# The integral of a function over $S^1$

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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Is the function over $S^1$ of period 1? Why? – Vladimir Feb 4 '12 at 15:39
The topology of $S^1$ identifies 0 and 1 so it forces functions to be periodic when viewed over $\mathbb{R}$. – Chris Janjigian Feb 4 '12 at 15:43
@Chris The period is uncertain? It depends on the context? – Vladimir Feb 4 '12 at 15:56
Well I mean in theory you could have a function with period less than one (take a function with period $\frac{1}{2}$ over $S^1$ then extend it over $\mathbb{R}$ to get a function with period $\frac{1}{2}$) The definition of the group $\mathbb{R}/ \mathbb{Z}$ says that if you take any $a \in [0,1)$ and add $n \in \mathbb{Z}$ to it, you stay in equivalence class of $a$. That says that $f(a + n) = f(a)$. Thus when viewed over $\mathbb{R}$ the function must have period 1. Again though it could also have period less than 1. – Chris Janjigian Feb 4 '12 at 16:09
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 Torus – AD. Feb 4 '12 at 17:48

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).
Why it is "of course equivalent to integration over any interval"? $\mathbb R/ \mathbb Z=\{a+\mathbb Z:a\in \mathbb R\}?$ What's the relationship between functions over $S^1$ and $\mathbb R$？ Why? I am quite confused about it. Help me please! – Vladimir Feb 4 '12 at 15:44
For any $t>0$ you may consider the group $t\mathbb{Z}$, and consider $[0,t]$ as $\mathbb{R}/t\mathbb{Z}$. – AD. Feb 4 '12 at 15:47
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.