Proving a trig infinite sum using integration

Question

How can I prove the following using integration and elementary functions?

Prove that:

$$\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \frac{\pi}{2} - \frac{\theta}{2}$$

$0 < \theta < 2\pi$

Answer 1

Let, $$S_1 = \sum_{n=1}^{\infty}\frac{\cos n\theta}{n}\\ S_2 = \sum_{n=1}^{\infty}\frac{\sin n\theta}{n}$$

Then $$S_1 + iS_2 = \sum_{n=1}^{\infty}\frac{\cos(n\theta)+i\sin(n\theta)}{n}=\sum_{n=1}^{\infty}\frac{e^{in\theta}}{n}$$

Now, from the Taylor expansion, $\ln (1+x) = x -\frac{x^2}{2}+\frac{x^3}{3} ...$ $$\implies -\ln(1-x) = x+ \frac{x^2}{2}+\frac{x^3}{3} ... = \sum_{n=1}^{\infty}\frac{x^n}{n}$$ $$\begin{align} \therefore S_1+iS_2 &= -\ln(1-e^{i\theta}) \\&=-\ln(1-\cos\theta-i\sin \theta) \\ &=-\ln(2\sin^2\theta/2 - 2i\sin(\theta/2)\cos(\theta/2)) \\ &=-\ln(2\sin\theta/2)-\ln(\sin\theta/2-i\cos\theta/2) \\ &=-\ln(2\sin\theta/2)+\ln(\sin\theta/2+i\cos\theta/2) \\ &=-\ln(2\sin\theta/2)+\ln(e^{i(\pi/2-\theta/2)}) \\ &=-\ln(2\sin\theta/2)+i(\pi/2-\theta/2) \end{align} $$

Taking the imaginary part of both sides, $$S_2 = \frac{\pi}{2} - \frac{\theta}{2}$$

Answer 2

$$ I=\sum_{n=1}^\infty \frac{\sin(n\theta)}{n} $$ Now, \begin{align} \frac{dI}{d\theta} &= \sum_{n=1}^\infty \cos(n\theta)\\ \frac{dI}{d\theta}\cos(\theta)&=\sum_{n=1}^\infty \cos(n\theta)\cos(\theta)\\ &=\sum_{n=1}^\infty \frac{\cos((n-1)\theta)+\cos((n+1)\theta)}2\\ &=\sum_{n=0}^\infty \frac{\cos(n\theta)}{2}+\sum_{n=2}^\infty\frac{\cos(n\theta)}{2}\\ &=\frac{1}2+\sum_{n=1}^\infty \frac{\cos(n\theta)}{2}+\sum_{n=1}^\infty\frac{\cos(n\theta)}{2}-\frac{\cos(\theta)}2\\ &=\frac{1-\cos(\theta)}{2}+\frac{dI}{d\theta}\\ \frac{dI}{d\theta}(\cos(\theta)-1) &= -\frac{\cos(\theta)-1}{2}\\ \frac{dI}{d\theta} &= -\frac12 \end{align} Noting that $$ I(\pi) = \sum_{n=1}^\infty \frac{\sin(n\pi)}n = 0 $$ we integrate around $\theta=\pi$ to get $$ I = -\frac\theta2 + \frac\pi2 = \frac\pi2-\frac\theta2 $$ Note that this doesn't strictly require that the $\cos$ sum converges, as we may alter the summation process to obtain convergence. What is important is which terms may be extracted for the purposes of the integration.

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There are two obvious ways to handle the nonconvergent nature of the sum for $\frac{dI}{d\theta}$.

Option 1: use $$I = \sum_{n=1}^\infty \frac{\sin(n\theta)}n z^n$$ and then take the limit as $z\to1^{-}$. For any $|z|<1$, the sum in the derivative will converge, and in the limit it will be $\frac12$.

Option 2: Change the order of summation. Let $$ I = \sum_{n=1}^\infty \frac{\sin(n\theta)}n \sum_{k=1}^\infty 2^{-k} $$ which is the same as multiplying by $1$. Now change the order of summation to $n+k=m$ first, as $$ I = \sum_{m=2}^\infty \sum_{k=1}^{m-1} \frac{\sin((m-k)\theta)}{m-k}2^{-k} $$ When summed in this order, the derivative converges.

Answer 3

In this answer, I show

$$ \begin{align} \sum_{k=1}^\infty\frac{\sin(2kx)}{k} &=\sum_{k=1}^\infty\frac{e^{i2kx}-e^{-i2kx}}{2ik}\\ &=\frac1{2i}\left(-\log(1-e^{i2x})+\log(1-e^{-i2x})\right)\\ &=\frac1{2i}\log(-e^{-i2x})\\[4pt] &=\frac\pi2-x\quad\text{for }x\in\left(0,\pi\right) \end{align} $$

which is, in essence, milind's answer. However, the question asks about integration. This sounds as if the question is asking to find the Fourier Series of $f(\theta)=\frac\pi2-\frac\theta2$. First, note that $f(\theta)$ is odd; that is, $$ \begin{align} f(2\pi-\theta) &=\frac\pi2-\frac{2\pi-\theta}2\\ &=\frac\theta2-\frac\pi2\\ &=-\left(\frac\pi2-\frac\theta2\right)\\[6pt] &=-f(\theta)\tag{1} \end{align} $$ Equation $(1)$ implies that $$ \begin{align} \color{#00A000}{\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta} &=\int_0^{2\pi}f(2\pi-\theta)\cos(n 2\pi-n\theta)\,\mathrm{d}\theta\\ &=-\color{#00A000}{\int_0^{2\pi}f(\theta)\cos(n\theta)\,\mathrm{d}\theta}\\[6pt] &=0\tag{2} \end{align} $$ because $\color{#00A000}{x}=-\color{#00A000}{x}\implies\color{#00A000}{x}=0$.

Now the question is $$ \begin{align} \int_0^{2\pi}f(\theta)\sin(n\theta)\,\mathrm{d}\theta &=\int_0^{2\pi}\left(\frac\pi2-\frac\theta2\right)\sin(n\theta)\,\mathrm{d}\theta\\ &=-\frac1n\int_0^{2\pi}\left(\frac\pi2-\frac\theta2\right)\,\mathrm{d}\cos(n\theta)\\ &=\left.-\frac1n\left(\frac\pi2-\frac\theta2\right)\cos(n\theta)\right]_0^{2\pi}\\ &\hphantom{=\,}+\frac1n\int_0^{2\pi}\cos(n\theta)\,\mathrm{d}\left(\frac\pi2-\frac\theta2\right)\\[4pt] &=\frac\pi{n}\tag{3} \end{align} $$ $(3)$ says that the Fourier series for $f(\theta)$ on $(0,2\pi)$ is $$ f(\theta)=\sum_{n=1}^\infty\frac{\sin(n\theta)}{n}\tag{4} $$