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MathGroup Archive 2004

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Re: Can this integration be done?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50109] Re: [mg50074] Can this integration be done?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 15 Aug 2004 03:14:43 -0400 (EDT)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

Clear[a,b,c,x, soln];

soln[a_,b_,c_] := Evaluate[Module[{x,int},
        int = Integrate[x/((a+x)*(b+(c+x^2)^2)),x];
        FullSimplify[Limit[int , x->Infinity]- (int /. x->0)]]];

soln[a,b,c]

(I*Pi*((c*(Sqrt[1/(c + I*Sqrt[b])] - 
         Sqrt[1/(c - I*Sqrt[b])]) + I*Sqrt[b]*
        (Sqrt[1/(c - I*Sqrt[b])] + 
         Sqrt[1/(c + I*Sqrt[b])]))*a^2 + 
     (c^2 + b)*(Sqrt[1/(c + I*Sqrt[b])] - 
       Sqrt[1/(c - I*Sqrt[b])])) + 
   2*a*(a^2 + c)*ArcCot[c/Sqrt[b]] + 
   a*Sqrt[b]*(4*Log[a] - Log[c^2 + b]))/
  (4*Sqrt[b]*((a^2 + c)^2 + b))

While this will not work for all values of the parameters, it can provide results:

Module[{a=1,b=1,c=1, s},
  {s=FullSimplify[soln[a,b,c]],Simplify[s-Integrate[x/((a+x)*(b+(c+x^2)^2)),{\
x,0,Infinity}]]==0}]

{(1/20)*(Pi + Sqrt[2*(-7 + 5*Sqrt[2])]*Pi - Log[2]), True}

Module[{a=1,b=2,c=1, s},
  {s=FullSimplify[soln[a,b,c]],FullSimplify[s-Integrate[x/((
        a+x)*(b+(c+x^2)^2)),{x,0,Infinity}]]==0}]

{(1/24)*(Sqrt[-10 + 6*Sqrt[3]]*Pi + 
    2*Sqrt[2]*ArcTan[Sqrt[2]] - Log[3]), True}

Module[{a=1,b=1,c=-1, s},
  {s=FullSimplify[
    soln[a,b,c]],FullSimplify[s-Integrate[x/((a+
        x)*(b+(c+x^2)^2)),{x,0,Infinity}]]==0}]

{(1/4)*(Sqrt[2*(-1 + Sqrt[2])]*Pi - Log[2]), True}


Bob Hanlon

> 
> From: Daohua  Song <ds2081 at columbia.edu>
To: mathgroup at smc.vnet.net
> Date: 2004/08/14 Sat AM 01:50:24 EDT
> To: mathgroup at smc.vnet.net
> Subject: [mg50109] [mg50074] Can this integration be done?
> 
> Dear Group,
>       I am wondering whether the following integration has an analytic 
> expression. Alough i believe so.
>       Integrate[x/((a + x)*(b + (c + x^2)^2)),{0,Infinity}].
>       When i set a=b=c=1,mathematica give me an answer.
> 
>       Any help is appreciated.
> sincerely
> Daohua
> 
> 


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