Re: Re: Some bugs in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg59330] Re: [mg59328] Re: Some bugs in Mathematica
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 6 Aug 2005 01:29:21 -0400 (EDT)
- References: <200508010505.BAA24522@smc.vnet.net> <200508030519.BAA06378@smc.vnet.net> <dcsc4q$q0d$1@smc.vnet.net> <200508050523.BAA13570@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 5 Aug 2005, at 07:23, akhmel at hotmail.com wrote:
>>
>
> Example 3 is the same as example 2, so you just substitute the upper
> limit of the integration and subtract the lower limit of integration.
>
> Sincerely,
>
> Alex
>
>
Obviously this is plain nonsense since with the limits you had in
your problem you are going to get 0 in the denominator. You could
however do this:
Limit[Integrate[1/(r^3*Sqrt[r^2 - a^2]*Sqrt[r^2 - b^2]),
{r, x, c}, GenerateConditions -> False], c -> a,
Direction -> 1]
x^2/(2*a^2*b^2*Sqrt[x^2 - a^2]*Sqrt[x^2 - b^2]) +
(Sqrt[1 - a^2/x^2]*Sqrt[1 - b^2/x^2]*
AppellF1[1, 1/2, 1/2, 2, a^2/x^2, b^2/x^2])/
(4*a^2*Sqrt[x^2 - a^2]*Sqrt[x^2 - b^2]) +
(Sqrt[1 - a^2/x^2]*Sqrt[1 - b^2/x^2]*
AppellF1[1, 1/2, 1/2, 2, a^2/x^2, b^2/x^2])/
(4*b^2*Sqrt[x^2 - a^2]*Sqrt[x^2 - b^2]) +
(Sqrt[-(1/a)]*Sqrt[-a]*Sqrt[(a^2 - b^2)/a^2]*
ArcTanh[b/a])/(2*a^2*b*Sqrt[a^2 - b^2]) +
(Sqrt[-(1/a)]*Sqrt[-a]*Sqrt[(a^2 - b^2)/a^2]*
ArcTanh[b/a])/(2*b^3*Sqrt[a^2 - b^2]) -
1/(2*a^2*Sqrt[x^2 - a^2]*Sqrt[x^2 - b^2]) -
1/(2*b^2*Sqrt[x^2 - a^2]*Sqrt[x^2 - b^2]) +
1/(2*Sqrt[x^2 - a^2]*Sqrt[x^2 - b^2]*x^2)
I have no idea this is correct or not and do not intend to spend time
on checking it. However, since Mathematica did not try to generate
conditions or check convergence the answer is likely to be valid only
for special values of the parameters (if at all).
- References:
- Some bugs in Mathematica
- From: akhmel@hotmail.com
- Re: Some bugs in Mathematica
- From: akhmel@hotmail.com
- Re: Some bugs in Mathematica
- From: akhmel@hotmail.com
- Some bugs in Mathematica