Hi,
I needed these two integrals for my work:
In [23]: integrate(erf(alpha*r)*exp(-alpha**2 * r**2), (r, 0, oo))
Out[23]:
⎧ ___
⎪ ╲╱ π │ ⎛ 2 ⎞│ π
⎪ ───── for │periodic_argument⎝polar_lift (α), ∞⎠│ ≤ ─
⎪ 4⋅α 2
⎪
⎪∞
⎨⌠
⎪⎮ 2 2
⎪⎮ -α ⋅r
⎪⎮ ℯ ⋅erf(α⋅r) dr otherwise
⎪⌡
⎪0
⎩
In [24]: integrate(erf(alpha*r)*exp(-alpha**2 * r**2)*r, (r, 0, oo))
Out[24]:
⎧ ___
⎪ ╲╱ 2 ⎛│ ⎛ 2 ⎞│ π │
⎪ ───── for ⎜│periodic_argument⎝polar_lift (α), ∞⎠│ ≤ ─ ∧ │p
⎪ 2 ⎝ 2
⎪ 4⋅α
⎪
⎨∞
⎪⌠
⎪⎮ 2 2
⎪⎮ -α ⋅r
⎪⎮ r⋅ℯ ⋅erf(α⋅r) dr
⎪⌡
⎩0
⎛ 2 ⎞│ π⎞ │ ⎛ 2
eriodic_argument⎝polar_lift (α), ∞⎠│ < ─⎟ ∨ │periodic_argument⎝polar_lift (α),
2⎠
otherwise
⎞│ π
∞⎠│ < ─
2
What does periodic_argument and polar_lift mean? It seems really
complicated. For comparison, Mathematica returns:
In [1]: Integrate[Erf[alpha*r]*Exp[-alpha^2*r^2], {r, 0, Infinity}]
Out [1]: ConditionalExpression[Sqrt[\[Pi]]/(4 alpha),
Re[alpha^2] > 0 && Re[alpha] > 0]
In [2]: Integrate[Erf[alpha*r]*Exp[-alpha^2*r^2]*r, {r, 0, Infinity}]
Out [2]: ConditionalExpression[1/(2 Sqrt[2] alpha^2),
Re[alpha^2] > 0 && Re[alpha] > 0]
Which is much simpler. So I think we should return something similar.
Ondrej
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