Are you sure this step:

to

is correct? You skipped a lot of steps there. I'm not sure I followed them all.

Just double-check your work, is all.

The idea is very elegant. Nice work!

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- July 22nd 2010, 08:23 AMAckbeet
Are you sure this step:

to

is correct? You skipped a lot of steps there. I'm not sure I followed them all.

Just double-check your work, is all.

The idea is very elegant. Nice work! - July 22nd 2010, 11:29 AMshawsend
- July 22nd 2010, 11:34 AMAckbeet
Which version of MMA do you have? Mine doesn't compute that (it's version 4.0). Neither does WolframAlpha.

- July 22nd 2010, 11:53 AMshawsend
I have ver. 7.0. Seems I don't even need to specify the assumption. The command:

InverseFourierTransform[Sech[m*(z/2)]^

(1/m), z, t]

gives:

- July 22nd 2010, 11:56 AMAckbeet
Ah, well. Those special functions, including the beta functions, are all related to the confluent hypergeometric functions. It's probably equivalent, through an amazing amount of difficult analysis. It even has similar arguments to the functions as what AlexisM has already computed.

- July 22nd 2010, 10:24 PMAlexisM

so I separate the lonely x which give me the 1/x and 1/2x including the 1/2 which is in factor.

and then I have - July 22nd 2010, 10:29 PMAlexisM
- July 23rd 2010, 01:36 AMAckbeet
Reply to AlexisM at post # 21:

Ah, got it. Looks good!