Re: Different results with FourierTransform[]

*To*: mathgroup at smc.vnet.net*Subject*: [mg97719] Re: Different results with FourierTransform[]*From*: m.r at inbox.ru*Date*: Thu, 19 Mar 2009 02:13:48 -0500 (EST)*References*: <gpl5sh$ou6$1@smc.vnet.net>

Wieland Brendel wrote: > Dear reader, > I somewhat stumbled over the following behaviour of mathematica: I tried > to calculate the fouriertransform of Tanh[x]. I did this in two ways: > > 1. Directly: > InverseFourierTransform[Tanh[x], x, p] > > 2. Indirectly: > InverseFourierTransform[Tanh[B x], x, p] > > where I set B -> 1 in the end. > > However, the result between the two approaches differs: Whereas in the > first approach I get a complex number (with both real and imaginary part > being non-zero for almost all values of p), the result in the second > approach yields NO real part; the imaginary part however is the same as > in the first approach. Is there any explanation for this behaviour? > > Thanks a lot in advance! I am really stuck with that... > Wieland The second one is correct, but in general you have to be careful with integral transforms with parameters, as there may be implicit assumptions present. Notice that InverseFourierTransform[Tanh[B x], x, p] == -I Sqrt[Pi/2] Csch[p Pi/(2*B)]/B cannot be right for both positive and negative B, because the transform has to be an odd function of B. Here are other ways to get the correct answer for InverseFourierTransform[Tanh[x], x, p]: In[1]:= {-I/p InverseFourierTransform[Tanh'[x], x, p], InverseFourierTransform[Tanh[x] E^(-a Abs[x]), x, p] /. a -> 0, 1/Sqrt[2 Pi] Integrate[(Tanh[x] - Sign[x]) E^(-I p x), {x, - Infinity, Infinity}, Assumptions -> Element[p, Reals]] + InverseFourierTransform[Sign[x], x, p]} // FullSimplify Out[1]= {-I Sqrt[Pi/2] Csch[p Pi/2], -I Sqrt[Pi/2] Csch[p Pi/2], -I Sqrt[Pi/2] Csch[p Pi/2]} Maxim Rytin m.r at inbox.ru