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## #504 bogus PV integral [defint(exp(a*x),x,0,inf)]

open
nobody
5
2006-04-10
2004-01-30
Barton Willis
No

(C1) defint(exp(a*x),x,0,inf);
Is a positive, negative, or zero?

zero;Principal Value
(D1) 0

This isn't a PV integral; the integral diverges when
a == 0.

(C2) build_info();

Maxima version: 5.9.0
Maxima build date: 19:10 2/9/2003
host type: i686-pc-mingw32
lisp-implementation-type: Kyoto Common Lisp
lisp-implementation-version: GCL-2-5.0

## Discussion

• Robert Dodier
2006-04-10

• labels: --> Lisp Core - Integration

• Harald Geyer
2008-03-25

Logged In: YES
user_id=929336
Originator: NO

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

• Dieter Kaiser
2009-05-27

The behavior has changed:

The sign of a is unknown. Maxima returns a noun form:

(%i3) defint(exp(a*x),x,0,inf);
(%o3) 'integrate(%e^(a*x),x,0,inf)

The sign of a is negative. We get the correct solution:

(%i4) assume(a<0)\$

(%i5) defint(exp(a*x),x,0,inf);
(%o5) -1/a

The sign of the parameter b is positive. The integral is divergent:

(%i6) assume(b>0)\$

(%i7) defint(exp(b*x),x,0,inf);
defint: integral is divergent.
-- an error. To debug this try debugmode(true);

Up to here the solutions are correct and consistent. We assume a paramter less or equal zero and get a divergent result. I think, this should give a noun form too:

(%i10) forget(a<0)\$
(%i11) assume(a<=0)\$

(%i12) defint(exp(b*x),x,0,inf);
defint: integral is divergent.
-- an error. To debug this try debugmode(true);

This is again correct. The integral is divergent when b is greater or equal zero:

(%i13) forget(b>0)\$
(%i14) assume(b>=0)\$

(%i15) defint(exp(b*x),x,0,inf);
defint: integral is divergent.
-- an error. To debug this try debugmode(true);

Perhaps we can change the one open case for a parameter a<=0. Then all cases will be correct.

Dieter Kaiser