Re: Integral question
- To: mathgroup at smc.vnet.net
- Subject: [mg73727] Re: Integral question
- From: "Michael Weyrauch" <michael.weyrauch at gmx.de>
- Date: Sun, 25 Feb 2007 04:46:00 -0500 (EST)
- References: <eropmh$9cb$1@smc.vnet.net>
Hello, I think that Mathematica's result is equivalent to yours: Just use the well-known relation ArcSin[z]=-I*Log[I*z+Sqrt[1-z^2]] to transform Mathematica's result into yours, and remember that you always can add an arbitray complex constant to the result of an indefinite integral. Regards Michael Weyrauch "Tulga Ersal" <tersal at umich.edu> schrieb im Newsbeitrag news:eropmh$9cb$1 at smc.vnet.net... > Dear Mathematica users, > > Let's consider the integral > > Integrate[1/Sqrt[v^2 - c*s^2], s] > > where v and c are positive reals. 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. > > 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. > > I'd appreciate your help. > > Thanks, > Tulga > > >