Trying to check an integral leading to the famous soliton, i find the
following problem with maxima:
(%i1) integrate(1/(t*sqrt(1-t)),t);
(%o1) log(sqrt(1 - t) - 1) - log(sqrt(1 - t) + 1)
Note this is not good because presumably the argument of the first log is
<0. The correct solution is as follows:
(%i2) X: -2*atanh(sqrt(1-t));
(%o2) - 2 atanh(sqrt(1 - t))
(%i3) diff(X,t);
1
(%o3) -------------
sqrt(1 - t) t
This, by inversion leads to the soliton:
(%i4) solve(x=X,t);
Is x positive, negative, or zero?
p;
x
(%o4) [sqrt(1 - t) = - tanh(-)]
2
(%i5) %^2;
2 x
(%o5) [1 - t = tanh (-)]
2
(%i6) solve(%,t);
2 x
(%o6) [t = 1 - tanh (-)]
2
(%i7) trigsimp(%);
1
(%o7) [t = --------]
2 x
cosh (-)
2
The last formula is the well known one soliton solution. If one performs the
same calculation starting from %o1 one gets 1/sh^2(x/2) which is very bad
since it diverges at 0.
By the way, the same result is obtained in a much direct way with maple:
niobe% maple
|\^/| Maple 9 (IBM INTEL LINUX)
._|\| |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc.
2003
\ MAPLE / All rights reserved. Maple is a trademark of
<____ ____> Waterloo Maple Inc.
| Type ? for help.
> int(1/(t*sqrt(1-t)),t);
1/2
-2 arctanh((1 - t) )
> solve(%=X,t);
2
1 - tanh(X/2)
> quit
--
Michel Talon