# Evaluate $\sum_{n=1}^{\infty} \frac{\sin \ n}{ n }$ using the fourier series

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 ?


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

• For this you have to believe the computations with divergent series that physicists use. – GEdgar Jan 8 '14 at 1:09
• @GEdgar You're right. Thanks. – Felix Marin Jan 8 '14 at 1:27

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.

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 $$\sum_{n\in\mathbb{Z}} \frac{\sin n}{n} = \pi.$$

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