2

How to evaluate the value of $$\sum_{n=0}^\infty \frac{\sin ((2n+1)\theta)}{(2n+1)}$$

My try:: I tried similar manipulating Proving a trig infinite sum using integration , but I am getting constant. $$\log ( 1 + e^{i \theta}) - \log (1 - e^{i \theta}) = 2 \sum_{n=1}^{\infty} \frac{e^{i (2n+1)}}{2n+1}$$ Comparing imaginary parts on both sides, I am getting $\displaystyle \sum_{n=0}^\infty \frac{\sin ((2n+1)\theta)}{(2n+1)} = \frac{\pi}{4}$?

Community
  • 1
Monkey D. Luffy
  • 1,632

2 Answers2

5

I'll post anyway since my answer is a little different, using $\mathrm{Im}(\log(z))=\mathrm{arg(z)}$: $$ \begin{align} \sum_{n=0}^\infty\frac{\sin((2n+1)\theta)}{2n+1} &=\mathrm{Im}\left(\sum_{n=0}^\infty\frac{e^{i(2n+1)\theta}}{2n+1}\right)\\ &=\mathrm{Im}\left(\frac12\log\left(\frac{1+e^{i\theta}}{1-e^{i\theta}}\right)\right)\\ &=\frac12\arg\left(\frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}\right)\\ &=\frac12\arg\left(i\cot(\theta/2)\right)\\[6pt] &=\frac\pi4\mathrm{sgn}(\cot(\theta/2))\\[6pt] &=\frac\pi4\mathrm{sgn}(\sin(\theta)) \end{align} $$

robjohn
  • 345,667
  • If you were to integrate both sides of the equation from $0$ to $\theta$, the left-hand side would become -$\sum_{n=0}^{\infty} \frac{\cos \left((2n+1)\theta \right)}{(2n+1)^{2}} + \frac{\pi^{2}}{8}. $ But what about the right-hand side of the equation? Can it be expressed in terms of the floor function? – Random Variable Nov 05 '20 at 14:19
  • Yes, but it is not simple: $$\frac\pi4\left(2\pi\left\lfloor\frac{\theta+\pi}{2\pi}\right\rfloor^2-2\pi\left\lfloor\frac{\theta}{2\pi}\right\rfloor^2-2\theta\left\lfloor\frac{\theta+\pi}{2\pi}\right\rfloor+2(\theta-\pi)\left\lfloor\frac{\theta}{2\pi}\right\rfloor+\theta\right)$$ – robjohn Nov 05 '20 at 15:02
  • No wonder I couldn't figure it out. I was hoping there was a relatively simple way to express that integral so that I could use it to evaluate something else. – Random Variable Nov 05 '20 at 15:24
  • It’s just a sawtooth wave, but expressing it in terms of floor gets messy. – robjohn Nov 05 '20 at 15:36
  • Yes, it's a simple-looking sawtooth wave, which is why I thought there had to be a simple way to express it in terms of the floor function. – Random Variable Nov 05 '20 at 15:59
  • The simplest formula seems to be $\frac\pi2\left|,\sin^{-1}(\sin(\theta/2)),\right|$ – robjohn Nov 05 '20 at 16:53
4

$$S(\theta) = \sum_{n=0}^{\infty} \dfrac{\sin((2n+1)\theta)}{2n+1} = \text{Imag}\left(\sum_{n=0}^{\infty} \dfrac{e^{i(2n+1)\theta}}{2n+1}\right) = \text{Imag}\left(\dfrac12 \ln \left(\dfrac{1+e^{i \theta}}{1-e^{i\theta}}\right)\right)$$ $$S(\theta) = \text{Imag}\left(\dfrac12 \ln \left(i \cot(\theta/2)\right)\right) = \dfrac12 \times \dfrac{\pi}2 = \begin{cases}\dfrac{\pi}4 & \text{if }\theta \in \left(2n \pi,(2n+1)\pi\right)\\ - \dfrac{\pi}4 & \text{if } \theta \in ((2n+1)\pi, (2n+2) \pi)\\ 0 & \text{if }\theta = n \pi \end{cases}$$

Edit: Another way is to look at the Fourier series of the function $$f(x) = \begin{cases}-\dfrac{\pi}4 & x \in(-\pi,0)\\ 0 & x = 0\\ \dfrac{\pi}4 & x \in (0,\pi) \end{cases}$$ And conclude from this. (Note that since the function is odd, the $\cos$ terms go away.)