# abs(x)cos(x) in Fourier space

I am working on some problems concerning Fourier Transform and I am facing something I don't understand.

I am trying to understand what is the representation of the function f(x)=abs(x)cos(x) in the Fourier space.

In wolfram alpha (sorry for this ad.), I type: Fourier(abs(x)*cos(x))

which gives me as result: g(w)= -((Sqrt[2/Pi] (1 + ω^2))/(-1 + ω^2)^2)

Where w is the frequency variable.

• My first question is: what is a 'frequency variable'?

Now, If I understand well, I believe that Fourier(abs(x)*cos(x)) gives the projection of the function f(x)=abs(x)*cos(x) into the Fourier space.

So, to check this, if I put some sampled values of a range of the function f(x)=abs(x)*cos(x), let's say: D={abs(0)*cos(0), abs(0.1)*cos(0.1), ......, abs(0.9)*cos(0.9)} Then, by applying the Discrete Fourier Transform on such data D, I should get an approximation on values of g(w) for w in D.

Meaning that: g(abs(0.5)*cos(0.5)) should be equal to DFT(D) at rank 6.

But, in Wolfram Alpha, when I compute the DFT of D: Fourier{0*cos(0),0.1*cos(0.1),0.2*cos(0.2),0.3*cos(0.3),0.4*cos(0.4),0.5*cos(0.5),0.6*cos(0.6),0.7*cos(0.7), 0.8*cos(0.8),0.9*cos(0.9)}

I get, at rank 6: -0.0957306 + 0. I

But g(0.5)=-((Sqrt[2/Pi] (1 + (0.5)^2))/(-1 + (0.5)^2)^2)=-1.77

I don't understand why wolfram alpha gives me this resultat for Fourier(f(x)). I am lost...

• Can anybody tells me where am I wrong in my thoughts? Many thanks...

The frequency variable is the argument of the Fourier transform $\omega$. Your Fourier transform is a tricky one, the original function being not absolutely or square integrable. But it can be defined as a limit, see below. We start with
• It looks like you want \func to get you the operator font. You can do this with \operatorname{Re} to get $\operatorname{Re}$ – Ross Millikan Apr 29 '14 at 22:52