On Mon, Mar 23, 2015 at 4:35 PM, Joachim Durchholz <[email protected]> wrote:
> (Following up by myself)
>
> Background: The test does integrate(exp(-s*x)*sinh(x), (x, 0, oo)).
>
> Old sort order gives me
>
> Piecewise(
>   (
>     -1/(s + 1)/2 - 1/(-s + 1)/2, #####1
>     And(
>       Ne(1/s, 1), #####2
>       Abs(periodic_argument(s, oo)) < pi/2,
>       Abs(periodic_argument(s, oo)) <= pi/2,
>       cos(Abs(periodic_argument(s, oo)))*Abs(s) - 1 > 0
>     )
>   ),
>   (
>     Integral(exp(-s*x)*sinh(x), (x, 0, oo)),
>     True
>   )
> )
>
> New sort order comes back with
> Piecewise(
>   (
>     -1/(2*(s + 1)) - 1/(2*(-s + 1)), #####1
>     And(
>       Abs(periodic_argument(s, oo)) < pi/2,
>       Abs(periodic_argument(s, oo)) <= pi/2,
>       Ne(s, 1), ####2
>       cos(Abs(periodic_argument(s, oo)))*Abs(s) - 1 > 0
>     )
>   ),
>  (
>     Integral(exp(-s*x)*sinh(x), (x, 0, oo)),
>     True
>   )
> )
>
> Points of interest:
>
> #####1 Different formulae, but same meaning. Haven't checked yet whether
> that's just due to different term ordering or to different choices made
> during Meijer.
>
> #####2 Now this is asserting Eq(1/s,1) for old and Eq(s,1) for new. These
> are semantically different for the case s=0 (i.e. integral of just sinh(x)):
> Old sort order gives a fail, new sort order gives a definite result.
> Now I'm wondering how this is accidentally improving Meijer heuristics. (Or
> is the difference mathematically irrelevant?) And I'm wondering where a
> changed sort order might accidentally worsen the heuristics.

If s = 0 in either case Ne(s, 1) and Ne(1/s, 1) will be False, so I
think in this case they are completely the same.

The version with s is obviously simpler, but we shouldn't rely on the
algorithm to give simple results, unless there are no other changes.

I would try to find some other example that this gives different
results for, or, preferably, get ahold of Tom Bachmann, and see if he
can shed any insight.

Aaron Meurer

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