In a paper I am reading, I have come across the following formula:

$$\int_\omega f\, \partial \Omega$$

in which $\Omega$ is a bounded region in $\mathbb{R}^n$, $f:\Omega \to \mathbb{R}$ a function, and $\omega \subset \Omega$.

What is the meaning of $\partial \Omega$ in the integral? Is this standard notation?


I am not actually so much interested in the concrete problem in which the above formula occurs. I just did not understand the use of the "curvy delta" symbol in the integrand, and presumed it would be some kind of standard notation I did not know. FWIW, here is a link to the paper, the formula in question is number (4) on page 268.

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    $\begingroup$ $\partial\Omega$ is inside the integral not under the integral. Under the integral is $\omega$ $\endgroup$ – user9413 Jul 4 '11 at 18:44
  • $\begingroup$ OK changed that... should be clear now. $\endgroup$ – koletenbert Jul 4 '11 at 18:48
  • $\begingroup$ Do you have a reference? Maybe it is clearer in context. At least, can you say what topic/subject the paper is about? $\endgroup$ – Willie Wong Jul 4 '11 at 18:54
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    $\begingroup$ I believe it was correct before you exchanged $\partial\Omega$ and $\omega$, could you please show us how it looked before? $\endgroup$ – leftaroundabout Jul 4 '11 at 19:50

If you compare the paragraph before Equation (5) in that paper, which states that for a function $\phi$ such that $\phi$ is positive on $\omega$ and negative outside $\omega$, you have

$$ \int_{\omega} \partial\Omega = \int_{\Omega} H(\phi) dx $$

where $H(\cdot)$ is the Heaviside function, one may infer that by $\partial\Omega$ the authors meant to express some sort of induced volume element on $\omega\subset \Omega$.

But this certainly is not mainstream/standard notation; I for one have never seen its like before.


Assuming what you originally had was $$ \int_{\partial\Omega}f\,\omega $$ this means simply: integrate $f\omega$ over the boundary of $\Omega$.

$\omega$ is a differential form here, an $n-1$-form if $\Omega\subset\mathbb{R}^n$. It can then be written as $$ \omega = \omega_1\cdot\mathrm{d}x_2\wedge\ldots\wedge\mathrm{d}x_n + \omega_2\cdot\mathrm{d}x_1\wedge\mathrm{d}x_3\wedge\ldots\wedge\mathrm{d}x_n + \ldots\ldots $$ where the $\omega_i$ are functions $\mathbb{R}^n\to\mathbb{R}$ (as $f$ also is). (Of course, you could swap the meaning of $\omega$ and $f$, but this is the more common variant.)

As a simple example, let $\Omega$ be the unit square in $\mathbb{R}^2$. Then, $\omega$ is a 1-form, let it be simply $$ \omega := \mathrm{d}x. $$ Let furthermore $$ f(x,y) := x\cdot y $$ then your integral is simply $$ \int_{\partial\Omega}f\omega = \int_{\partial\Omega}xy\;\mathrm{d}x $$ We yet have to figure out what $\partial\Omega$ is, but in this case it's simply a path consisting of four straight segments, two in $x$- and two in $y$-direction. As it is an integral over $\mathrm{d}x$, we only need the parts in $x$-direction, for one of these, $y=0$, for the other $y=1$. So $$ \int_{\partial\Omega}f\omega = \int\limits_0^1x\cdot 0\;\mathrm{d}x + \int\limits_0^1x\cdot 1\;\mathrm{d}x = \frac12. $$

  • $\begingroup$ Thanks for your effort... it makes total sense, however the formula given in the paper actually is $\int_\omega f\, \partial \Omega$. My original edit (as related in the comment above) was just to change the phrasing "under the integral" to "in the integral". I will now add some more context to the original question. $\endgroup$ – koletenbert Jul 4 '11 at 22:10
  • $\begingroup$ This is really strange, especially since $\Omega$ is widespreadly used for integration domains as well as $\omega$ is for differential forms. $\int_\omega f\partial\Omega$ really does not make any sense then, and I have no idea what else it could mean. Without having looked it up in the paper, I'm inclined to think there has just been an editing error, but I could be completely wrong. $\endgroup$ – leftaroundabout Jul 4 '11 at 23:22

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