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Jul 17, 2015 at 6:48 vote accept Jeff
Jul 16, 2015 at 23:12 comment added Jeff solved, you are right. there is no contradiction.
Jul 16, 2015 at 20:15 comment added Carlo Beenakker try $\gamma=3$ or $\gamma=4$, where the integral can be done exactly for finite $R$; then take the limit $R\rightarrow\infty$ and you get 1/2, what contradiction might there be?
Jul 16, 2015 at 19:23 history edited Carlo Beenakker CC BY-SA 3.0
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Jul 16, 2015 at 15:32 comment added Jeff There is a contradiction with this proof. I posed a question to highlight this contradiction. You may see math.stackexchange.com/questions/1363393/… and let me know your comments.
Jul 16, 2015 at 12:55 review Suggested edits
Jul 16, 2015 at 12:59
Jul 15, 2015 at 17:50 history edited Carlo Beenakker CC BY-SA 3.0
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Jul 15, 2015 at 17:44 history undeleted Carlo Beenakker
Jul 15, 2015 at 17:42 history deleted Carlo Beenakker via Vote
Jul 15, 2015 at 17:39 history answered Carlo Beenakker CC BY-SA 3.0