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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
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