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Nov 26, 2020 at 14:01 history bounty ended CommunityBot
Nov 24, 2020 at 10:11 comment added Paul For me there is still some mistake here, first $s =\int_0^r \frac{2 t^{\beta+1}}{\beta} e^{-\frac{1}{2t^\beta}}$ which is not very important for the rest. But at the end $J(s)\sim \frac{\beta}{2} sr^{-\beta}$ which change the form of the metric... I am going to edit my post in this sense
Nov 24, 2020 at 7:06 comment added ubik Sorry for my very stupid mistake ! I had another compution. The metric is not conical !
Nov 24, 2020 at 7:05 history edited ubik CC BY-SA 4.0
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Nov 23, 2020 at 17:00 comment added Paul re-doing your computation I found $s= \frac{r}{\beta+2}e^{-\frac{1}{2r^\beta}}- \int_0^r \frac{t^{\beta+2}}{\beta(\beta+2)} \frac{d^2}{dt^2} \left(e^{-\frac{1}{2r^\beta}} \right)$, the problem is that the last term divided by $s$ diverge at 0.
Nov 22, 2020 at 8:32 history answered ubik CC BY-SA 4.0