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Comment on POW-10: Another hard integral? by A.

A. — Sun, 02 Dec 2012 21:17:02 +0000

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

Comment on POW-10: Another hard integral? by A.

A. — Sun, 02 Dec 2012 11:38:35 +0000

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].

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Comment on Stolz-Cesàro Theorem by Luqing Ye

Luqing Ye — Wed, 19 Sep 2012 10:41:12 +0000

Abel’s lemma can be used to prove integration by parts.Stolz’s theorem can be used to prove L’Hospital’s law.