ERFI Imaginary Error Function
ERFI.1 Introduction
Let x be a complex variable of C \ {∞}.The function Imaginary Error Function
(noted erfi) is defined by the following second order differential equation
(ERFI.1.1)
−2x
∂y(x) ∂ 2 y(x)
+
= 0.
∂x
∂x2
The initial conditions of ERFI.1.1 are given at 0 by
erfi(0) = 0,
(ERFI.1.2)
∂ erfi(x)
2
(0) = √ .
∂x
π
ERFI.2 Series and asymptotic expansions
ERFI.2.1 Taylor expansion at 0.
ERFI.2.1.1 First terms.
2
1
1
1
2
√ x9 +
erfi(x) = √ x + √ x3 + √ x5 + √ x7 +
π
3 π
5 π
21 π
108 π
(ERFI.2.1.1.1)
1
1
1
√ x11 +
√ x13 +
√ x15 + O x16 .
660 π
4680 π
37800 π
ERFI.2.1.2 General form.
(ERFI.2.1.2.1)
erfi(x) =
∞
X
u(n)xn .
n=0
The coefficients u(n) satisfy the recurrence
(ERFI.2.1.2.2)
−2nu(n) + n2 + 3n + 2 u(n + 2) = 0.
Initial conditions of ERFI.2.1.2.2 are given by
(ERFI.2.1.2.3)
u(0) = 0,
2
u(1) = √ .
π
ERFI.2.2 Asymptotic expansion at ∞.
1
2
ERFI IMAGINARY ERROR FUNCTION
ERFI.2.2.1 First terms.
(−2)
erfi(x) ≈ ex
xy0 (x) + ser
1,1,
0, [0,−i]
,
where
x2
+ √ + 2...
2 π
+ . . .
y1 (x) = terms y0 (x) = π
− 12
1,1,
0, [0,−i]
ERFI.2.2.2 General form.
ERFI.2.2.2.1 Auxiliary function y0 (x). The coefficients u(n) of y0 (x) satisfy the
following recurrence
−2nu(n) + u(n − 2) −4 + 3n + (n − 2)2 = 0
whose initial conditions are given by
u(0) = π
− 21
u(1) = 0
This recurrence has the closed form solution
Γ n + 12
,
u(2n) =
π
u(2n + 1) = 0.
ERFI.2.2.2.2 Auxiliary function y1 (x). The auxiliary function y1 (x) has the
exact form
y1 (x) = −i
ERFI.3 Graphs
ERFI.3.1 Real axis.
ERFI.3 GRAPHS
ERFI.3.2 Complex plane.
3