You can use the old assumptions:
In [*10*]: x = Symbol('x')
In [*11*]: y = Symbol('y')
In [*12*]: a = Symbol('a')
In [*13*]: integrate(cos(x*y), (x, 0, a))
Out[*13*]:
⎧sin(a⋅y)
⎪──────── for y > -∞ ∧ y < ∞ ∧ y ≠ 0
⎨ y
⎪
⎩ a otherwise
In [*14*]: y = Symbol('y', nonzero=*True*)
In [*15*]: integrate(cos(x*y), (x, 0, a))
Out[*15*]:
sin(a⋅y)
────────
y
The new assumptions and refine can handle nonzero. The point about
relations not being implemented is that there is not currently a way to use
the assumption that e.g. x > y but that's not needed here so this should
work:
In [*16*]: y = Symbol('y')
In [*17*]: integrate(cos(x*y), (x, 0, a))
Out[*17*]:
⎧sin(a⋅y)
⎪──────── for y > -∞ ∧ y < ∞ ∧ y ≠ 0
⎨ y
⎪
⎩ a otherwise
In [*18*]: refine(_, Q.nonzero(y))
Out[*18*]:
⎧sin(a⋅y)
⎪──────── for y > -∞ ∧ y < ∞ ∧ y ≠ 0
⎨ y
⎪
⎩ a otherwise
I guess that this just isn't implemented yet in refine.
Oscar
On Monday, 10 August 2020 11:06:02 UTC+1, My Name wrote:
>
> I do this:
>
> import sympy
> sympy.srepr(sympy.integrate(S('cos(x * y)'), S('(x, 0, a)')))
>
> It returns this:
>
> "Piecewise(ExprCondPair(Mul(Pow(Symbol('y'), Integer(-1)),
> sin(Mul(Symbol('a'), Symbol('y')))), And(StrictGreaterThan(Symbol('y'),
> -oo), StrictLessThan(Symbol('y'), oo), Unequality(Symbol('y'),
> Integer(0)))), ExprCondPair(Symbol('a'), true))"
>
> How do I find the value of the Piecewise expression when y is nonzero? I
> have tried using sympy.refine with Q.nonzero, but that has not worked.
> Indeed the docs for refine warn, "Relations in assumptions are not
> implemented (yet)". Does that mean there's no way to find the value of the
> integral assuming y is nonzero?
>
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