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Aug 20, 2019 at 23:08 comment added Zhe To be fair, that's not the function you're actually integrating...
Aug 20, 2019 at 20:47 comment added Buck Thorn Hmmm, while I see what you are doing, I wonder if you aren't allowed to use L'Hopitals rule.
Aug 20, 2019 at 16:53 comment added Zhe $\frac{x^{2}}{x}$ is effectly just $x$ except at $x = 0$ where it is not defined, since it is $\frac{0}{0}$.
Aug 20, 2019 at 6:58 comment added Buck Thorn Good explanation, the volume shell element dV goes to zero at the origin. But I don't get why you should write that $r^2/r$ is not defined at the origin? It doesn't diverge, it is zero (because the volume of the shell is zero).
Aug 19, 2019 at 15:14 history edited Zhe CC BY-SA 4.0
clear up language of integrals of unbounded functions
Aug 19, 2019 at 15:12 comment added Zhe @IvanNeretin Argh. You're absolutely right. I will clean up the language here.
Aug 19, 2019 at 15:07 comment added Ivan Neretin Actually, some integrals of unbounded functions converge all right.
Aug 19, 2019 at 14:53 vote accept Raphaël
Aug 19, 2019 at 14:52 history answered Zhe CC BY-SA 4.0