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 gives 0.

I think it makes sense, both mathematically and from a practical point
of view, to always make integrals over zero-measure domains equal to
0.

Aaron Meurer

On Fri, Feb 2, 2018 at 6:26 PM, Chris Smith <smi...@gmail.com> wrote:
> 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 condition depends on x and y whereas in the 2nd
> case it only
> depends on y...and in the first case since it depends on x and x's range for
> the 2nd expression
> is from 3 to 3 (zero width) it can't contribute to the integration. So is it
> right to ignore the value
> of y?
>
> And what if there were a DiracDelta function there instead of 2 and the
> integration went to 4 instead
> of 3...then there *would* be a contribution...but this is presently ignored
> by the current routine:
>
> ```
>>>> integrate( Piecewise((DiracDelta(x),x<2),(0,True)), (x,-1,1))
> 1
>>>> integrate( Piecewise((DiracDelta(x),Eq(x,0)),(0,True)), (x,-1,1))
> 0
> ```
>
> /c
>
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