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Consider the function $z(s)\in\mathbb{C}$ defined as $z(s)=\int_0^s \exp\left[i(q u+\lambda(u))\right]du$ for some $q\in\mathbb{Q}-\mathbb{Z}$ and $\lambda(s)$ a $2\pi$-periodic real differentiable function. Are there nontrivial $\lambda(s)$ for which $z(s)$ has a closed form? Special functions are entirely acceptable but should be properly cited.

This question is related to some other recent problems of mine [1] [2]; background on the construction used here can be found in an old AMM article [3]. To sum up: Let $\kappa(s)$ be the curvature of an arc-length parametrized closed curve $z(s)\in\mathbb{C}$. We may then express $z(s)$ as above if we identify $\lambda'(s)=\kappa(s)-q$ i.e. the deviation from the mean curvature.

If $q=m/n\in \mathbb{Q}-\mathbb{Z}$ in lowest terms, then $z(s)$ will be $2\pi n$-periodic. Moreover, the fact that $z(s)$ is arc-length parametrized (note that $|z'(s)|=1$) means that the perimeter is simply $2\pi n$. So the perimeter of a wide class of curves may be specified quite generically. (Note that in general the resulting curve will have self-intersections; a non-intersecting example is preferable.)

What should come next is to compute the area of the curve $C$ determined by some convenient $\lambda(s)$ (preferably one with some free parameters). But this amounts to the integral $\frac{1}{2i}\oint_C \overline{z}\,dz$, which despite its simple statement does not appear easily approachable for generic $\lambda(s)$. Thus I wanted to look for a suitable choice of $\lambda(s)$ such that $z(s)$ could be computed explicitly. So far, though, searching through tables of integrals has lead nowhere. Does anyone know a tractable case?

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    $\begingroup$ A minor simplification, unless I'm missing something: $\kappa(u)$ is integrable and $2\pi$-periodic with a zero average iff $\lambda(u)$ is $2\pi$-periodic and differentiable with $\lambda'(u) = \kappa(u)$, so you really just seek a $2\pi$-periodic differentiable function $\lambda:\mathbb R \to \mathbb R$ such that $\int_0^s \exp\bigl\{i[qu+\lambda(u)]\bigr\}\,du$ has a closed form. $\endgroup$ – Antonio Vargas Aug 26 '14 at 4:50
  • $\begingroup$ I deleted my answer because Antonio Vargas has already said the same thing. $\endgroup$ – Ian Mateus Aug 26 '14 at 5:02
  • $\begingroup$ @AntonioVargas: After thinking about it, your statement of the question seems more to the point and I'll use it in the question instead. Thanks! $\endgroup$ – Semiclassical Aug 26 '14 at 15:26

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