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I'm losing my mind here. Consider $\Omega=S_1$, a circle, but could be any (1D) closed manifold. Now $\phi$ is my angle, parametrizing the circle. I define the angle to be $\phi\in [0,2\pi[$ and I want to calculate the following integral in two ways: $$\int_\Omega \phi\partial_\phi c,$$ where $c$ is a constant. Direct evaluation gives me 0.

Integration by parts (or Green Identity) gives me instead $$\int_{\partial\Omega}\phi c\, -\int_\Omega c.$$ If we were on a line, this would be 0. But since there is no boundary, we instead just get $2\pi c$!

Can someone explain what I am overlooking?

Confuse-ray30
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    $f(\phi)=\phi$ is not a well defined function on $S^1$ because $f(2\pi)\ne f(0)$. – Pranay Jan 27 '25 at 17:10
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    The function $H(\phi) = \phi$ has a jump of magnitude $2\pi$ at the point where $\phi = 0 \equiv 2\pi$ which must be taken into account when integrating by parts. – Hans Engler Jan 27 '25 at 17:15
  • @Pranay I thought this is ok: Integrating over disc. functions is well defined. – Confuse-ray30 Jan 27 '25 at 17:28
  • @HansEngler How do I properly account for discontinuities when integrating by parts? Is it that I need to split up my domain into where my functions are piecewise cont., and this introduces the boundaries $0,2\pi$ again? Since I am now less confused, I would probably think that the disc. at the same point should be regarded as two separate points during integration. – Confuse-ray30 Jan 27 '25 at 17:30
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    Few people will probably do integration-by-parts to calculate the circumference of the unit circle but why not? As @HansEngler pointed out, the angle has a jump. The integration-by-parts formula was recently discussed here. Its term $F(b)G(b)-F(a)G(a)$ (whether we call it "jump" or "boundary") will resolve your paradox. – Kurt G. Jan 27 '25 at 18:42
  • @KurtG. Ill take a look, thanks! – Confuse-ray30 Jan 27 '25 at 18:56
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    Integrating over the disc won't work because the function $\phi$ is not defined at the center point. – Andreas Blass Jan 27 '25 at 21:09

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