Finding $\varphi(\theta)$ such that $\sum_{n\geq 0} e^{in\theta}/(n+1) = \sum_{n \geq 0} \hat{\varphi} e^{in\theta}$

Question

I have a fourier-series $$ \sum_{n = 0}^\infty \frac{1}{n+1} e^{in\theta} $$ How can I find a function $\varphi(\theta)$ which has the Fourier-coefficients $1/(n+1)?$ In other words, finding $\varphi$ such that $$ \frac{1}{n+1} = \hat{\varphi}(n) = \frac{1}{2\pi} \int_{0}^{2\pi} \varphi(\theta) e^{-in\theta}\,\mathrm{d}\theta $$ Is there a systematic approach, or is it all down to guess-work?

Answer 1

This is too long for Comments, so I write this here.

Since $ie^{-i\theta }(\pi-\theta )$ has Fourier coeficients $$ c_n=\int_0^{2\pi}ie^{-i\theta }(\pi-\theta )e^{-in\theta }d\theta =\frac{1}{n+1}\quad (n\ne -1, n=0,1,\pm 2,\pm 3,...), $$ $$ c_{-1}=0, $$

and $ie^{-i\theta }(\pi-\theta )$ is continuous in $0<\theta<2\pi$, we see by the general theory of Fourier series that $$ \sum_{n=-\infty,n\ne -1}^\infty \frac{1}{n+1}e^{in\theta}=ie^{-i\theta}(\pi-\theta). $$ But when the series $$ \sum_{n=-\infty,n\ne -1}^\infty \frac{1}{n+1}e^{in\theta} $$

is given, how can we get it's closed form without using any results of Fourier series?
Since $$ \sum_{n=-\infty,n\ne -1}^\infty \frac{1}{n+1}e^{in\theta} =2ie^{-i\theta }\sum_{n=1}^\infty \frac{\sin n \theta }{n}, $$ the problem reduces how to sum the series $$ \sum_{n=1}^\infty \frac{\sin n \theta }{n} $$ by direct methods.
Perhaps residue methods is useful.

EDIT: See, for instance, Schaum's Outlines, Complex Analysis by Spiegel and others, McGraw-Hill, Problem 7.65 at page 237. It states $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}n\sin n\theta }{n^2+\alpha ^2}=\frac{\pi}{2}\frac{\sinh \alpha\theta }{\sinh \alpha \pi},\, -\pi<\theta <\pi.$$
When $\alpha =0$ we have $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}\sin n\theta }{n}=\frac{\pi}{2}\quad ( -\pi<\theta <\pi), $$ which leads to$$ \sum_{n=1}^\infty \frac{\sin n\theta }{n}=\frac{\theta -\pi}{2}\quad (0<\theta <2\pi). $$

Answer 2

All such problems become power series problems, which gives you a method to try. And it works for this case. For $|z| < 1$, $$ \frac{d}{dz}\sum_{n=0}^{\infty}\frac{1}{n+1}z^{n+1}=\sum_{n=0}^{\infty}z^{n}=\frac{1}{1-z} \\ \sum_{n=0}^{\infty}\frac{1}{n+1}z^{n+1}=-\log(1-z)+C. $$ By Abel's theorem, you may take limits to obtain the series on the boundary of the disk: $$ \sum_{n=0}^{\infty}\frac{1}{n+1}e^{i(n+1)\theta}=-\log(1-e^{i\theta})+C, \;\;\; 0 < \theta < 2\pi, $$ assuming you use the branch of the log that is continuous on $0 < \theta < 2\pi$. The constant $C$ is determined by setting $\theta=\pi$: $$ \sum_{n=0}^{\infty}\frac{1}{n+1}(-1)^{n+1}=-\log(2)+C. $$ Verify: For $n > 0$, integration by parts works, but you have to use a special anti-derivative of $e^{in\theta}$ that will promote convergence at $\theta=0,2\pi$ of the evaluation terms: \begin{align} \int_{0}^{2\pi}e^{-in\theta}\log(1-e^{i\theta})d\theta & = \left.\frac{e^{-in\theta}-1}{-in}\log(1-e^{i\theta})\right|_{0}^{2\pi} +\int_{0}^{2\pi}\frac{e^{-in\theta}-1}{-in}\frac{1}{1-e^{i\theta}}e^{i\theta}d\theta \\ & = \frac{1}{-in}\int_{0}^{2\pi}\frac{e^{-in\theta}-1}{e^{-i\theta}-1}d\theta \\ & = \frac{1}{-in}\int_{0}^{2\pi}(e^{-i(n-1)\theta}+e^{-i(n-2)\theta}+\cdots+1)d\theta \\ & = \frac{2\pi}{-in} \end{align} The coefficients for $n < 0$ are $0$ because $\log(1-z)$ is holomorphic inside the disk. For $n=0$ you must take into account $C$ as derived above, which can be ignored for $n > 0$, as it was above.