python绘图集成结果，怎么做像Mathematica做的

Question

我想把这个Mathematica代码重写成Python，但是我当时很困惑，请帮帮我！谢谢。也许集成的结果与数组不同，但是我不知道！你知道吗

IMAGE: mathematica code which I want to rewrite in python

我当时对python代码感到困扰。。。。你知道吗

# -*- coding: utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl
import math
import scipy as sp
from pylab import *
#from scipy import *
from scipy.integrate import quad, dblquad, tplquad

plt.figure('God Bless : fig1')
plt.title(r'fig1_')
plt.xlabel(r'z')
plt.ylabel('$L_0$(erg $s^{-1}$)')
#plt.axis([0,10,-2,2])


Limit = 1.e48
c  = 2.997e10
Om = 0.27
H0 = 70e5/(1.e6*3.86e18) 
z  = np.arange(0,10, 0.01) 
def dLz(z):
  return 1./(1.-Om + Om*(1.+z)**3.)**(1./2)
val, abserr = quad(dLz, 0, 10)
print ("integral value =", val, ", absolute error =", abserr)    
dL     = val
Fmin   = 2.0e-8 
Llimit = 4.0*np.pi*dL**2*Fmin
plt.plot(z, Limit)


plt.savefig('fig_1.eps', dpi=300)
plt.show()

Answer 1

我想就是这样！哈利路亚！！你知道吗

#     -  DETERMIN  zimax       -

Limit = 1.e48
c  = 2.997e10
Om = 0.27
H0 = 70e5/(1.e6*3.86e18) 
Fmin   = 2.0e-8 
z  = np.arange(0,10, 0.1)
def dLz(z):
  return (c/H0)*(1.+z)/(1.-Om + Om*(1.+z)**3.)**(1./2)
x_lower = 0
vals = []
Llimits = []
for x_upper in z :  
  val, abserr = quad(dLz, x_lower, x_upper)
  vals.append(val)
  print ("integral value =", val, ", absolute error =", abserr)
  dL     = val
  Llimit = 4.0*np.pi*dL**2*Fmin
  Llimits.append(Llimit) # add to array
plt.plot(z, Llimits, 'b ' )

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