Does there exist a simple smooth closed curve $\gamma:S^1\to \mathbb C$ such that $$ \int_{0}^{2\pi} e^{\gamma(e^{it})} \, |\gamma'(e^{it} )|\,dt =0?$$
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4$\begingroup$ What about $\gamma(z)\equiv z$? $\endgroup$– Iosif PinelisCommented Oct 28, 2022 at 17:17
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$\begingroup$ @IosifPinelis but this does not make sense since I am asking for $\gamma$ to be a closed contour. Here $S^1$ is the torus. $\endgroup$– AliCommented Oct 28, 2022 at 18:00
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1$\begingroup$ @Ali: Your notation is a bit confusing since the symbol $t$ would suggest a real parameter, but then your integration is over $S^1$. Similarly, $\gamma'(t)$ is easily interpreted for real $t$, but not for a complex $t$ when $\gamma$ is just smooth (not holomorphic). $\endgroup$– Christian RemlingCommented Oct 28, 2022 at 19:11
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$\begingroup$ Perhaps, I don't understand your notations/definitions. Usually, $S^1$ denotes the unit circle. In you comment, you say that $S^1$ is "the torus". Then $S^1$ must be the one-dimensional "torus" (that is, the unit circle), in order for $\gamma$ to be a curve indeed. So, it seems that then the identity map of $S^1$ is a simple smooth closed curve (if anything is a simple smooth closed curve in your setting). $\endgroup$– Iosif PinelisCommented Oct 28, 2022 at 19:12
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$\begingroup$ The integral is just supposed to be the integral of $e^z$ over a simple closed curve with respect to its lebesgue induced measure. $\endgroup$– AliCommented Oct 28, 2022 at 19:22
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1 Answer
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After the change of the variable $w=\gamma(t)$ your integral becomes $$\int_\gamma e^wds,$$ where $ds$ is the length element, $ds=\sqrt{dx^2+dy^2}$, and $\gamma$ is a Jordan curve of zero index about the origin. Now notice that on every vertical segment of length $2\pi k$, where $k$ is an integer, this integral is zero. Now construct a simple closed curve, of zero index about the origin, consisting of 8 straight segments, 4 vertical and 4 horizontal. All vertical segments have length $2\pi k_j$ and all hosizontal segments lie on the lines $y=\pi/2+\pi k_j$, and the curve is symmetric with respect to complex conjugation. Such a curve is easy to draw, and the integral along it is 0. Indeed it is zero on each vertical segment, while contributions from horizontal segments cancel.
Let me add how to make the curve smooth. We will slightly modify the original curve, keeping the vertical segments unchanged, so their contribution to the integral is zero. We will also keep the symmetry with respect to the complex conjugation, so that imaginary part of the integral will be 0.
First we choose the vertical segments $\mathrm{Re} z=x_k$ so that for some segments $\cos x_j<0$ and for some other $\cos x_j>0$, so that the contribution to the integral of some segments is positive and for other segments it is negative. Then, to make the curve smooth, we replace some small pieces of horizontal segments near the corners by pieces which make the curve smooth. Since the contribution of some of these pieces is negative and others positive, it is clear that the pieces can be chosen so that their contribution cancels.
answered Oct 28, 2022 at 19:26
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$\begingroup$ The OP requested that the curve be smooth. Is your curve smooth? $\endgroup$ Commented Oct 28, 2022 at 22:27
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$\begingroup$ It is piecewise smooth. But it is possible to round the corners, and keep integral zero. $\endgroup$ Commented Oct 29, 2022 at 1:59
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$\begingroup$ I am wondering how this can be done: "to round the corners, and keep integral zero". $\endgroup$ Commented Oct 30, 2022 at 2:00
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$\begingroup$ On the other hand, if the smoothness condition is disregarded, it is unclear to me why you need $8$ (rather than $4$) straight segments and why you need the curve to be of zero index. $\endgroup$ Commented Oct 30, 2022 at 14:00
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$\begingroup$ @Josif Pinelis: you are right: 4 straight segments is enough. On how to round the corners, I will add to my ans. $\endgroup$ Commented Oct 30, 2022 at 15:14