Question #3675d

1 Answer
Nov 21, 2017

This integral converged to pi/4

Explanation:

I=int_0^oo cos(2t)/t*dt

I use Feynman's trick for evaluating integral,

I(a)=int_0^oo (e^(-at)*cos(2t))/t*dt

I note that original integral can be reached after setting b=0.

After differentiating under integral sign according to a,

I'(a)=int_0^oo -e^(-at)*cos2t*dt

=1/a*[e^(-at)*cos2t]_0^oo+2/a*int_0^oo e^(-at)*sin2t*dt

=-1/a+2/a*int_0^oo e^(-at)*sin2t*dt

=-1/a-2/a^2*[e^(-at)*sin2t]_0^oo-4/a^2*int_0^oo e^(-at)*cos2t*dt

=-1/a-2/a^2*0-4/a^2*I'(a)

(1+4/a^2)*I'(a)=-1/a

I'(a)=-a/(a^2+4)

After integrating both sides,

I(a)=-1/2*arctan(a/2)+C

Now, I choose a strategic value b = b_0 in order to make our integrand vanish so that I(b_0) =0. In this case, took b=oo, so that I(oo)=0

Hence -1/2*arctan(oo/2)+C=0, so C=pi/4

Thus I=I(0)=-1/2*arctan(0/2)+pi/4=pi/4