PROBLEM. Let $ \theta(t)$ and $ \phi(t)$ be two real analytic non-constant functions $ [0,2\pi]\rightarrow \mathbb{R}$ . I am trying to prove the following claim

If the integral $ $ \int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt=0 $ $ for all $ n\in\mathbb{N}_0$ than the first derivative $ \theta’$ and $ \phi$ are periodic of common period $ 2\pi/l$ with $ 1\neq l\in\mathbb{N}$ .

Note that this is equivalent to $ F(\lambda):=\int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt=0$ for all $ \lambda \in \mathbb{R}$ . In fact, $ F(\lambda)$ is analytic in $ \lambda$ and its being constantly equal to 0 is equivalent to the vanishing of all its derivatives $ F^{(n)}(0)=\int_0^{2\pi} e^{i\theta(t)} (\phi(t))^n dt$ . Geometrically this means that the curve obtained by integrating the (tangent) vector function $ (\cos(\theta+\lambda\phi),\sin(\theta+\lambda\phi))$ over $ [0,2\pi]$ is closed $ \forall \lambda$ .

Just in case, a back-up less general claim for which I would like to see a clean solution is

If, in the hypotesis above, $ \phi$ is a polynomial, then $ \phi$ is constantly $ 0$ .

OBSERVATION. If $ \theta’$ and $ \phi$ are periodic of common period $ \frac{2\pi}{l}$ with $ 1\neq l \in \mathbb{N}$ and $ \int_0^{\frac{2\pi}{l}} e^{i\theta}\neq 0$ then the converse implication is true. In fact, in this setting $ \theta=c\cdot t+\theta_p(t)$ with $ c=\frac{2\pi}{l}(\theta(\frac{2\pi}{l})-\theta(0))$ and $ \theta_p$ periodic of period $ \frac{2\pi}{l}$ . Then $ $ \begin{align} \int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt &=& \sum_{j=0}^{l-1} \int_{j \frac{2\pi}{l}}^{(j+1) \frac{2\pi}{l}} e^{i(c\cdot t+\theta_p(t)+\lambda\phi(t))} dt \ &=& \sum_{j=0}^{l-1} e^{i\cdot j \cdot \frac{2\pi}{l}} \int_{0}^{\frac{2\pi}{l}} e^{i(c\cdot t+\theta_p(t)+\lambda\phi(t))} dt, \end{align} $ $ where the last equality is obtained by repetedly applying the substitution $ t’=t-\frac{2\pi}{l}$ . Since we know $ \sum_{j=0}^{l-1} e^{i\cdot j \cdot \frac{2\pi}{l}} \int_{0}^{\frac{2\pi}{l}} e^{i\theta(t)}dt=\int_0^{2\pi} e^{i\theta(t)} dt=0$ then also the integral above must be $ 0$ . In the following picture the curve associated to $ \theta(t)=t + \cos( 12 t)$ deformated in the direction $ \cos(3 t)$ . In this case $ l=3$ and the curve is closed $ \forall \lambda$ .

$ \theta(t)=t + \cos( 12 t)$ deformated in the direction $ \cos( 3 t)$ . In this case $ l=3$ and the curve is closed $ \forall \lambda$ .”>

IDEA. If $ \theta$ monotone one can substitute $ s=\theta(t)$ in the integral and get $ $ \int_{\theta(0)}^{\theta(2\pi)} e^{i s} \frac{(\phi(\theta^{-1}(s)))^n}{\theta'(\theta^{-1}(s))} ds=0. $ $ In this case the idea behind the hypotesis becomes apperent: $ \phi(\theta^{-1}(s))$ is periodic of non-trivial period iff $ \phi$ and $ \theta’$ have the common period property. It seems here that looking at the Fourier expansion of our functions on $ [\theta(0),\theta(2\pi)]$ could be a good idea: the condition we have means indeed that, $ \forall n$ , the first harmonic of the function $ \frac{(\phi(\theta^{-1}(s)))^n}{\theta'(\theta^{-1}(s))}$ is $ 0$ . Fourier coefficients of a product are obtained by convolutions and therefore the condition above becomes, $ \forall n$ : $ $ \sum_{k_n=-\infty}^{+\infty} \sum_{k_{n-1}} … \sum_{k_{2}}\sum_{k_{1}} \widehat{\frac{1}{\theta’}}(1-\sum_{i=1}^{n} k_i) \prod_{i=1}^{n} \widehat{\phi}(k_i)=0. $ $ Is this approach viable? Can one from here exploit the fact that a function is periodic of non-trivial period iff there exists $ k$ such that only harmonics multiple of $ k$ are different from 0? Other way round, do non-zero harmonics of coprime orders imply a contradiction with our constraints? As for a toy example, if $ \theta(t)=t$ ,$ \theta'(s)=1$ and $ \phi(s)=\cos(2s)+\cos(3s)$ already $ \widehat{f^2}(1)= 2 \widehat{f}(3)\widehat{f}(-2) \neq 0$ ; in the general setting interaction of coefficients is not straightforward.

NOTE: This question originated from Orthogonality relation in $ L^2$ implying periodicity. As suggested in the comments to the previous post, since the target of the question changed over time and edits were major, here I hope I gave a clearer and more consistent presentation of my problem.

Thank you for your time.