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MathGroup Archive 2004

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Re: Can this integration be done?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50116] Re: [mg50074] Can this integration be done?
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 15 Aug 2004 03:15:13 -0400 (EDT)
  • References: <200408140550.BAA15300@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Question: what's the analytic expression for Integrate[1/t,{t,1,x}]?

If you say Log[x] you're half right, but what IS Log[x], and how do you compute it?

There actually is no closed-form expression for Integrate[1/t,{t,1,x}]; we've simply NAMED it Log[x].

So... that's a very simple integral, isn't it? Why should YOUR integral have a closed-form solution?

Here's an antiderivative (apparently):

antiderivative = Integrate[x/((a + x)*(b + (c + x^2)^2)), x]

(-2*(-1)^(1/4)*(Sqrt[b] -
      I*c)*Sqrt[Sqrt[b] + I*c]*
     (I*a^2 + Sqrt[b] + I*c)*
     ArcTan[((-1)^(3/4)*x)/
       Sqrt[Sqrt[b] - I*c]] -
    Sqrt[Sqrt[b] - I*c]*
     (2*(-1)^(1/4)*(Sqrt[b] +
        I*c)*(a^2 +
        I*Sqrt[b] + c)*
       ArcTan[((-1)^(1/4)*x)/
         Sqrt[Sqrt[b] +
           I*c]] +
      a*Sqrt[Sqrt[b] + I*c]*
       (2*(a^2 + c)*ArcTan[
          Sqrt[b]/(c + x^2)] +
        Sqrt[b]*
         (4*Log[a + x] -
          Log[b + (c + x^2)^
           2]))))/(4*Sqrt[b]*
    Sqrt[Sqrt[b] - I*c]*
    Sqrt[Sqrt[b] + I*c]*
    (a^4 + b + 2*a^2*c + c^2))

If you compute this at x = 0 and Infinity, then subtract, does that give the right integral? It depends on whether there's a "branch" of the antiderivative that's continuous on the positive real line. But that depends, in a complicated way, on the values of a, b and c. Look at the Log and ArcTan terms, and consider all the possible branch cuts you might run into.

Still, you're welcome to subtract these two results:

antiderivative /. {x -> 0}

-((a*(2*(a^2 + c)*ArcTan[
        Sqrt[b]/c] +
      Sqrt[b]*(4*Log[a] -
        Log[b + c^2])))/
    (4*Sqrt[b]*(a^4 + b +
      2*a^2*c + c^2)))

Limit[antiderivative, x -> Infinity]

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

Neither is defined for all a, b, and c, even if branch cuts DON'T interfere.

Bobby

On Sat, 14 Aug 2004 01:50:24 -0400 (EDT), Daohua Song <ds2081 at columbia.edu> wrote:

> 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
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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