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Re: an Integrate question

"Antonio Carlos Siqueira" <acsl at> wrote in message 
news:dbie3r$brm$1 at
> 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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