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I am a beginner with Fourier series and I have to evaluate the sum

$$\sum_{n =1}^{\infty}{\sin\left(n\right) \over n}$$

I don't know which function I have to take to evaluate the fourier series ... Someone can give me a hint ?

Thanks in advance!

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{n = 1}^{\infty}{\sin\pars{n} \over n} = \half\pars{\,\sum_{n = -\infty}^{\infty}{\sin\pars{n} \over n} - 1}.\quad}$ See $\large\tt details$ over here .

\begin{align} \sum_{n = -\infty}^{\infty}{\sin\pars{n} \over n}&= \int_{-\infty}^{\infty}{\sin{x} \over x}\sum_{n = -\infty}^{\infty}\expo{2n\pi x\ic} \,\dd x = \int_{-\infty}^{\infty}\half\int_{-1}^{1}\expo{\ic kx}\,\dd k \sum_{n = -\infty}^{\infty}\expo{-2n\pi x\ic}\,\dd x \\[3mm]&= \pi\sum_{n = -\infty}^{\infty}\int_{-1}^{1}\dd k \int_{-\infty}^{\infty}\expo{\ic\pars{k - 2n\pi}x}\,{\dd x \over 2\pi} = \pi\sum_{n = -\infty}^{\infty}\int_{-1}^{1}\delta\pars{k - 2n\pi}\,\dd k \\[3mm]&= \pi\sum_{n = -\infty}^{\infty}\Theta\pars{{1 \over 2\pi} - \verts{n}} = \pi\,\Theta\pars{1 \over 2\pi} = \pi \end{align}

Then, $$\color{#0000ff}{\large% \sum_{n = 1}^{\infty}{\sin\pars{n} \over n} = \half\pars{\pi - 1}} $$

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    $\begingroup$ For this you have to believe the computations with divergent series that physicists use. $\endgroup$ – GEdgar Jan 8 '14 at 1:09
  • $\begingroup$ @GEdgar You're right. Thanks. $\endgroup$ – Felix Marin Jan 8 '14 at 1:27
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What is the function that has its fourier coefficients $A_n=0$ and $B_n = \frac{1}{n}$, i.e $$ f(x) = \sum_{n=1}^{\infty} \frac{1}{n} \sin(n x)$$

Once you have figured out $f(x)$, find $f(1)$.

By the way $f(x)$ is a "standard" function in engineering analysis.

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By the way, note that $\frac{\sin n}{n}$ is exactly the n-th Fourier coefficient of the function $ \sqrt{\frac{\pi}{2}}\chi_{[-1,1]}(x). $ Since $\chi_{[-1,1]}(x)$ has a compact support, one can use the Poisson formula: $$ \sum_{n\in\mathbb{Z}} \widehat{\chi_{[-1,1]}}(n) = \sqrt{2\pi} \sum_{n \in \mathbb{Z}} \chi_{[-1,1]}(2\pi n) = \sqrt{2\pi}, $$ and get $$ \sum_{n\in\mathbb{Z}} \frac{\sin n}{n} = \pi. $$

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Hint: $$\sum_{n=1}^\infty \frac{\sin (n)}{n}= \text{Im }\sum_{n=1}^\infty \frac{e^{in}}{n}=\text{Im } \sum_{n=1}^\infty \int_0^1 x^{n-1} \mathrm{d} x \bigg|_{x=e^i}=\text{Im } \int_0^1 \frac{\mathrm{d} x}{1-x} \bigg|_{x=e^i}=\text{Im Log }(1-e^i) $$

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