Re: Integral question
- To: mathgroup at smc.vnet.net
- Subject: [mg73726] Re: Integral question
- From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
- Date: Sun, 25 Feb 2007 04:45:29 -0500 (EST)
- Organization: NewsReader.Com Subscriber
- References: <eropmh$9cb$1@smc.vnet.net>
Tulga Ersal <tersal at umich.edu> wrote: > Dear Mathematica users, > > Let's consider the integral > > Integrate[1/Sqrt[v^2 - c*s^2], s] > > where v and c are positive reals. Might it also be that v^2 > c*s^2 is known? > When I calculate the integral by hand, I get > > ArcSin[(Sqrt[c]*s)/v]/Sqrt[c] > > However, if I evaluate the integral in Mathematica, I get > > (I*Log[(-2*I)*Sqrt[c]*s + 2*Sqrt[-(c*s^2) + v^2]])/Sqrt[c] > > which not only does not look like what I have found by hand, but > also, for v=4, c=2, s=1, for example, gives > > 0.255525 + 1.47039 i > > as opposed to just 0.255525. Note that Mathematica's answer is also a correct antiderivative. Its answer and yours merely differ by a constant of integration, namely, by I Log[2 v]/Sqrt[c] > If you evaluate the integral in Mathematica with the said values (v=4, > c=2) > > Integrate[1/Sqrt[4^2 - 2*s^2], s] > > you get the answer > > ArcSin[s/(2*Sqrt[2])]/Sqrt[2] > > which agrees with what I have found by hand. > > My question is: How can I get what I found by hand using Mathematica > without having to assign values to v and c before evaluating the > integral? I tried adding the assumptions v>0, c>0, but it didn't help. Unfortunately, using additional assumptions such as v^2 > c*s^2 doesn't seem to help either. At the moment, the closest I can come to getting what you want requires a change of variable, letting u = Sqrt[c] s: In[3]:= 1/Sqrt[c]*Assuming[v^2 > u^2, FullSimplify[Integrate[1/Sqrt[v^2 - u^2], u]]] Out[3]= ArcTan[u/Sqrt[-u^2 + v^2]]/Sqrt[c] In[4]:= % /. u -> Sqrt[c] s Out[4]= ArcTan[(Sqrt[c]*s)/Sqrt[(-c)*s^2 + v^2]]/Sqrt[c] I hope that someone else can do better. David