)+i\sum_{n=1}^{N}\left(\widehat{f}(n)-\widehat{f}(n)\right)\sin(n\theta)\rightarrow\sum_{n=-\infty}^{\infty}\widehat{f}(n)e^{in\theta}, \end{align*} and the result follows.