慢速收敛积分的数值积分
我基本上对这种类型的数值积分有一个问题：
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad
# Parameter
T = 94
Eps_inf = 3.05
Eps_s = 19.184544603857724
tau_0 = 1.27*10**-16
tau_CD = 0.34580390675331274
gamma = 0.49
C_pek = 1/Eps_inf - 1/Eps_s
hbar = 6.582119569*10**-16
# Functions
def func(w):
return -4 * 1/(np.pi * C_pek) * Eps(w).imag/abs(Eps(w))**2
def Eps(w):
return Eps_inf + (Eps_s - Eps_inf)/(1 + (1j * w * tau_CD))**gamma
w = np.logspace(-9,80,100000)
y = func(w)
IntegrandS = quad(lambda w: func(w),0,np.inf,limit=100000)[0]
print(f'quadResult: {IntegrandS}')
这给了我警告：积分可能是发散的，或者缓慢收敛的。
并且函数确实在慢慢收敛： 
如果我输入积分的上限只是一个像1e15这样的大数字，它会给我一个结果，但该结果永远不会收敛到越来越高的积分限制。
无论如何，有没有办法处理这个函数，以便四元函数（或任何其他集成方法，我也尝试过 scipys trapz，给我同样的问题）可以处理这个函数？
谢谢！
I basically have a problem with numerical integrations of this type:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad
# Parameter
T = 94
Eps_inf = 3.05
Eps_s = 19.184544603857724
tau_0 = 1.27*10**-16
tau_CD = 0.34580390675331274
gamma = 0.49
C_pek = 1/Eps_inf - 1/Eps_s
hbar = 6.582119569*10**-16
# Functions
def func(w):
return -4 * 1/(np.pi * C_pek) * Eps(w).imag/abs(Eps(w))**2
def Eps(w):
return Eps_inf + (Eps_s - Eps_inf)/(1 + (1j * w * tau_CD))**gamma
w = np.logspace(-9,80,100000)
y = func(w)
IntegrandS = quad(lambda w: func(w),0,np.inf,limit=100000)[0]
print(f'quadResult: {IntegrandS}')
This gives me the warning: The integral is probably divergent, or slowly convergent.
And the function is indeed slowly converging:
If I put in the upper limit of the integration just a big numer like 1e15 it gives me a result but that result will never converge for higher and higher integration limits.
Is there anyway to treat this function, so that the quad function (or any other integration method, i also tried scipys trapz, giving me the same problem) can deal with this function?
Thanks!
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
积分是发散的。这可以通过将
func(w)的渐近行为视为w来解释→ 无穷大。
func(w)依赖于w的部分是商Eps(w).imag/abs(Eps(w))**2.当
w→ Infinity 时，Eps(w)的实部接近Eps_inf，虚部趋于 0，因此w→ ∞，abs(Eps(w))**2→Eps_inf**2。即，分母接近正常数。因此，该商的重要部分是分子，Eps(w).imag。当
w→ ∞时，Eps(w).imag收敛到0，但这并不意味着func(w)的积分会收敛。为了收敛，Eps(w).imag必须“足够快”收敛到 0。当w→ ∞ 时，Eps(w).imag的行为类似于D * w**-gamma，其中D> 是一个独立于w的常量。从x0到 Infini（对于某些x0> 0）的x**p形式的积分仅当 p <; 时才收敛。 -1。根据您的函数，该幂为-gamma，即 -0.49。所以你的函数衰减太慢，积分无法收敛。您可以检查一下，如果将
gamma更改为大于 1 的值，quad会收敛。例如，当 gamma 为 1 时，积分发散。在这种情况下，这里有两次使用
quad的尝试。当
gamma< 1，quad（正确地）报告积分可能发散。The integral is divergent. This can be explained by looking at the asymptotic behavior of
func(w)asw→ ∞.
The part of
func(w)that depends onwis the quotientEps(w).imag/abs(Eps(w))**2.As
w→ ∞, the real part ofEps(w)approachesEps_infand the imaginary part goes to 0, so asw→ ∞,abs(Eps(w))**2→Eps_inf**2. That is, the denominator approaches a positive constant. So the important part of that quotient is the numerator,Eps(w).imag.As
w→ ∞,Eps(w).imagconverges to 0, but that does not mean the integral offunc(w)will converge. For convergence,Eps(w).imagmust converge to 0 "fast enough". Asw→ ∞,Eps(w).imagbehaves likeD * w**-gamma, whereDis a constant that is independent ofw. An integral of the formx**pfromx0to ∞ (for somex0> 0) is convergent only when p < -1. With your function, that power is-gamma, which is -0.49. So your function decays too slowly for the integral to converge.You can check that if you change
gammato a value greater than 1,quadconverges. E.g.The integral is divergent when gamma is 1. Here are two attempts to use
quadin this case.When
gamma< 1,quadreports (correctly) that the integral is probably divergent.