[Date Index]
[Thread Index]
[Author Index]
Re: integration
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92850] Re: integration
*From*: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
*Date*: Wed, 15 Oct 2008 05:36:58 -0400 (EDT)
*References*: <gcchbh$s2c$1@smc.vnet.net> <48EA5916.2050008@gmail.com>
On Wed, Oct 15, 2008 at 5:19 AM, Gobithaasan <gobithaasan at gmail.com> wrote:
> Greetings...
> Thanks Jean-Marc Gulliet,
> 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 Jean...
>
> Gobithaasan
Please, could you post the expression you used and its result. On my
system, using the above assumptions, Mathematica returns the integral
unevaluated, which is in agreement with what I already noticed when 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]
Regards,
-- Jean-Marc
Prev by Date:
**Re: Exclude O[] from Series[] for Solve[] in Mathematica**
Next by Date:
**Mathematica 6.0.0 on Suse 11.0**
Previous by thread:
**Re: integration**
Next by thread:
**Re: integration**
| |