python - 二维高斯函数的集成(python)

Question

我已经使用 Area、sigmax 和 sigmay 参数定义了二维高斯分布(独立变量之间没有相关性)。当我在两个变量中从 (-inf, inf) 进行积分时，我只会在 sigmax 和 sigmay 为 1 时获得面积。

import numpy as np
import scipy.integrate as sci

class BeamDistribution(object):
    def __init__(self, Ipeak, sigmax, sigmay):
        print Ipeak, sigmax, sigmay
        self.__Ipeak = Ipeak
        self.__sigmax = sigmax
        self.__sigmay = sigmay

    def value(self, x, y):
        factor = self.__Ipeak/(2.*np.pi*self.__sigmax * self.__sigmay)
        factorx = np.exp(-x**2/(2.*self.__sigmax**2))
        factory = np.exp(-y**2/(2.*self.__sigmay**2))
        return factor*factorx*factory

    def integral(self, a, b, c, d):
        integration = sci.dblquad(self.value, a, b, lambda x: c, lambda x: d,
                                  epsrel = 1e-9, epsabs = 0)
#        sci.quad_explain()
        return integration

    def __call__(self, x, y):
        return self.value(x, y)


if __name__ == "__main__":
    Ipeak = 65.0e-3
    sigmax = 0.2e-3
    sigmay = 0.3e-3
    limit = np.inf
    my_beam_class = BeamDistribution(Ipeak, sigmax, sigmay)
    total = my_beam_class.integral(-limit, limit, -limit, limit)
    print "Integrated total current ",total," of Ipeak ", Ipeak

    my_beam_class = BeamDistribution(Ipeak, 1, 1)
    total = my_beam_class.integral(-limit, limit, -limit, limit)
    print "Integrated total current ",total," of Ipeak ", Ipeak

输出是

0.065 0.0002 0.0003
Integrated total current  (7.452488478001055e-32, 6.855160478762106e-41)  of Ipeak  0.065
0.065 1 1
Integrated total current  (0.4084070449667172, 1.0138233535120856e-11)  of Ipeak  0.065

知道为什么会这样吗？我想它应该很简单，但经过数小时的查看，我看不出有什么问题。

Answer 1

集成高斯核的正确方法是使用 Gauss-Hermite 正交，参见 here

它在 Python 中作为 SciPy 模块实现 http://docs.scipy.org/doc/numpy-1.10.1/reference/generated/numpy.polynomial.hermite.hermgauss.html

代码，高斯核积分应该是√(π)

import math
import numpy as np

a,w = np.polynomial.hermite.hermgauss(32)

print(a)
print(w)

def f(x):
    return 1.0

s = 0.0
for k in range(0,len(a)):
    s += w[k]*f(a[k])

print(s - math.sqrt(math.pi))

二维案例

Ipeak = 0.065
sigmax = 0.2e-3
sigmay = 0.3e-3
sqrt2 = math.sqrt(2.)

def h(x, y):
    return Ipeak*1.0/math.pi

s = 0.0
for k in range(0, len(a)):
    x = sqrt2 * sigmax * a[k]
    t = 0.0
    for l in range(0, len(a)):
        y = sqrt2 * sigmay * a[l]
        t += w[l] * h(x, y)
    s += w[k]*t

print(s)