search

rtoy / maxima

A Clone of Maxima's repo
Other
0 stars 0 forks source link

bogus PV integral [defint(exp(a*x),x,0,inf)] #2749

Open rtoy opened 4 months ago

rtoy commented 4 months ago

Imported from SourceForge on 2024-07-06 20:34:08 Created by willisbl on 2004-01-30 14:49:56 Original: https://sourceforge.net/p/maxima/bugs/504

(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

rtoy commented 4 months ago

Imported from SourceForge on 2024-07-06 20:34:09 Created by robert_dodier on 2006-04-10 04:09:08 Original: https://sourceforge.net/p/maxima/bugs/504/#c033

rtoy commented 4 months ago

Imported from SourceForge on 2024-07-06 20:34:12 Created by hgeyer on 2008-03-25 03:00:44 Original: https://sourceforge.net/p/maxima/bugs/504/#67f8

Logged In: YES user_id=929336 Originator: NO

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

rtoy commented 4 months ago

Imported from SourceForge on 2024-07-06 20:34:15 Created by crategus on 2009-05-27 00:08:50 Original: https://sourceforge.net/p/maxima/bugs/504/#72e0

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