MathGroup Archive 2008

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg92606] Re: integration
  • From: RG <gobithaasan at gmail.com>
  • Date: Wed, 8 Oct 2008 06:24:17 -0400 (EDT)
  • References: <gcchbh$s2c$1@smc.vnet.net> <gcfiaj$giu$1@smc.vnet.net>

On Oct 7, 7:48 pm, Jean-Marc Gulliet <jeanmarc.gull... at gmail.com>
wrote:
> RG wrote:
> > I have been trying to simplify(integrate) the following function, but
> > M6 seems to give a complex answer which i cann't understand.. please
> > help.
>
> > x[s_]=\!\(
> > \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[
> > \*FractionBox[\(r\ t\ \((\(-\[Kappa]0\) + \[Kappa]1 + r\ \[Kappa]1)\)
> > + \((1 + r)\)\ S\ \((\[Kappa]0 - \[Kappa]1)\)\ \((\(-Log[S]\) + Log[S
> > + r\ t])\)\),
> > SuperscriptBox[\(r\), \(2\)]]] \[DifferentialD]t\)\)
>
> First, notice that if we use the *InputForm* of the above expression, we
> can easily add assumptions on the parameters of the integral (or we
> could use *Assuming*), for instance that S, r, and s are real and r != =
0
> or s > 0.
>
> However, it seems that the above integral has no solution if the
> parameter S is positive. On the other hand, ff we allow S to be negative
> (or complex) then the integral has a symbolic complex solution.
>
> In[49]:= Integrate[
>   Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
>        S*(\[Kappa]0 - \[Kappa]1)*
>              (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
>     Assumptions -> S > 0]
>
> Out[49]= Integrate[
>   Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
>        r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2=
], {t,
>     0, s}, Assumptions -> S > 0]
>
> In[46]:= Integrate[
>   Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
>        S*(\[Kappa]0 - \[Kappa]1)*
>              (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
>     Assumptions -> {Element[{S, r, s}, Reals], r != 0, s > 0}]
>
> Out[46]= If[r S > 0 || s + S/r <= 0, (1/(
>   2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1)))
>   r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>
> [... output partially deleted ...]
>
>         r^2)] Sin[(S \[Kappa]0)/r^2 - (S \[Kappa]1)/r^2 - (
>         S \[Kappa]1)/r]),
>   Integrate[
>    Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
>         r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/
>     r^2], {t, 0, s},
>    Assumptions ->
>     r != 0 && s > 0 && r S <= 0 && r (r s + S) > 0 &&
>      S \[Element] Reals]]
>
> You can manipulate further the integral thanks to *FullSimplify* and
> some assumptions on the parameters.
>
> Assuming[r S > 0 || s + S/r <= 0,
>   FullSimplify[
>    1/(2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1))
>     r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>
>          [... input partially deleted ...]
>
>          S \[Kappa]1)/r])]]
>
> HTH,
> -- Jean-Marc

Dear All, i think M6 would be able to give a simplified answer, which
is more understandable without the appearance of imaginary numbers in
the answer. The assumption of the integral should be:
[1]{k1,k2,r,s,S} are real  numbers
[2] r > -1
[3] S > 0
[4] 0<= s<= S

I tried doing with these assumption, but the imaginary part still
exists.. Is there anyway to ask M6 to give the right assumption for
imaginary-free answer? Thank you very much sirs...


  • Prev by Date: Re: Why is Mathematica assuming k==l and how do I make it not to?
  • Next by Date: Re: Axis in two scales
  • Previous by thread: Re: integration
  • Next by thread: Re: integration