按照@Andras Deak的回答，可以解析地计算出high-x展开式，然后使用一些简单的平滑方法在它和scipy函数之间进行插值。实际上在高x展开中有两个项会被取消，所以你得小心一点。在

我得到的答案是：import numpy as np

import matplotlib.pyplot as plt

from scipy.special import wofz

def Z(x):

return wofz(x)

## first derivative of wofz (analytically)

def Zp(x):

return -2/1j/np.pi**0.5 - 2*x*Z(x)

def dawsn_expansion(x):

# Accurate to order x^-9, or, relative to the first term x^-8

# So when x > 100, this will be as accurate as you can get with

# double floating point precision.

y = 0.5 * x**-2

return 1/(2*x) * (1 + y * (1 + 3*y * (1 + 5*y * (1 + 7*y))))

def dawsn_expansion_drop_first(x):

y = 0.5 * x**-2

return 1/(2*x) * (0 + y * (1 + 3*y * (1 + 5*y * (1 + 7*y))))

def dawsn_expansion_drop_first_two(x):

y = 0.5 * x**-2

return 1/(2*x) * (0 + y * (0 + 3*y * (1 + 5*y * (1 + 7*y))))

def blend(x, a, b):

# Smoothly blend x from 0 at a to 1 at b

y = (x - a) / (b - a)

y *= (y > 0)

y = y * (y <= 1) + 1 * (y > 1)

return y*y * (3 - 2*y)

def g(x):

"""Calculate `x + (1-2x^2) D(x)`, where D(x) is the dawson function"""

# For x < 50, use dawsn from scipy

# For x > 100, use dawsn expansion

b = blend(x, 50, 100)

y1 = x + (1 - 2*x**2) * special.dawsn(x)

y2 = dawsn_expansion_drop_first(x) - dawsn_expansion_drop_first_two(x) * 2*x**2

return b*y2 + (1-b)*y1

def Zpp(x):

# only return the imaginary component

return -4j/np.pi**0.5 * g(x)

x = np.logspace(0, 5, 2000)

dx = 1e-3

plt.plot(x, (Zp(x+dx) - Zp(x-dx)).imag/(2*dx))

plt.plot(x, Zpp(x).imag)

ax = plt.gca()

ax.set_xscale('log')

ax.set_yscale('log')

从而产生以下曲线图：

蓝线是数值导数，绿线是使用展开式的导数。后者实际上在大x下有更好的行为