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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
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