i noticed F.T. has its share of curiosity:
http://mathworld.wolfram.com/FourierTransform1.html

let us do this without using duality property.
we want the Fourier transform of f(x)=1:

F(k)= [Lt X->inf] [int -X,X] exp(-i*2*pi*k*x) dx

= [Lt X->inf] sin(2*pi*k*X)/(pi*k)

if we wish to argue F(k)= delta(k), then by definition
of delta function, for k # 0, F(k)=0.
[ http://mathworld.wolfram.com/DeltaFunction.html ]

that means for k # 0,
[Lt X->inf] sin(2*pi*k*X) = 0 !

which says for X->inf, a sinusoid goes to zero, which
is not quite right by the definition of sinusoidal
function. it should remain oscillating between [-1,1] as
X-> infinity by definition...


-Akhila Raman