# Why is Cos[Abs[x]] Fourier transformed to 0 in mathematica?

I found that

FourierTransform[Cos[Abs[x]], x, ξ, FourierParameters -> {1, -1}]


gives 0, while

FourierTransform[Cos[x], x, ξ, FourierParameters -> {1, -1}]


yields π DiracDelta[-1 + ξ] + π DiracDelta[1 + ξ]

How does this happen?

• What version of Mathematica are you using? I get the second answer for both statements using Windows 10, Mathematica 10.4.
– JimB
Apr 8, 2016 at 15:07
• @JimBaldwin, i'm using version 9. i'll check the new version, thanks~ Apr 8, 2016 at 15:28
• @JimBaldwin, but on wolframalpha FourierTransform[Cos[Abs[t]],t,w] still results 0 Apr 8, 2016 at 15:33
• @andre, the second result is surely ok, but why 0 in the first one? Apr 8, 2016 at 15:34
• When I try this on v10.0, the first code returns unevaluated rather than evaluating to zero, but interestingly enough, Assuming[x \[Element] Reals, FourierTransform[Cos[Abs[x]], x, \[Xi], FourierParameters -> {1, -1}]] returns 0. Apr 8, 2016 at 15:44

FourierTransform[Cos[Abs[x]], x, ξ, FourierParameters -> {1, -1}]