FourierCosTransform
- To: mathgroup at smc.vnet.net
- Subject: [mg70284] FourierCosTransform
- From: dimmechan at yahoo.com
- Date: Wed, 11 Oct 2006 01:54:10 -0400 (EDT)
Consider the following function
g[s_, y_] := 1/(E^(y*s)*s)
Then (the integral exists in the Hadamard sense)
FourierCosTransform[g[s, y], s, x]
(-EulerGamma - Log[1/x^2 + 1/y^2] + Log[1/y^2])/Sqrt[2*Pi]
FullSimplify[%, {x > 0, y > 0}]
-((EulerGamma + Log[1 + y^2/x^2])/Sqrt[2*Pi])
FullSimplify[FourierCosTransform[%, x, s], y > 0]
(-1 + E^((-s)*y))/s
As far as I know, the last result should be equal to g[s,y].
Also based on a distributional approach (Sosa and Bahar 1992)
the FourierCosTransform of g[s,y] should be
-((EulerGamma + Log[x^2+ y^2])/Sqrt[2*Pi]).
Can someone pointed me out was it is going here?
Following the next procedure I succeed in getting a result
similar with this of Sosa and Bahar (1992).
Integrate[g[s, y]*Cos[s*x], {s, e, ee}, Assumptions -> 0 < e < ee];
Normal[FullSimplify[Series[%, {e, 0, 2}], e > 0]];
Normal[Block[{Message}, FullSimplify[Series[%, {ee, Infinity, 2}], ee >
0]]];
DeleteCases[%, (a_)*Log[e], Infinity];
DeleteCases[%, _[_, _[ee, _], _], Infinity];
Limit[%, e -> 0];
% /. -Log[a_] - Log[b_] :> -Log[a*b];
FullSimplify[%]
(1/2)*(-2*EulerGamma - Log[x^2 + y^2])
I really appreciate any kind of help.
Regards
Dimitris