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Let $S^1=\mathbb R/\mathbb Z,$ I was wondering how to calculate the integral of a function over $S^1$ and why. Like, $\int_{S^1}1 dx=?$ Given an "appropriate" function $f$, what is $\int_{S^1}f(x)dx?$

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    Maybe you should think through what your real question is? To me $S^1$ is the circle group (however since I do not know an other common notation is $\mathbb{T}$, which I prefer as$a$1-dimensional *T*orus2012-02-04

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A function on the circle group is the same thing as a periodic function on $\mathbb{R}$. In particular $\int_{S^1}f(x)dx=\int_0^1 f(x)dx$ in the case when we look upon $S^1=\mathbb{R}/\mathbb{Z}$ (this is of course equivalent to integration over any interval).

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    For any t>0 you may consider the group $t\mathbb{Z}$, and consider $[0,t]$ as $\mathbb{R}/t\mathbb{Z}$.2012-02-04
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The integral over $S^1$ is exactly what you think it is. If you have a copy of Rudin Real & Complex around, you may want to run through the section on the trigonometric system and see how he handles integrating over the torus. Basically you can think of it as being [0,1) topologized so that 0 and 1 are identified.

You can also phrase your question as "what is the Haar measure on $\mathbb{R}/ \mathbb{Z}$?" Since this group in its natural topology is compact any translation invariant measure is unique up to a constant. Since the Lebesgue measure is translation invariant on $S^1$ we know that the Haar measure is the Lebesgue measure. It is customary to choose the scaling constant to be 1 so that the integral is normalized.