Another problem of the week has been long overdue. Here’s one that may appeal to those who liked POW-5 (maybe we should have an integration problem every fifth time?):

Evaluate

Please submit solutions to **topological[dot]musings[At]gmail[dot]com** by **Friday, October 10**, 11:59 pm (UTC); do **not** submit solutions in Comments. Everyone with a correct solution will be inducted into our Hall of Fame! We look forward to your response.

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October 12, 2008 at 12:56 am

Solution to POW-10: Another hard integral? « Todd and Vishal’s blog[…] by Nilay Vaish: The answer to POW-10 is . […]

December 2, 2012 at 11:38 am

A.Yes, I know it’s 4 years later, but this isn’t exactly a “hard” integral, but rather a classic application of “integrals with parameters”.

First, make the change of variables t=arctan(x), obtaining “integral from 0 to Infinity of arctan(t) / t*(1+t^2)”.

Next consider the parameter-dependent integral I(a)=”integral from 0 to Infinity of arctan(a*t) / t*(1+t^2)”. Compute its derivative I'(a) as “integral from 0 to Infinity of 1 / (1+t^2)*(1+a^2*t^2)”. Split it into 2 easy integrals, and get I'(a)=”Pi/2*(1+a)”. Next, integrate back I'(a), obtaining (you fix the constant produced by integration by evaluating at a=0) I(a)=”Pi/2 * Log[a]”.

Finally, make a=1 and obtain the desired result: Pi/2 * Log[2].

December 2, 2012 at 9:17 pm

A.I apologize for the mistype, it should be “Log[a+1]” instead of “Log[a]”.

March 8, 2019 at 9:08 pm

cusaduthese integrals are very important observed prof dr mircea orasanu and prof drd horia orasanu and followed that thus can be approached Elliptic Integrals