find the error function of root t in terms of expansion
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Series expansion for integral including error function
taylor-expansion error-function
What is the series expansion of f for small q?
U(q) =qeq2erfcq I(q,q′) =∫
2π
0
dϕ
2π
U(
√
q2+q′2−2qq′cosϕ
)−U(q′) f(q) =∫
∞
0
dq′I(q,q′)
Even better, perhaps you can integrate this analytically to find f [or I(q,q′)]? When I numerically integrate, the results are not grid dependent for small q: i.e. I get consistent results with double the gridpoints and/or double the grid maximum, so it seems to be well-behaved (and the result looks approximately quadratic in q). Here is my attempt to find the series expansion of f by differentiating under both integrals, then integrating over ϕ analytically:
Dn(q′) ≡[
∂nI(q,q′)
∂qn
]q=0 D0(q′) =0 D1(q′) =∫
2π
0
dϕ
2π
[
U′(
√
q2+q′2−2qq′cosϕ
)
2
√
q2+q′2−2qq′cosϕ
(2q−2q′cosϕ)]q=0 =−U′(q′)∫
2π
0
dϕ
2π
cosϕ=0 D2(q′) =∫
2π
0
dϕ
2π
[sin2ϕ
U′(q′)
q′
+cos2ϕU″(q′)] =
U′(q′)
2q′
+
U″(q′)
2
=(2q′3+4q′+
1
2q′
)eq′2erfcq′−
3+2q′2
√
π
D3(q′) =0 f(q) =
q2
2
∫
∞
0
dq′D2(q′)+O(q4)
D2(q′)=
1
2
q′−1+O(1) so the last integral diverges logarithmically for small q′. Also it seems that D2n(q′)=O(q′1−2n) so the higher terms get worse.
I don't know why the Taylor series looks like a sum of divergent terms: perhaps the series expansion of f is in non-integer powers of q rather than a Taylor series?
In case it is helpful here is some Mathematica:
U[q_] := q Exp[q^2] Erfc[q]
Dint[qd_, n_] := Simplify[Integrate[(D[U[Sqrt[q^2 + qd^2 - 2 q qd Cos[phi]]] -U[qd], {q, n}]) /. q -> 0, {phi, 0, 2 Pi}]/(2 Pi), qd >= 0]
so that Dint[qd, 2] gives the above expression for D2(q′). Integrate[Dint[qd,2], qd] gives an expression in terms of 2F2 and that expression is divergent for q′→0.
EDIT: If I try to evaluate I> from joriki's answer, I note that q is small and x is large, so qx can be anything, but ϵ=q(
√
1+x2−2xcosϕ
−x)=q[−cosϕ+sin2ϕ/2x+O(x−2)] is small, so
I> =q∫
∞
1
dx∫
2π
0
dϕ
2π
U(qx+ϵ)−U(qx) =
∞
∑
n=1
qn+1
n!
∫
∞
1
dxU(n)(qx)∫
2π
0
dϕ
2π
(
√
1+x2−2xcosϕ
−x)n =
q2
2
∫
∞
1
dxU′(qx)∫
2π
0
dϕ
2π
(
√
1+x2−2xcosϕ
−x) +
q3
6
∫
∞
1
dxU″(qx)∫
2π
0
dϕ
2π
(
√
x2+1−2xcosϕ
−x)2+O(q4) =
q2
2
∫
∞
1
dxU′(qx)f(x)+
q3
6
∫
∞
1
dxU″(qx)[1−2xf(x)]+O(q4) f(x) =
(x+1)E[4x/(x+1)2]+(x−1)E[−4x/(x−1)2]
π
−x=
1
4x
+O(x−3)
so, since U′(q)=−
2q
√
π
+(1+2q2)eq2erfc(q), the first x integral diverges logarithmically and the next one is worse. Expanding in ϵ gives an expansion in q which is divergent. Perhaps someone has an idea of how to proceed to get an expansion in 1/q for small q.