Let $\theta(t)$ and $\phi(t)$ be two real $C^1$ functions $[0,2\pi]\rightarrow \mathbb{R}$. Let us assume $\theta$ has the properties $$ \int_0^{2\pi} e^{i\theta(t)} dt=0. $$ Geometrically this means that the curve obtained by integrating the (tangent) vector function $(\cos(\theta),\sin(\theta))$ over $(0,2\pi)$ is closed. I would like to prove the following claim

If the integral $$ \int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt=0 $$ for all $\lambda \in \mathbb{R}$ than $\phi$ is periodic of period $2\pi/l$ with $1\neq l\in\mathbb{N}$.

Note that the function $F(\lambda):=\int_0^{2\pi} e^{i(\theta(t)+\lambda\phi(t))} dt$ is analytic in $\lambda$ and therefore it is constantly equal to $0$ iff its derivatives $F^{(n)}(0)=\int_0^{2\pi} e^{i(\theta(t))} (\phi(t))^n dt$ vanish for all $n\in \mathbb{N}$.

OBSERVATION. If $\frac{d}{dt}\theta$ 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 $0$.

  • $\begingroup$ Is it not true that if $\phi$ is constant then $\theta$ can be any function satisfying the original conditions? $\endgroup$ – Nik Weaver Jun 13 at 7:56
  • 1
    $\begingroup$ Or if $\phi$ has even period, any $\theta$ with $\theta(t+\pi) = -\theta(t)$ would work. $\endgroup$ – Nik Weaver Jun 13 at 8:05
  • $\begingroup$ For the first comment, yes. This is providing a rotation of the original curve by changing the parameter I start measuring the turning angle from. You are right, my formulation include also this trivial case, which is something I do not want to consider. I am going to edit it. $\endgroup$ – Leonardo Jun 13 at 8:08
  • $\begingroup$ Concerning the second comment, good point. Can one at least deduce the periodicity of $\phi$ of some non-trivial period $2\pi/l$? $\endgroup$ – Leonardo Jun 13 at 9:00
  • 2
    $\begingroup$ Yes, creating a new post is generally preferred over changing the question after getting the answer you didn't want. But before you do that you might spend a little time thinking about the problem on your own, now that you understand why the original question had a negative answer... $\endgroup$ – Nik Weaver Jun 13 at 9:57

Your Answer

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

Browse other questions tagged or ask your own question.