I agree with Leonid, there is no consistent way to define integral
[0,anything) of the delta function. It is a theorem, in fact.
Aaron, Sympy seniors,
What was the reasoning to define DiracDelta as a function? How difficult
would it be to insert GeneralizedFunction class above the Function class
By definition (e.g.,
https://en.wikipedia.org/wiki/Dirac_delta_function#As_a_measure),
DiracDelta would have integral one over any set that contains 0, even
if that set has only one point.
So one has to ask, when we integrate over (x, a, b), do we mean closed
interval [a, b], open (a, b), or half-o
The condition is an And, so it requires both to be true. Thus, it is
only 2 at one value, x=3, which means it doesn't contribute to the
integral.
I'm not really sure how to interpret the Dirac Delta integral. It
seems to me that it should be the same as integrate(DiracDelta(x), (x,
0, 0)), which g
I'm also wondering if the following behavior is right:
```
>>> Piecewise((1, x < 1), (2, Eq(x, 3) & (y < x)), (3, True)).integrate((x,
0, 3))
7
>>> Piecewise((1, x < 1), (2, y < 3), (3, True)).integrate((x, 0, 3))
Piecewise((5, (y >= -oo) & (y < 3)), (7, True))
```
In the first case, the 2nd con
