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Aug 25, 2018 at 19:27 comment added R.G.J @Frobenius : thanks for elaborating on my answer. I thought it would be sufficient!
Aug 25, 2018 at 8:44 comment added Voulkos $\uparrow$ My answer is based on your short one.
Aug 24, 2018 at 17:20 comment added gh3 Yep, got that. My problem is that in doing the derivation I end up with a factor of $p$ in front of the gaussian, still integrating from negative infinity to positive infinity, which means the integral evaluates to $0$.
Aug 24, 2018 at 17:04 comment added K7PEH Minor comment: qualified by $a>0$.
Aug 24, 2018 at 16:28 history answered R.G.J CC BY-SA 4.0