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 $\forall \lambda$.

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.

highlynontrivial way. Just rotate 3 opposite pairs in some independent ways for a while recovering the original star after 30 degree rotation and then move the 2 equilateral triples in some independent ways arriving at the 60 degree rotation. From pairs $\ell$ can be only $2$, but from triples it can be only $3$, so no $\ell$ exists in the end. $\endgroup$ – fedja Jun 17 '19 at 22:03