#4721: [with patch; needs work] Indefinite integration for piecewise functions
-------------------------+--------------------------------------------------
Reporter: pbutler | Owner: burcin
Type: enhancement | Status: new
Priority: minor | Milestone: sage-3.4.1
Component: calculus | Resolution:
Keywords: |
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Changes (by wdj):
* summary: [with patch; partial positive review] Indefinite integration
for piecewise functions => [with patch; needs
work] Indefinite integration for piecewise
functions
Comment:
Sorry, I didn't see this earlier or I would've reviewed it.
This does not seem to work as advertised. There was a long thread in sage-
devel which I though settled some design issues. One of them was that the
FTC should be true for this indefinite integral. For example, the indef
int F of the function f defined on (-4,3)
{{{
sage: f1(y) = -1
sage: f2(y) = y + 3
sage: f3(y) = -y - 1
sage: f4(y) = y^2 - 1
sage: f5(y) = 3
sage: f =
Piecewise([[(-4,-3),f1],[(-3,-2),f2],[(-2,0),f3],[(0,2),f4],[(2,3),f5]])
sage: F = f.integral(y)
}}}
should have the property that F(3)-F(-4) is the area under the curve.
This *is* true, as the following shows:
{{{
sage: f.integral(y,(-4,3))
19/6
sage: F(3)-F(-4)
19/6
}}}
This is not tested. In my option, this needs to be added to the docstring.
*However*, what the docstring says is "If definite=True is given, returns
the definite integral."
I don't know what the output
{{{
sage: f.integral(definite=True)
2*(y^2 - 1) + y + 2*(-y - 1) + 5
}}}
means. I would expect it to be 19/6. Don't you assume the function is 0
outside
(-4,3)? In any case, the definite integral seems incorrect, and is not
consistenet
with the old behaviour, as you said it would be in the thread referred to
above.
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/4721#comment:8>
Sage <http://sagemath.org/>
Sage - Open Source Mathematical Software: Building the Car Instead of
Reinventing the Wheel
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