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Re: integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg92848] Re: integration
  • From: Gobithaasan <gobithaasan at gmail.com>
  • Date: Wed, 15 Oct 2008 05:36:36 -0400 (EDT)
  • References: <gcchbh$s2c$1@smc.vnet.net> <48EA5916.2050008@gmail.com> <48EA758D.7000502@gmail.com>

Greetings...
Thanks Jean-Marc Gulliet,
I think Mathematica 6 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 Mathematica 6 to give the right assumption for 
imaginary-free answer? Thank you very much Jean...

Gobithaasan



Jean-Marc Gulliet wrote:
> Jean-Marc Gulliet 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])]]
>
> It took a long time, but the last expression returned the following 
> result (which is valid only for r S > 0 || s + S/r <= 0):
>
> (1/(2 r))E^(-((I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/
>   r^2)) (S ExpIntegralE[-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
>       r^2), -((I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2)] -
>    S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(
>     1 + (I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)
>      ExpIntegralE[-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2), -((
>       I (r s + S) (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2)] +
>    E^((2 I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/
>     r^2) (S ExpIntegralE[(I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2, (
>         I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2] - (S^3)^((
>        I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2) (r s + S)^(
>        1 - (I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
>         r^2) ((\[Kappa]0 - (1 + r) \[Kappa]1)^2)^((
>        I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
>        r^2) (S^2 (r s + S)^2 (\[Kappa]0 - (1 + r) \[Kappa]1)^2)^(-((
>         I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
>         r^2)) ((r s + S) (S + r Conjugate[s]))^((
>        I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)
>         ExpIntegralE[(I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2, (
>         I (r s + S) (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2]))
>
> Regards,
> -- Jean-Marc
>


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