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user_id=588346
Interestingly, if you exponentialize the expression first,
it gets the right answer. But still asks, unnecessarily,
the sign of 'a'.
-------------------------------------------------------------
Maxima version: 5.9.1
Maxima build date: 16:35 2/10/2005
host type: i686-pc-linux-gnu
lisp-implementation-type: GNU Common Lisp (GCL)
lisp-implementation-version: GCL 2.6.6
-------------------------------------------------------------
integrate(1/cosh(a*x)^2,x,-inf,inf);
Is a positive, negative, or zero?
p;
(%o3) 0
Correct answer is 2/a.
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user_id=588346
Interestingly, if you exponentialize the expression first,
it gets the right answer. But still asks, unnecessarily,
the sign of 'a'.
Logged In: YES
user_id=588346
Interestingly, if you exponentialize the expression first,
it gets the right answer. But still asks, unnecessarily,
the sign of 'a'.
Logged In: YES
user_id=28849
I think this integral fails because polelist is unable to
find the roots of z^(4*a)+2*z^(2*a)+1.
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user_id=929336
Originator: NO
Still observed in 5.14.0 and cvs HEAD from 2008-03-25
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user_id=1797506
Originator: NO
src/defint.lisp rev 1.57:
o mtoinf: exponentializing before numden avoids losing multiplicities
fixes [ 1309432 ] integrate(1/cosh(a*x)^2,x,-inf,inf);
o ztorat: nn* -> s
fixes integrate(x^3/(1 + + 4*x^2 + 6*x^4 + 4*x^6 + x^8 ), x, 0, inf);
[ 1668087 ] integrate(1/cosh(x/R)^4,x,-inf,inf) = 0
(%i3) integrate(1/cosh(x*3)^2,x,-inf,inf);
(%o3) 2/3
(%i4) integrate(1/cosh(x*a)^2,x,-inf,inf);
Is a positive, negative, or zero?
p;
(%o4) 2/a
(%i5) integrate(1/cosh(x*a)^2,x,-inf,inf);
Is a positive, negative, or zero?
n;
Is %e^a-1 positive, negative, or zero?
n;
(%o5) -2/a
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