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I recently ran into a kind of integral that I've never encountered before.

How should the following integral be expressed as a "normal" double integral?

$\iint \mathrm d f(u,v)$ where $f:\mathbb{R}^2\rightarrow \mathbb{R}$

The only thing I can think of that might be correct is $\iint \frac{\partial f}{\partial u} \frac{\partial f}{\partial v} \mathrm du \ \mathrm dv$ but I cannot motivate this and I've had no luck with my reference books nor with Google.


The following comes from

$$\hat V(t) = n(1-\hat F_n(t))^2 \int_0^t\int_0^t \frac{d\hat H_n(u,v)}{Y_n(u) Y_n(v)}$$

where $\hat F_n(t) \in \mathbb{R}$ is a distribution function, $Y_n(t) \in \mathbb{N}$ is a count and $\hat H_n$ is defined as $$\hat H_n(s,t) = \frac{1}{n}\sum_{i=1}^m\sum_{j=1}^{K_i}\sum_{l=1}^{K_i} \left\{ I_{(Z_{ij}\le s,\Delta_{ij}=1)} - \int_0^s I_{(Z_{ij}\ge u)} d\hat\Lambda_n(u) \right\} $$ $$ \times \left\{ I_{(Z_{il}\le t,\Delta_{il}=1)} - \int_0^t I_{(Z_{il}\ge u)} d\hat\Lambda_n(u) \right\}$$

where $I\in \{0,1\}$ is the indicator function and $\hat\Lambda_n \in \mathbb{R}$ is the so-called Nelson estimator. $Z_{ij} \in \mathbb{R}$ is a time and $\Delta_{ij} \in \{0,1\}$ is a censoring indicator.

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There seems to be a problem with the notation. $df(u, v) = \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v} dv$, which is a $1$-form. Taking a double integral of a $1$-form doesn't make sense. – Shaun Ault Oct 17 '11 at 23:12
Precisely what I thought, hence this question. The place I found this notation was an article in the Journal of Multivariable Analysis though, so I hope they knew what they were writing about. – Anton Oct 17 '11 at 23:14
Could you scan the page and post it? I'm just baffled. – Shaun Ault Oct 17 '11 at 23:55
Quoted parts of the source and added a DOI link to the source itself. Will expand upon the source material if required. – Anton Oct 18 '11 at 9:01
Thanks for posting the details, but I'm still at a loss. Sorry! – Shaun Ault Oct 19 '11 at 2:44

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