Comment #10 on issue 2624 by [email protected]: Sympy 0.7.1 can't
integrate Gaussians
http://code.google.com/p/sympy/issues/detail?id=2624
The ideal solution would be to figure out ways to make it simpler
in the algorithm itself.
I agree, but I don't see how to. integrate(exp(-(x-1)**2), (x, -oo, oo))
comes out nice because the input is nice. [As you saw above, there the
integral is split into two exp(-x**2) integrals, which both come out neat].
On the other hand integrate(exp(2*x - x**2 - 1), (x, -oo, oo)) does not
suggest anything sensible, so it is just split at zero. We happen to be
able to do the integral, but I don't know how one could recognise in
advance that the answer is going to be so neat.
Tom what exactly is the functionality in gsoc-3 that isn't in gsoc-2?
Mostly polishing:
- some bug fixes
- improved heuristics (i.e. we can do more integrals, and do them faster)
- some performance improvements
- lerch transcendent, polylog etc; and recognition of those in hyperexpand()
- exponential integrals (recognised in both meijerint and hyperexpand)
Tom would have to answer if gsoc-3 can be ready by the 23rd
Ready for show-off certainly. Ready for review as well, I think. Reviewed
seems unlikely to me :-).
2) What has to be done to handle the integral from
http://www-m3.ma.tum.de/bornemann/Numerikstreifzug/Chapter9/MeijerG.pdf?
Nothing:
In [3]: alpha = Symbol('alpha', positive=True)
In [4]: meijerint_definite((2-x)**alpha*sin(alpha/x), x, 0, 2)
Out[4]:
⎛ ⎛ α 1 α │ 2⎞ ⎞
⎜ ⎽⎽⎽ ╭─╮3, 0 ⎜ ─ + ─, ─ + 1 │ α ⎟ ⎟
⎜╲╱ π ⋅α⋅Γ(α + 1)⋅│╶┐ ⎜ 2 2 2 │ ──⎟ ⎟
⎜ ╰─╯2, 4 ⎜ │ 16⎟ ⎟
⎜ ⎝0, 0, 1/2 -1/2 │ ⎠ ⎟
⎜───────────────────────────────────────────────────────, True⎟
⎝ 4 ⎠
[This is not quite the same answer as in the paper, but they agree
numerically.]
To be fair:
- I just fixed a bug (which, incidentially, I had noticed *yesterday* last
thing before end of work, and before seeing your comment)
- if you run it through integrate(), it's slow, even with meijerg=True. The
reason is that for integrals over finite intervals we first try to compute
an antiderivative [since most often meijerg does not work in that case].
And meijerint_indefinite tries hard to find an antiderivative, but does not
succeed.
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