Compute for all integer the value of

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- Feb 17th 2009, 01:56 PMKrizalidImproper integral with log, floor and ceil function
Compute for all integer the value of

- Feb 18th 2009, 07:27 PMAjreb
very hard problem!! could u post a solution?

- Feb 18th 2009, 07:28 PMKrizalid
I can do that, within few hours.

- Feb 19th 2009, 11:08 AMKrizalid
For each it's and the integrand is hence the integral becomes Since is an integer, then the inner floor function is reduntant so the integral is and we're done.

- Feb 21st 2009, 06:09 PMAjreb
thanks a lot for the solution!! but it's kind of hard to me to understand, can you explain any further how did you get that the integral equals ?

- Feb 21st 2009, 07:23 PMKrizalid

Thus, for and we have hence the integral is and we're done.