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I have frequently seen this symbol used in advanced books in physics:

$$\oint$$

What does the circle over the integral symbol mean? What kind of integral does it denote?

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up vote 8 down vote accepted

It's an integral over a closed line (e.g. a circle), see line integral.

In particular, it is used in complex analysis for contour integrals (i.e closed lines on a complex plane), see e.g. example pointed out by Lubos.

Also, it is used in real space, e.g. in electromagnetism, in Faraday's law of induction (part of the Maxwell equations, written in an integral form):

$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A} $$ saying that the generated voltage (an integral of electric field along a circle) is the same as the time derivative of the magnetic flux.

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So it is only a normal line integral where the line C is closed? – R. M. Sep 28 '12 at 17:54
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@R.M. It is indeed just a normal integral. The circle is there to remind us that the domain of integration, whether it be 1D or 2D or whatever, is closed, in just the same way we could put multiple integral signs to remind us how many dimensions we're in. – Chris White Sep 28 '12 at 19:38
    
@ChrisWhite Perfect, thank you! – R. M. Sep 28 '12 at 19:52
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The first animation on the Wikipedia page on line integrals was amazingly helpful. – Reid Sep 28 '12 at 20:44

It basically means you are integrating things over a loop. For e.g. a circle with an element $\text{d} \textbf{l}$ if you do $\oint{\text{d} \textbf{l}}$ it will give you circumference of the circle.

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It's an integral over a closed contour (which is topologically a circle). An example from Wikipedia: $$ \begin{align} \oint_C {1 \over z}\,dz & {} = \int_0^{2\pi} {1 \over e^{it}} \, ie^{it}\,dt = i\int_0^{2\pi} 1 \,dt \\ & {} = \Big[t\Big]_0^{2\pi} i=(2\pi-0)i = 2\pi i, \end{align} . $$

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