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