Message Boards Message Boards

GROUPS:

Why does Mathematica return this Fourier transform?

Posted 2 months ago
537 Views
|
5 Replies
|
5 Total Likes
|

Is my integration wrong? I get a different result regardless of which definition of the Fourier transform I use. The difference is the sign of the ikb term in the exponent.

enter image description here

5 Replies

The documentation states that Mathematica for its FourierTransform uses Exp[I k x] in the integral, rather than Exp[-I k x].

Thanks, Gianluca. Thumbs up. This community is a great help. I never knew there were so many Fourier transforms. If I had known years ago, I could have found definitions to justify answers marked wrong on my calculus exams.

Is there a version not listed in the documentation that uses the Fourier transform defined in my textbook? It's the inverse of the default definitions.

I don't know, but we can recover our favourite definition with a change in sign:

myFourierTransform[f_, x_, k_] := FourierTransform[f, x, -k]

Jay,

If you look at the documentation for FourierParameters I posted in your other thread, you control the exponent with the second parameter.

enter image description here

In the case that you like the default definition but prefer the sign change in the exponent

FourierParameters-> {0,-1} 

should do it.

Regards.

Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract