## #793 integrate(1/cosh(a*x)^2,x,-inf,inf);

closed
nobody
5
2008-07-26
2005-09-30
Anonymous
No

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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

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integrate(1/cosh(a*x)^2,x,-inf,inf);

Is a positive, negative, or zero?

p;
(%o3) 0

Correct answer is 2/a.

## Discussion

• Stavros Macrakis - 2005-11-09

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Interestingly, if you exponentialize the expression first,
it gets the right answer. But still asks, unnecessarily,
the sign of 'a'.

• Stavros Macrakis - 2005-11-09

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Interestingly, if you exponentialize the expression first,
it gets the right answer. But still asks, unnecessarily,
the sign of 'a'.

• Raymond Toy - 2006-02-15

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I think this integral fails because polelist is unable to
find the roots of z^(4*a)+2*z^(2*a)+1.

• Robert Dodier - 2006-04-10
• labels: --> Lisp Core - Integration

• Harald Geyer - 2008-03-25

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Still observed in 5.14.0 and cvs HEAD from 2008-03-25

• Dan Gildea - 2008-07-26

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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

• Dan Gildea - 2008-07-26
• status: open --> closed