[Date Index] [Thread Index] [Author Index]

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
>
>
>