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