以数组值为上限的数组数值积分

Question

所以我有以下代码：

import numpy as np
import scipy.special as spec
import scipy.integrate as integrate

a = array
b = constant * a
c = b*a/constant
e3 = spec.expn(3,c)

现在，我需要计算 e3 从 0 到 a 的积分。由于 e3 是 c 的函数，而 c 又是 a 的函数，所以我很难解决这个问题。 我尝试使用 loop、integrate.quad 或 both 来解决此问题，但我不断收到各种错误。
我在将 sutff 集成到 python 方面很新，所以任何帮助都将不胜感激。

Answer 1

这里有一个可能的解决方案，请检查我是否正确理解了问题以及是否有用...

import scipy.special as spec
from scipy.integrate import quad
import numpy as np

array = np.array([0, 1, 2, 3, 4, 5])  # array[i] >= 0 for all i in [0, len(array) - 1]. For example, array = np.array([1, 2, 3, 4, 5])
constant = 3 # >= 0. For example, constant = 3


a = array
b = constant * a # multiply numpy array by a scalar (it is possible with numpy arrays (: )
c = b * a / constant

e3 = spec.expn(3, c)

# Some usefull explanations:

print('Output:', e3) # return a ndarray with the value of the integral e3 en each point specified in c (numpy array)
# Output: [1.09691967e-01 2.76136095e-03 1.04786279e-05 5.96855753e-09 4.97790975e-13]

# For evaluate the integral e3 from 0 to a[len(a) - 1]:
result1 = e3[len(a) - 1] - e3[0] 
print('Output:', result1)
# Output: -0.499999999950223

# Other thing is the integral of e3 from 0 to a[len(a) - 1]. You can use quad (return the value and an estimted error)
# See https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.quad.html
result2 = quad(lambda x: spec.expn(3, x), 0, c[len(a) - 1])
print('Output:', result2)
# Output: (0.3333333333328521, 4.440892098500626e-16)

# Note that integration limits are floats not arrays.