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MathGroup Archive 2005

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Re: an Integrate question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58918] Re: an Integrate question
  • From: "Drago Ganic" <drago.ganic at in2.hr>
  • Date: Sat, 23 Jul 2005 05:32:52 -0400 (EDT)
  • References: <dbie3r$brm$1@smc.vnet.net><dbkkiq$s49$1@smc.vnet.net> <dbninb$hlj$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Antonio,
We have a more precise analytical solution in Mathematica 5.2 than in 5.1. 
Here is the solution I got with 5.2:

In[3]:=
Integrate[E^(I*h*x)/(E^(m*Sqrt[k2 + x2])*
(Abs[x] + Sqrt[k2 + x2])), {x, 0, Infinity}]

Out[3]=
Piecewise[{{((1/2)*(2*Gamma[0, (-I)*h*Sqrt[k2 + x2]] -
2*Log[(-I)*h] + 2*Log[(-I)*h*Sqrt[k2 + x2]] -
Log[k2 + x2]))/E^((I*h + m)*Sqrt[k2 + x2]),
Im[h] > 0},
{(Pi*MeijerG[{{0}, {1/2}}, {{0, 0}, {1/2}},
I*h*Sqrt[k2 + x2]])/E^(m*Sqrt[k2 + x2]),
h \[Element] Reals && Im[h] <= 0}},
Integrate[E^(I*h*x - m*Sqrt[k2 + x2])/
(Abs[x] + Sqrt[k2 + x2]), {x, 0, Infinity},
Assumptions -> h \[NotElement] Reals && Im[h] <= 0]]
and using assumptions we get just one solution:

In[6]:=
Assuming[{h, m} \[Element] Reals && Re[k2] <= 0),
Integrate[E^(I*h*x)/(E^(m*Sqrt[x2 + k2])*(Abs[x] + Sqrt[x2 + k2])),
{x, 0, Infinity}]]

Out[6]=
(Pi*MeijerG[{{0}, {1/2}}, {{0, 0}, {1/2}},
I*h*Sqrt[k2 + x2]])/E^(m*Sqrt[k2 + x2])

I did not verify the solution. But I don't understand the condition:

 h \[Element] Reals && Im[h] <= 0

because   h \[Element] => Im[h] == 0 ?!?  Your case should be covered with 
the solution given.

Gretings,
Drago Ganic

"Antonio Carlos Siqueira" <acsl at dee.ufrj.br> wrote in message 
news:dbninb$hlj$1 at smc.vnet.net...
> Dear Mathgroup
> First of all thanks to Carl Woll and Andreas Dieckman for providing the
> answer to the integral.   Digging my results a little bit I found
> myself with a little different integral. I am trying to calculate some
> electromagnetic field in conductive soil, the integral looks like this:
> Integrate[Exp[-m Sqrt[x2+k2]]*> Exp[I h
> x]/(Abs[x]+Sqrt[x2+k2]),{x,0,Infinity}]
> where h and m are real and k is either pure imaginary or has a negative
> real part.
> Any ideas on how to solve such integral analytically?
> After some tests I believe this integral is a job for NIntegrate,
> nevertheless it is worth asking if anyone knows any trick to transform
> this integral in something that can be solved analytically.
> Nice Regards
> Antonio
> 



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