What's with Fourier Transforms in Mathematica?

Questions
1

What’s the issue with Fourier Transforms in Mathematica? Why does it always have this DiracDelta?

In[1]:= << "FourierSeries`"

In[2]:= f[x_] := (2 b/\[Pi])^(3/4) E^-br^2

In[3]:= ft[k_] = FourierTransform[f[x], x, k]

Out[3]= 2 b^(3/4) E^-br^2 (2/\[Pi])^(1/4) DiracDelta[k]

In[4]:= ft[k_] = 
 FourierTransform[f[x], x, k, FourierParameters -> {-1, 1}]

Out[4]= b^(3/4) E^-br^2 (2/\[Pi])^(3/4) DiracDelta[k]
2

The expressions being transformed are constants. They do not depend on x. The delta function is from the Fourier transform of 1.

3

OK, but why should Mathematica always give a DiracDelta when evaluating Fourier Transforms of functions?

For instance, FourierTransform[Cos[bx], x, k] gives

Sqrt[2 \[Pi]] Cos[bx] DiracDelta[k]

This is irritating – is there a way of suppressing this?

4

That one doesn’t depend on x either. Did you forget to put a space between b and x?

5

Hi Yaj,

In this example, the argument is also constant because the bx is a single symbol. for b*x you need either the asterisk or, more commonly, a space between b and x. Mathematica outputs the delta when it is in fact correct. Consider:

In[1]:= FourierTransform[Exp[- x^2/b^2], x, k]

Out[1]= E^(-(1/4) b^2 k^2)/(Sqrt[2] Sqrt[1/b^2])

In[2]:= FourierTransform[UnitStep[x - 1] - UnitStep[x - 2], x, k]

Out[2]= -((I E^(I k) (-1 + E^(I k)))/(k Sqrt[2 \[Pi]]))

Best, David

6

How to suppress a correct result? - A truly interesting question! In your example you are transforming for x a function which does not depend on x but on bx. You probably meant:

FourierTransform[Cos[b x], x, k]
(* Out:  Sqrt[\[Pi]/2] DiracDelta[-b+k]+Sqrt[\[Pi]/2] DiracDelta[b+k]   *)

Again it contains this irritating Dirac-Deltafunction, but nevertheless this answer is correct. What would you like to get?

EDIT: @John: When I wrote the above I had not seen your response - sorry!

7

Thank you David.