It is known that one can sometimes derive certain Fourier series without alluding to the methods of Fourier analysis. It is often done using complex analysis. There is a way of deriving the formula $$\sum_{k = 1}^\infty \frac{\sin(kz)}{k} = \frac{\pi - z}{2}$$ using complex analysis for some $z$. In other words, it can be shown that $\sum_{k = 1}^\infty \sin(kz)/k$ converges to $(\pi - z)/2$ for some $z$. My question is: How can we show that the formula above holds for $z \in (0, 2\pi)$ without using Fourier analysis?


Using blue's suggestion I concocted the following proof.

Let $\log z$ be the principal value of the logarithm (with the branch cut along the negative real axis). Recall that $e^{iz} - e^{-iz} = 2i\sin z$ and $\text{Arg}(z) = \text{Arg}(z^*)$. Furthermore, we have \begin{align*} \log(1 - e^{iz}) &= -\sum_{k = 1}^\infty \frac{e^{ikz}}{k} \\ &= \log|1 - e^{iz}| + i\theta \\ &= \log\left|2\sin\frac{z}{2}\right| + i\theta, \end{align*} where $\theta = \text{Arg}(1 - e^{iz})$. Now, write $1 - e^{iz} = 1 - \cos z - i\sin z$ and let $z \in (0, \pi)$. Then $$\tan \theta = \frac{\sin z}{1 - \cos z} = \frac{\cos(z/2)}{\sin(z/2)} = \tan\left(\frac{\pi}{2} - \frac{z}{2}\right).$$ Hence, $\theta = \pi/2 - z/2$. Moreover, since $z \in (0, \pi)$, we have $\theta \in (0, \pi/2)$. Using Euler's formula $e^{iz} = \cos z + i \sin z$ and the fact that the sine function is odd, we see that equating the imaginary parts gives $$\sum_{k = 1}^\infty \frac{\sin kz}{k} = \frac{\pi - z}{2}, \quad z \in (-\pi, 0) \cup (0, \pi).$$ Lastly, observe that $(-\pi, 0) \cup (0, \pi)$ can be replaced by $(0, 2\pi)$ due to the periodicity of the sine function.

  • 1
    $\begingroup$ This one is closely related to your question and it is easily evaluated by means of Abel-Plana Formula. $\endgroup$ – Felix Marin Jun 28 '14 at 2:03

In order to make sense of sines with complex arguments, you need to decompose them into complex exponentials. After that notice you're working with two series, each a Taylor expansion of the natural logarithm function (written appropriately). Proceed from there.

  • $\begingroup$ Thanks. I just added a proof using your suggestion. Is that what you meant? $\endgroup$ – glebovg Jun 28 '14 at 0:04
  • $\begingroup$ Is there a shorter way? $\endgroup$ – glebovg Jun 28 '14 at 0:52
  • $\begingroup$ @glebovg Yes that is what I meant. Dunno if it can be shortened, but it's the thing that immediately came to mind when you asked for non-Fourier arguments. $\endgroup$ – blue Jun 28 '14 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.