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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
    Robert Dodier
    2006-04-10

    • labels: --> Lisp Core - Integration
     
  • Harald Geyer
    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
    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