period of the curvature and the period of the corresponding curve $\frac1{2\pi}\int_0^{\rho_k}k(s)\mathrm ds\in\Bbb Z$ The paper " When Is a Periodic Function the Curvature of a Closed Plane Curve?", May 2008 The American Mathematical Monthly 115(5):405-414
The paper says: A Closed Plane Curve, when the (minimum) period of the (signed) curvature and the period of the corresponding curve do not coincide, if and only if
$$\frac1{2\pi}\int_0^{\rho_k}k(s)\mathrm ds\in\Bbb Q-\Bbb Z$$
where $\rho_k$ is the (minimum) period of the signed curvature.
Please refer to Proposition 2.1 and   the beginning of part 4 of the paper：if  $\frac1{2\pi}\int_0^{\rho_k}k(s)\mathrm ds\in Z$, then in this case the (minimum) period of the curvature and the period of the corresponding curve would coincide if the latter eventually closes up.

But the paper dones't prove this. How to show the claim is correct? Thanks
the paper : Wherher take a look at it, doesn't affect the topic here
When Is a Periodic Function the Curvature of a Closed Plane Curve?
 A: Suppose the plane curve closes up and has period $\rho$.  Then we know that the (minimum) period $\rho_\kappa$ of the curvature divides evenly into $\rho$, say $\rho = n \rho_k$.  We want to show $\rho = \rho_k$, i.e., $n = 1$.  By equation (2.3) of the paper, it suffices to show $$\int_s^{s+\rho_k} \exp \left( i \int_0^u \kappa(t) \, dt \right) \, du = 0.$$
Let $m = \frac{1}{2\pi} \int_0^{\rho_\kappa} \kappa(s) \, ds \in \mathbb{Z}$.  Performing the same algebraic manipulations as in (2.6) and (2.7), we find
$$\begin{equation}
\tag{$\dagger$}
\int_s^{s+\rho} \exp \left(i \int_0^u \kappa(t) \, dt \right) \, du = \left\{ \sum_{j=0}^{n-1} \exp(2\pi i m j) \right\} \int_s^{s+\rho_\kappa} \exp \left(i \int_0^u \kappa(t) \, dt \right) \, du.
\end{equation}$$
The key point now is that $\sum_{j=0}^{n-1} \exp(2\pi i m j) \neq 0$.  In fact, the sum is equal to $n$.  However, the left hand side of ($\dagger$) is equal to zero by the assumption that the curve closes up.  Therefore, the other factor on the right is zero, which is what we wanted to show.
