Timeline for Deriving Newtonian capacity of sphere from Brownian motion

12 events

when toggle format	what		by	license	comment
Jan 10, 2015 at 22:59	comment	added	Thomas Kojar		The S-P theorem allowed $\int_{a}^{b}\int_{0}^\infty \sin[u r]f(u)\,drdu={\cal P}\int_{a}^{b}\frac{1}{u}f(u)du=\int_{a}^{b}\frac{1}{u}f(u)du$ where $f(u)=(1-e^{-u^{2}/2t})(tu)^{-1}$ and P was ignored because as Beenakker commented, $f(u)u^{-1}$ is not singular at u=0 (as the exponent grows faster).
Jan 10, 2015 at 20:09	vote	accept	Thomas Kojar
Jan 9, 2015 at 20:49	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 18 characters in body
Jan 9, 2015 at 20:36	comment	added	Carlo Beenakker		I added a more detailed derivation of this step.
Jan 9, 2015 at 20:36	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 721 characters in body
Jan 9, 2015 at 19:31	comment	added	Thomas Kojar		why is $\int_{r_{0}}^{\infty} \int_{0}^{\infty}\frac{1-e^{-u^{2}}t/2}{u}sin(u(r-r_{0}))drdu$ a finite value? The $\int_{r_{0}}^{\infty} sin(u(r-r_{0}))dr$ is not converging.
Jan 9, 2015 at 19:13	vote	accept	Thomas Kojar
Jan 9, 2015 at 19:32
Jan 7, 2015 at 14:35	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 17 characters in body
Jan 7, 2015 at 14:25	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 131 characters in body
Jan 7, 2015 at 6:58	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 62 characters in body
Jan 7, 2015 at 6:46	history	edited	Carlo Beenakker	CC BY-SA 3.0	added 164 characters in body
Jan 7, 2015 at 6:40	history	answered	Carlo Beenakker	CC BY-SA 3.0