%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns
from scipy import integrate


"""
Explanation: Integration Exercise 2
Imports
End of explanation
"""


def integrand(x, a):
    return 1.0/(x**2 + a**2)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return 0.5*np.pi/a

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))


assert True # leave this cell to grade the above integral

"""
Explanation: Indefinite integrals
Here is a table of definite integrals. Many of these integrals has a number of parameters $a$, $b$, etc.
Find five of these integrals and perform the following steps:

Typeset the integral using LateX in a Markdown cell.
Define an integrand function that computes the value of the integrand.
Define an integral_approx funciton that uses scipy.integrate.quad to peform the integral.
Define an integral_exact function that computes the exact value of the integral.
Call and print the return value of integral_approx and integral_exact for one set of parameters.

Here is an example to show what your solutions should look like:
Example
Here is the integral I am performing:
$$ I_1 = \int_0^\infty \frac{dx}{x^2 + a^2} = \frac{\pi}{2a} $$
End of explanation
"""


def integrand(x, a):
    return (1-np.cos(a*x))/x**2

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return 0.5*np.pi*a

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))


assert True # leave this cell to grade the above integral

"""
Explanation: Integral 1
Here is the integral I am performing:
$$ I_1 = \int_0^\infty \frac{1-cos(ax)}{x^2}dx=\frac{\pi a}{2} $$
End of explanation
"""


def integrand(x, a):
    return np.sqrt(a**2 - x**2)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, a, args=(a,))
    return I

def integral_exact(a):
    return 0.25*np.pi*(a**2)

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))


assert True # leave this cell to grade the above integral

"""
Explanation: Integral 2
Here is the integral I am performing:
$$ I_2 = \int_0^a \sqrt{a^2-x^2}dx=\frac{\pi a^2}{4} $$
End of explanation
"""


def integrand(x, a):
    return np.exp(-a*(x**2))

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return 0.5*np.sqrt(np.pi/a)

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))


assert True # leave this cell to grade the above integral

"""
Explanation: Integral 3
Here is the integral I am performing:
$$ I_3 = \int_0^\infty e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}} $$
End of explanation
"""


def integrand(x, a):
    return x/(np.exp(x)+1)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, np.inf, args=(a,))
    return I

def integral_exact(a):
    return (np.pi**2)/12

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))


assert True # leave this cell to grade the above integral

"""
Explanation: Integral 4
Here is the integral I am performing:
$$ I_4 = \int_0^\infty \frac{x}{e^x+1}dx=\frac{\pi^2}{12} $$
End of explanation
"""


def integrand(x, a):
    return np.log(1+x)/(x)

def integral_approx(a):
    # Use the args keyword argument to feed extra arguments to your integrand
    I, e = integrate.quad(integrand, 0, 1, args=(a,))
    return I

def integral_exact(a):
    return (np.pi**2)/12

print("Numerical: ", integral_approx(1.0))
print("Exact    : ", integral_exact(1.0))

assert True # leave this cell to grade the above integral

"""
Explanation: Integral 5
Here is the integral I am performing:
$$ I_5 = \int_0^1 \frac{ln(1+x)}{x}dx=\frac{\pi^2}{12} $$
End of explanation
"""


