Comment #7 on issue 1893 by [email protected]: integrate(log(x) *
x**(k-1) * exp(-x) / gamma(k), (x, 0, oo)) hangs
http://code.google.com/p/sympy/issues/detail?id=1893
We can do the definite integral, albeit with a few tricks:
In [1]: integrate(exp(-x)*log(x)*x**y, (x, 0, oo), meijerg=True)
Out[1]:
⎧ ╭─╮2, 1 ⎛ -y 1, 1 │ ⎞ ╭─╮0, 3 ⎛-y, 1, 1 │ ⎞
⎪- │╶┐ ⎜ │ 1⎟ + │╶┐ ⎜ │ 1⎟ for 0 < re(y) +
1
⎪ ╰─╯3, 2 ⎝0, 0 │ ⎠ ╰─╯3, 2 ⎝ 0, 0 │ ⎠
⎪
⎪ ∞
⎨ ⌠
⎪ ⎮ y -x
⎪ ⎮ x ⋅ℯ ⋅log(x) dx otherwise
⎪ ⌡
⎪ 0
⎩
In [4]: hyperexpand(_.args[0].args[0], allow_hyper=True)
Out[4]:
2 ┌─ ⎛y + 1, y + 1 │
⎞ 2 ┌─ ⎛y + 1, y + 1 │ ⎞
Γ(y + 1) ⋅ ├─ ⎜ │ -1⎟ Γ(-y -
1) ⋅ ├─ ⎜ │ -1⎟
2╵ 2 ⎝y + 2, y + 2 │
⎠ 2╵ 2 ⎝y + 2, y + 2 │ ⎠
y⋅Γ(y)⋅polygamma(0, y) + Γ(y) - ────────────────────────────────── +
───────────────────────────────────
2 2
Γ(y +
2) Γ(-y)
In [5]: simplify(_)
Out[5]:
(y⋅polygamma(0, y) + 1)⋅Γ(y + 2)
────────────────────────────────
y⋅(y + 1)
In [10]: expand_func(_)
Out[10]: (y⋅polygamma(0, y) + 1)⋅Γ(y)
That is hyperexpand() is not currently clever enough to do these
simplifications on its own, and decides the g-functions are better than the
mess at the intermediate stages. Actually this sort of cancellation happens
a lot with integrals involving log(x) [since we write it as log(x)*H(x-1) +
log(x)*H(1-x)], I should try to make hyperexpand clever enough...
I'm not sure how the indefinite integral is computed by mathematica. Note
that my computation above is essentially the mellin transform of
exp(-x)*log(x). Since it contains polygamma(0, y) we see that
exp(-x)*log(x) is not a g-function, so the standard tricks don't work.
Playing around with the mellin transform I can show that there exists an
antiderivative of the form -[log(x)*G1(x) + G2(x)] but I have no idea how
right now how to make this into an algorithm...
[I can spell out the computation if you want.]
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