       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:
> {k1,k2,r,s,S} are real  numbers
>  r > -1
>  S > 0
>  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:= 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= 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

```

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