Re: an Integrate question

*To*: mathgroup at smc.vnet.net*Subject*: [mg58829] Re: an Integrate question*From*: "Carl K. Woll" <carlw at u.washington.edu>*Date*: Wed, 20 Jul 2005 00:29:26 -0400 (EDT)*Organization*: University of Washington*References*: <dbie3r$brm$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"Antonio Carlos Siqueira" <acsl at dee.ufrj.br> wrote in message news:dbie3r$brm$1 at smc.vnet.net... > Dear MathGroup > I am trying to find an analytical answer for an integral such as: > (1) Integrate[Exp[-m Sqrt[x^2+b^2]]/(x+Sqrt[x^2+b^2]),{x,0,Infinity}] > where b and m can be Complex. > I know that > (2) Integrate[1/(E^(m x)*(x+Sqrt[x^2+b^2])),{x,0,Infinity}] has an > analytical answer although Mathematica cannot find and it is: > (3) Pi/(2 b m) (StruveH[1,b m]-BesselY[1, b m]) - 1/(b^2 m^2) > The response (3) checks with the answer I get from NIntegrate even when > b and m are complex. > So I was wondering if any of you guys could help me figure out whether > or not an analytical result for (1) is possible. > BTW I can get the numerical results in (2) using NIntegrate. > So any suggestions or comments on how I can solve (1) are highly > appreciated. > I forgot to say, I am using Mathematica 5.1 on windows. > Nice Regards > Antonio > If you do the transformation x -> Sqrt[y^2+2 b y] your integral transforms to Exp[-b m]/b^2 Integrate[Exp[-m y] ((b + y)^2/Sqrt[2 b y + y^2]-(b+y)), {y, 0, Infinity}]] which Mathematica can integrate, yielding a mess. After some massaging, the result I get is: int[b_,m_]:=(Exp[-b m] (-1 - b m + b Exp[b m] m (3 b m BesselK[0, b m] + (1 - 2 b m) BesselK[1, b m]) + b m Sqrt[Pi] HypergeometricU[1/2, 0, 2 b m]))/(b^2 m^2) When I evaluate this mess with various values of b and m and compare to the original integral evaluated using NIntegrate they agree. Carl Woll Wolfram Research