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