The quantity asked about: #lim_(x->oo)x^alpha(cosx/x)^2#. Note that, as stated, this is not an integral, rather an expression. I assume that #x# is the quantity to take the infinite limit of.
Multiply out:
#lim_(x->oo)x^(alpha-2)cos^2x#
#cos^2x# periodically oscillates in the range #[0,1]# over the real numbers, so its behaviour at infinity is not going to be a potential problem here regarding convergence.
As #x->oo#, #x^(alpha-2)# has the following behaviours for differing #alpha#:
If #alpha>2#, then #alpha-2>0# and #x^(alpha-2)->oo#.
If #alpha=2#, then #x^(alpha-2)=x^0=1#.
If #alpha<2#, then #alpha-2<0# and #x^(alpha-2)->0#
So
If #alpha>2#, #x^alpha(cosx/x)^2->oo# as #x->oo#.
If #alpha=2#, #x^alpha(cosx/x)^2->cos^2x# as #x->oo#.
If #alpha<2#, #x^alpha(cosx/x)^2->0# as #x->oo#.
