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Mar 14, 2023 at 0:36 comment added Iosif Pinelis @user479223 : This follows (in view of Tonelli's theorem) by the interchange of the integration and the (double) summation, taking also into account that $d((x-r_k)_+^\beta)=\beta (x-r_k)_+^{\beta-1}\,dx$.
Mar 13, 2023 at 22:09 comment added user479223 I have kind of a stupid question. How do you get $\int_0^1 f dg =\sum_{j=1}^\infty \sum_{k=1}^\infty 2^{-j}2^{-k} I_{j,k}$?
Feb 14, 2023 at 12:19 vote accept user479223
Feb 14, 2023 at 4:00 history answered Iosif Pinelis CC BY-SA 4.0