#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#
