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Mathematics
Calculus
Is Leibniz integral rule allowed in this potential improper integral?
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[QUOTE="Beelzedad, post: 6182013, member: 662144"] Electric potential at a point inside the charge distribution is: ##\displaystyle \psi (\mathbf{r})=\lim\limits_{\delta \to 0} \int_{V'-\delta} \dfrac{\rho (\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV'## where: ##\delta## is a small volume around point ##\mathbf{r}=\mathbf{r'}## ##\mathbf{r}## is coordinates of field point ##\mathbf{r'}## is coordinates of source point ##\rho (\mathbf{r'})## is the density of charge distribution Taking the gradient of potential: ##\displaystyle \nabla \psi (\mathbf{r}) =\nabla\ \left[ \lim\limits_{\delta \to 0} \int_{V'-\delta} \dfrac{\rho (\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV' \right] =\lim\limits_{\delta \to 0} \int_{V'-\delta} \rho (\mathbf{r'})\ \nabla \left( \dfrac{1}{|\mathbf{r}-\mathbf{r'}|} \right) dV'## In the last step, we have applied Leibniz integral rule (basic form). The validity of this technique for improper integrals is discussed below: The following passage from the book "[URL='http://library1.org/_ads/48AEFC4C49281110D30CB1A8F1FFE9A9']Foundations of Potential Theory page 15[/URL]1" says the technique is not valid. But it says the equation ##\mathbf{E}=-\nabla \psi## still holds at points inside source regions ##V'##. It also gives a "little" proof of the argument.[ATTACH type="full" alt="243911"]243911[/ATTACH] [ATTACH type="full" alt="243912"]243912[/ATTACH] [ATTACH type="full" alt="243913"]243913[/ATTACH] [ATTACH type="full" alt="243914"]243914[/ATTACH] [ATTACH type="full" alt="243915"]243915[/ATTACH] [B][I](continued below)[/I][/B] [/QUOTE]
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Forums
Mathematics
Calculus
Is Leibniz integral rule allowed in this potential improper integral?
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