Re: Different results with FourierTransform[]

• To: mathgroup at smc.vnet.net
• Subject: [mg97616] Re: [mg97588] Different results with FourierTransform[]
• From: Richard Hofler <rhofler at bus.ucf.edu>
• Date: Tue, 17 Mar 2009 04:58:20 -0500 (EST)
• References: <200903160924.EAA25551@smc.vnet.net>

```Hello Wieland,

It appears that the answer you get depends on when the replacement rule
B -> 1 is evaluated.

In[4]:= sol1 = InverseFourierTransform[Tanh[x],x,p]

Out[4]= -(I/(2 p Sqrt[2 \[Pi]]))-3/4 I Sqrt[\[Pi]/2] Csch[(p
\[Pi])/2]-(5 HarmonicNumber[-(1/2)-(I p)/4])/(8 Sqrt[2 \[Pi]])-(3
HarmonicNumber[-(1/2)+(I p)/4])/(8 Sqrt[2 \[Pi]])+(5 HarmonicNumber[-((I
p)/4)])/(8 Sqrt[2 \[Pi]])+(3 HarmonicNumber[(I p)/4])/(8 Sqrt[2 \[Pi]])

In[5]:= sol2 = InverseFourierTransform[Tanh[B x],x,p]/.B->1

Out[5]= -I Sqrt[\[Pi]/2] Csch[(p \[Pi])/2]

In[6]:= sol3 = InverseFourierTransform[Tanh[B x]/.B->1,x,p]

Out[6]= -(I/(2 p Sqrt[2 \[Pi]]))-3/4 I Sqrt[\[Pi]/2] Csch[(p
\[Pi])/2]-(5 HarmonicNumber[-(1/2)-(I p)/4])/(8 Sqrt[2 \[Pi]])-(3
HarmonicNumber[-(1/2)+(I p)/4])/(8 Sqrt[2 \[Pi]])+(5 HarmonicNumber[-((I
p)/4)])/(8 Sqrt[2 \[Pi]])+(3 HarmonicNumber[(I p)/4])/(8 Sqrt[2 \[Pi]])

In[7]:= sol1 == sol3

Out[7]= True

HTH,
Richard

Richard Hofler

-----Original Message-----
From: Wieland Brendel [mailto:wielandbrendel at gmx.net]
Sent: Monday, March 16, 2009 5:25 AM
To: mathgroup at smc.vnet.net
Subject: [mg97616] [mg97588] Different results with FourierTransform[]

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

```

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