A Regular Sine Series with All $k$

Consider the Fourier series of the one-cycle unit-amplitude sawtooth wave, which is just the linear function $\theta/\pi$ between $-\pi$ and $\pi$. As is well known it is given by the sine series

$$S_{\rm s} = -\, \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\, \sin(k\theta).$$

The corresponding FC series is then

$$\bar{S}_{\rm s} = -\, \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\, \cos(k\theta),$$

the complex $S_{v}$ series is given by

$$S_{v} = -\, \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\, v^{k},$$

and the complex power series $S_{z}$ is given by

$$S_{z} = -\, \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\, z^{k}.$$

The ratio test tells us that the disk of convergence of $S_{z}$ is the unit disk. If we consider the inner analytic function $w(z)$ within this disk we observe that $w(0)=0$, as expected. We have for the function and its derivative

$$\begin{eqnarray*} w(z) & = & -\, \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(... ...\\ & = & -\, \frac{2}{\pi z} \sum_{k=1}^{\infty} (-z)^{k}. \end{eqnarray*}$$

This is a geometrical series and therefore we may write the derivative in closed form,

$$\begin{eqnarray*} \frac{dw(z)}{dz} & = & -\, \frac{2}{\pi z}\, \frac{-z}{1+z} \\ & = & \frac{2}{\pi}\, \frac{1}{1+z}. \end{eqnarray*}$$

This function has a simple pole at the point where the one-cycle sawtooth wave is discontinuous. This is a hard singularity of degree $1$. It is now very simple to integrate in order to obtain the inner analytic function $w(z)$ in closed form, remembering that we should have $w(0)=0$,

$$w(z) = \frac{2}{\pi}\, \ln(1+z).$$

This function has a logarithmic singularity at the point where the one-cycle sawtooth wave is discontinuous. This is a borderline hard singularity. If we now rationalize the argument of the logarithm in order to write it in the form

$$1+z = R\,{\rm e}^{\mbox{\boldmath$\imath$}\alpha},$$

then the logarithm is given by

$$\ln\!\left(1+z\right) = \ln(R)+\mbox{\boldmath$\imath$}\alpha,$$

which allows us to identify the real and imaginary parts of $w(z)$. We may write for the argument

$$\begin{eqnarray*} R\,{\rm e}^{\mbox{\boldmath$\imath$}\alpha} & = & 1 + \rh... ... \mbox{\boldmath$\imath$} \left[ \rho \sin(\theta) \right]. \end{eqnarray*}$$

From this it follows, after some algebra, that we have

$$\begin{eqnarray*} R(\rho,\theta) & = & \sqrt{1+2\rho\cos(\theta)+\rho^{2}}, ... ...tan\! \left[\frac{\rho\sin(\theta)}{1+\rho\cos(\theta)}\right]. \end{eqnarray*}$$

The part which is of interest now is the imaginary one, which is related to the series $S_{\rm s}$,

$$\Im[w(z)] = \frac{2}{\pi}\, \arctan\! \left[\frac{\rho\sin(\theta)}{1+\rho\cos(\theta)}\right].$$

The $\rho\to 1$ limit of this quantity is the function $f_{\rm s}(\theta)$. If we consider the case $\theta=\pm\pi$, we immediately get zero, because the argument of the arc tangent is zero, and therefore we get $\alpha=0$. Therefore we have

$$\begin{eqnarray*} f_{\rm s}(\pm\pi) & = & \lim_{\rho\to 1} \Im[w(\rho,\pm\pi)] \\ & = & 0. \end{eqnarray*}$$

This is the correct value for the Fourier-series representation of the one-cycle sawtooth wave at this point. For $\theta\neq\pm\pi$ we get

$$\begin{eqnarray*} f_{\rm s}(\theta) & = & \lim_{\rho\to 1} \Im[w(z)] \\ &... ...i} \arctan\! \left[\frac{\sin(\theta)}{1+\cos(\theta)}\right]. \end{eqnarray*}$$

In order to solve this we write everything in terms of $\theta/2$, and thus obtain

$$\begin{eqnarray*} f_{\rm s}(\theta) & = & \frac{2}{\pi} \arctan\! \left[\fr... ...an\!\left[\tan(\theta/2)\right] \\ & = & \frac{\theta}{\pi}. \end{eqnarray*}$$

This is the correct value for the one-cycle unit-amplitude sawtooth wave. The FC function is related to the real part

$$\Re[w(z)] = \frac{1}{\pi}\, \ln\!\left[1+2\rho\cos(\theta)+\rho^{2}\right].$$

In the $\rho\to 1$ limit this function is singular at $\theta=\pm\pi$. Away from this logarithmic singularity we may write

$$\begin{eqnarray*} \bar{f}_{\rm s}(\theta) & = & \lim_{\rho\to 1} \Re[w(z)] ... ... & = & \frac{1}{\pi}\, \ln\!\left[4\cos^{2}(\theta/2)\right]. \end{eqnarray*}$$

We see therefore that the FC function to the one-cycle sawtooth wave is a function with a logarithmic singularity at the point where the original function is discontinuous. The derivative of this conjugate function is easily calculated, and turns out to be

$$\frac{d\bar{f}_{\rm s}(\theta)}{d\theta} = \frac{1}{\pi}\, \frac{\cos(\theta)-1}{\sin(\theta)}.$$

This is a quite a simple function, with a simple pole at $\theta=\pm\pi$. Note that the logarithmic singularity of the FC function $\bar{f}_{\rm s}(\theta)$ is an integrable one, as one would expect since the Fourier coefficients are finite.

Note that we get a free set of integration formulas out of this effort. Since the coefficients $a_{k}$ can be written as integrals involving $\bar{f}_{\rm s}(\theta)$, we have at once that

$$\int_{-\pi}^{\pi}d\theta\, \ln\!\left[4\cos^{2}(\theta/2)\right] \cos(k\theta) = -\, \frac{2\pi(-1)^{k}}{k},$$

for $k>0$.