Python Plot an Integral with discrete and continuous parameters

Question

I have the following function:

def sbeta(beta,gamma,y):
    k = 2/(np.pi)**2.
    return k * np.sqrt(1 - (np.sqrt(1-y**2.)*np.sin(beta)*np.cos(gamma) - y*np.cos(beta))**2.)

where beta is a constant and y is defined between -1 and 1:

beta = 23.4
y = numpy.linspace(-1, 1, 100)

I want to plot the integral of this function for gamma evaluated from 0 to 2pi:

def integral(beta,gamma,y):
    for i in range(len(y)):
        I = integrate.quad(sbeta, 0., 2*np.pi, args=(beta, y[i]))
        print(I)
        plt.plot(y[i],I[0])

gamma = np.linspace(0., 2*np.pi, 10)
integral(beta,gamma,y)
plt.show()

There are no errors at this point, but I don’t think this is correct. I would like to compute the integral for gamma as variable from 0 to 2pi but y is a discrete array. How to compute and plot this for 10 y values from -1 and 1? Should I use scipy.integrate.cumtrapz?

Asked By: Arya

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Source

Answer 1

This is my attempt to the answer:

import numpy as np
import matplotlib.pyplot as plt
from scipy import integrate


def f(gamma, y):
    k = 2/(np.pi)**2.
    beta = 23.4*(np.pi/180.)
    return k * np.sqrt(1 - (np.sqrt(1-y**2.)*np.sin(beta)*np.cos(gamma) - y*np.cos(beta))**2.)


y = np.linspace(-1., 1.,10)
low = 0.
high = 2*np.pi

Ivals = []
for i in range(len(y)):
    I = integrate.quad(f, low, high, args=(y[i]))
    plt.scatter(y[i],I[0])
    Ivals.append(I)
plt.show()

which shows the following plot:

Can anyone confirm if this is correct?

Answered By: Arya

Python Plot an Integral with discrete and continuous parameters

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