Check out my other channel OWLS MATH! / @owlsmath Check out my other channel OWLS SCHOOL OF MATH! / @owlsschoolofmath9732 Integral practice: owlsmath.neoci... Website: owlsmath.neoci... #math #integrals #integrationtechniques
This is another of those problems where looking at it as two integrals you end up with the difference of two logarithmically divergent expressions. You can see that there's a problem if you put u = pi x in the first one, which transforms it into the second, giving zero. Since arctan(pi x) > arctan (x) for all x in [0, pi/2] that clearly can't be right. The problem is that the terms in the integrand both tend to pi/(2x) for large x, hence the divergence. So I tried writing y = arctan(a x) - arctan(x), so tan(y) = x(1-a)/(1 + a x^2) so the integrand becomes 1/x arctan(x(1-a)/(1 + a x^2) ). I couldn't do much with that, but it's clear that it's convergent, since the large x behaviour ~ 1/x arctan( (1-a)/(a x) ) ~ 1/x^2 * (1-a)/a. Then I gave up and used the Feynman thingy as the title suggested. ☺
@@Samir-zb3xk thanks. yes it's a nice formula! I think I have a few more like this coming up but with some variation. I feel like one didn't quite work the same way because f(x) didn't converge at infinity but not sure.