Riemann积分有两个局限：一是积分区间[a, b]必须有限；二是被积函数f(x)在积分区间[a, b]上有界。而反常积分（Improper Integral）就是对Riemann积分这两个局限的扩展，前者扩展后被称为“无穷积分”，后者扩展后被称为“瑕积分”。
  本节的内容属于“微积分的扩展”，课程中介绍的本就不太详细，主要讨论反常积分的“敛散性”我就不作扩展了，仅用Python辅助，展示课后练习和单元测试的解答。

#Exercise 7-7-1
from sympy import *
init_printing()
x = Symbol('x')
f = 1 / (1 + E ** x) ** 2
f, integrate(f, (x, 0, +oo))

$$\left ( \frac{1}{\left(e^{x} + 1\right)^{2}}, \quad - \frac{1}{2} + \log{\left (2 \right )}\right )$$

#Exercise 7-7-2
x = Symbol('x')
f = 1 / (x ** 2 + 4 * x + 9)
f, integrate(f, (x, -oo, +oo))

$$\left ( \frac{1}{x^{2} + 4 x + 9}, \quad \frac{\sqrt{5} \pi}{5}\right )$$

#Exercise 7-7-3
x = Symbol('x')
f = log(x) / (1 + x) ** 2
f, integrate(f, (x, 0, +oo))

$$\left ( \frac{\log{\left (x \right )}}{\left(x + 1\right)^{2}}, \quad 0\right )$$

#Exercise 7-7-4
x = Symbol('x')
f = 1 / sqrt(abs((1 - x)))
f1 = 1 / sqrt(1 - x)
f2 = 1 / sqrt(x - 1)
f, integrate(f1, (x, 0, 1)),  integrate(f2, (x, 1, 2))

$$\left ( \frac{1}{\sqrt{\left|{x - 1}\right|}}, \quad 2, \quad 2\right )$$

#Exercise 7-7-4
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
#x = np.array(filter(lambda t:t != 1, np.linspace(0, 2))).flatten()
x = np.linspace(0, 2)
t = np.asarray([1]).repeat(x.size)
y = 1 / np.sqrt(np.abs((t - x)))
plt.plot(x,y)
plt.show()

Exercise 7-7-5

$$\int_0^1 \sqrt{ \frac {x}{1-x}} \mathrm{d} x = ?$$
解：本题直接用sympy计算不出来，观察它的被积函数，其中含有2次根号，很容易相等“三角函数平方和/差公式”，令$x = sin^2(t)$，得：
$$\int_0^1 \sqrt{ \frac {x}{1-x}} \mathrm{d} x = \int_0^{\frac {\pi}{2}} tan(t) \mathrm{d} [sin^2(t)] = 2\int_0^{\frac {\pi}{2}} sin^2(t) \mathrm{d} t$$

#Exercise 7-7-3
x = Symbol('x')
f = sin(x) ** 2
2 * integrate(f, (x, 0, pi / 2))

$$\frac{\pi}{2}$$

Exercise 7-7-6
$$\int_0^{+\infty} \frac {\mathrm{d} x}{(x+1)\sqrt{x^2 + 1}} = ?$$
解：同样地，用三角函数换元法
$$\int_0^{+\infty} \frac {\mathrm{d} x}{(x+1)\sqrt{x^2 + 1}} = \int_0^{\frac {\pi}{2}} \frac {cos^2(t) \mathrm{d} [tan(t)]}{sin(t)+ cos(t)} \\ = \int_0^{\frac {\pi}{2}} \frac {\mathrm{d} t}{sin(t)+ cos(t)} = \int_0^{\frac {\pi}{2}} \frac {\mathrm{d} t}{\sqrt{2}sin(t + \frac{\pi}{4})}$$

#Exercise 7-7-6
x = Symbol('x')
f = 1 / (sqrt(2) * sin(x + pi / 4))
integrate(f, (x, 0, pi / 2))

$$\frac{\sqrt{2}}{2} \log{\left (\sqrt{2} \left(\frac{\sqrt{2}}{2} + 1\right) \right )} - \frac{\sqrt{2}}{2} \log{\left (\sqrt{2} \left(- \frac{\sqrt{2}}{2} + 1\right) \right )}$$

Exercise 7-7-1b

$$\int_2^{+\infty} \frac {\mathrm{d} x}{xln^p(x)}$$
若上式收敛，求p的范围
解：运用“比阶判别法的极限形式”
$$\lim_{x \rightarrow + \infty} x^qf(x) = \lim_{x \rightarrow + \infty} \frac {x ^ {q-1}}{ln^p(x)}$$
要是这个极限存在，则 $p \geq q - 1 > 1 - 1 = 0$

Exercise 7-7-2b

$$\int_2^{+\infty} \frac {\mathrm{d} x}{x + \sqrt{x^2 + 2}}$$
解：先看被积函数的性质
$$f(x) = \frac {1}{x + \sqrt{x^2 + 2}} = \frac {1}{2}(\sqrt{x^2 + 2} - x)$$
$$f'(x) = \frac {1}{2}(\frac {x}{\sqrt{x^2 + 2}} -1) = \frac {x - \sqrt{x^2 + 2}}{2\sqrt{x^2 + 2}} < 0$$
$$\lim_{x \rightarrow + \infty} f(x) = 0$$
因此，f(x)单调递减，且有下界，必收敛
运用“比阶判别法的极限形式”
$$\lim_{x \rightarrow + \infty} x^qf(x) = \lim_{x \rightarrow + \infty} \frac {x^p}{x + \sqrt{x^2 + 2}}$$
p=1时，这个极限值存在且不等于0，故该无穷积分发散


Exercise 7-7-3b
$$\int_0^1 ln (x) \mathrm{d} x$$
解：运用瑕积分收敛判断$(x-a)^p$
$$\lim_{x \rightarrow 0^+} x^p * ln(x)$$
当 p = -1 < 1 时，这个极限存在，故该瑕积分收敛


Exercise 7-7-4b
$$\int_0^1 \frac {e^x}{\sqrt{1-x}} \mathrm{d} x$$
解：运用瑕积分收敛判断$(x-a)^p$
$$\lim_{x \rightarrow 0^+} x^p * \frac {e^x}{\sqrt{1-x}} = lim_{x \rightarrow 0^+} x^p$$
当 0 < p < 1 时，这个极限存在，故该瑕积分收敛

#Exercise 7-7-5b
from sympy import *
init_printing()
x = Symbol('x')
for n in range(-1, 3):f = x ** (n) * E ** (-x)F = integrate(f, x)print(limit(F, x, +oo))

0
0
0
0


Unit Test 4-1
解：本题考察被积函数与原函数的关系。可以利用变上限积分将被积函数与原函数联系起来，假设其中一个积分限 a = 0

$$F(-x) = \int_0^{-x} f(t)\mathrm{d}t = \int_0^{x} f(-u)\mathrm{d}(-u) = -\int_0^{x} f(-u)\mathrm{d}u$$
很明显，f(x)若为奇函数，F(x)将为偶函数。

#Unit Test 4-5
from sympy import *
init_printing()
x = Symbol('x')
f = cos(2 * x) / (1 + 1 / 2 * sin(2 * x))
integrate(f, x)

$$1.0 \log{\left (0.5 \sin{\left (2 x \right )} + 1.0 \right )}$$


Unit Test 4-6
$$\int \frac{arctan \frac{1}{x}}{1 + x^2} \mathrm{d} x = ?$$
解：三角函数换元，令$x = tan(t), t = arctan(x)$
$$\int \frac{arctan \frac{1}{x}}{1 + x^2} \mathrm{d} x = \int \frac{arctan \frac{1}{tan(t)}}{1 + tan^2(t)} \mathrm{d} [tan(t)] \\ = \int arctan \frac{1}{tan(t)} \mathrm{d} t = \int (\frac{\pi}{2} - t) \mathrm{d} t$$


Unit Test 4-7
$$\int \frac{\mathrm{d} x}{e^x(1 + e^{2x})} = ?$$
解：先分解部分分式（partial fraction），再换元。其中分式分解也可以用sympy执行，如下代码
$$f(x) = \frac{1}{e^x(1 + e^{2x})} = \frac{1}{e^x} - \frac{e^x}{1 + e^{2x}}$$

#Unit Test 4-7
from sympy import *
init_printing()
x = Symbol('x')
expr = 1 / (x * (1 + x ** 2))
apart(expr)

$$- \frac{x}{x^{2} + 1} + \frac{1}{x}$$

t = E ** x
f = 1 / t
g = 1 / (1 + x ** 2)
integrate(f, x), integrate(g, x), integrate(f, x) - integrate(g, x).subs(x, E ** x) 

$$\left ( - e^{- x}, \quad \operatorname{atan}{\left (x \right )}, \quad - \operatorname{atan}{\left (e^{x} \right )} - e^{- x}\right )$$

#Unit Test 4-8
from sympy import *
init_printing()
x = Symbol('x')
f = 1 / (x ** 2 + 2 * x + 5)
integrate(f, x)

$$\frac{1}{2} \operatorname{atan}{\left (\frac{x}{2} + \frac{1}{2} \right )}$$


Unit Test 4-9
$$\int \frac{x\mathrm{d} x}{(1 + x^2) \sqrt{1 - x^2}} = ?$$
解：先凑微分，再三角函数替换。
$$\int \frac{x\mathrm{d} x}{(1 + x^2) \sqrt{1 - x^2}} = \int \frac{\mathrm{d} sin^2(t)}{2[1 + sin^2(t)] \sqrt{1 - sin^2(t)}} \\ = \int \frac{sin(t)cos(t)\mathrm{d} t}{[1 + sin^2(t)] cos(t)} = -\int \frac{\mathrm{d} cos(t)}{2 - cos^2(t)}$$
注意，本题经过了两次换元，求出原函数后要换回去，如下代码

#Unit Test 4-9
from sympy import *
init_printing()
x = Symbol('x')
f = - 1 / ( 2 - x ** 2)
integrate(f, x).subs(x, cos(x)).subs(x, asin(x))

$$\frac{\sqrt{2}}{4} \log{\left (\sqrt{- x^{2} + 1} - \sqrt{2} \right )} - \frac{\sqrt{2}}{4} \log{\left (\sqrt{- x^{2} + 1} + \sqrt{2} \right )}$$


Unit Test 4-10
$$\int \frac{\sqrt{x^2 - 9}}{x^2} \mathrm{d} x = ?$$
解：先有理化，再三角换元
$$\int \frac{\sqrt{x^2 - 9}}{x^2} \mathrm{d} x = \int \frac{u}{u^2 + 9} \mathrm{d} \sqrt{u^2 + 9} \\ = \int \frac{3tan(\theta)}{9tan^2 \theta + 9} \mathrm{d} \sqrt{9tan^2 \theta + 9} \\ = \int \frac{tan(\theta)}{sec^2 \theta} \mathrm{d} sec (\theta) = \int \frac{sin^2\theta}{cos (\theta)} \mathrm{d} \theta \\ = \int \frac{1}{cos (\theta)} \mathrm{d} \theta - \int cos (\theta) \mathrm{d} \theta$$

#Unit Test 4-10
from sympy import *
init_printing()
x = Symbol('x')
f = 1 / cos(x) - cos(x)
integrate(f, x).subs(x, atan(1 / 3 * x)).subs(x, sqrt(x ** 2 - 9))

$$- \frac{1.0}{\sqrt{x^{2}}} \sqrt{x^{2} - 9} - \frac{1}{2} \log{\left (\frac{1.0}{\sqrt{x^{2}}} \sqrt{x^{2} - 9} - 1 \right )} + \frac{1}{2} \log{\left (\frac{1.0}{\sqrt{x^{2}}} \sqrt{x^{2} - 9} + 1 \right )}$$

#Unit Test 4-11
from sympy import *
init_printing()
x = Symbol('x')
f = 1 / (1 + x + x ** 2)
integrate(f, x).subs(x, 2 ** x) / log(2)

$$\frac{2 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3}}{3} 2^{x} + \frac{\sqrt{3}}{3} \right )}}{3 \log{\left (2 \right )}}$$


Unit Test 4-12
$$I_n = \int \frac{\mathrm{d} x}{x^n \sqrt{x^2 + 1}}$$
求上式的递推公式
解：很显然，要应用“分部积分法”
$$I_{n-2} = \int \frac{\mathrm{d} x}{x^{n-2} \sqrt{x^2 + 1}} = \int \frac{\mathrm{d} \sqrt{x^2 + 1}}{x^{n-1} } \\ = \frac{\sqrt{x^2 + 1}}{x^{n-1} } - \int \sqrt{x^2 + 1}\mathrm{d}x^{1-n} \\ = \frac{\sqrt{x^2 + 1}}{x^{n-1} } + (n-1) \int \frac{\sqrt{x^2 + 1}}{x^n} \mathrm{d}x \\ = \frac{\sqrt{x^2 + 1}}{x^{n-1} } + (n-1) \int \frac{x^2 + 1}{x^n\sqrt{x^2 + 1}} \mathrm{d}x \\ = \frac{\sqrt{x^2 + 1}}{x^{n-1} } + (n-1)(I_{n} + I_{n-2})$$


Unit Test 4-13
$$\int \frac{\mathrm{d} x}{x(x^{10} + 1)} = ?$$
解：很显然，这是一个“Partial Fraction” （部分分式）的积分，要先进行分式分解
$$f(x) = \frac{1}{x(x^{10} + 1)} = \frac{1}{x} - \frac{x}{5x^2+5} - \frac{x(4x^6-3x^4+2x^2-1)}{5x^8-5x^6+5x^4-5x^2+5}$$
然后根据第二三项的特点进行“凑微分和换元”，凑成$x^2$

#Unit Test 4-13
from sympy import *
init_printing()
x = Symbol('x')
f = 1 / (x ** 11 + x)
apart(f), integrate(f, x)

$$\left ( - \frac{x \left(4 x^{6} - 3 x^{4} + 2 x^{2} - 1\right)}{5 x^{8} - 5 x^{6} + 5 x^{4} - 5 x^{2} + 5} - \frac{x}{5 x^{2} + 5} + \frac{1}{x}, \quad \log{\left (x \right )} - \frac{1}{10} \log{\left (x^{10} + 1 \right )}\right )$$

#Unit Test 4-14
from sympy import *
init_printing()
x = Symbol('x')
f = (x ** 2 + 1) * E ** (2 * x)
integrate(f, x)

$$\frac{e^{2 x}}{4} \left(2 x^{2} - 2 x + 3\right)$$

#Unit Test 4-15
from sympy import *
init_printing()
x = Symbol('x')
f = x ** 2 * acos(x)
integrate(f, x)

$$\frac{x^{3}}{3} \operatorname{acos}{\left (x \right )} - \frac{x^{2}}{9} \sqrt{- x^{2} + 1} - \frac{2}{9} \sqrt{- x^{2} + 1}$$


Unit Test 4-16
$$\int \frac{\mathrm{d} x}{\sqrt{x} + \sqrt[3]{x}} = ?$$
解：先有理化
$$\int \frac{\mathrm{d} x}{\sqrt{x} + \sqrt[3]{x}} = \int \frac{\mathrm{d} t^6}{t^3 + t^2} = \int \frac{6x^5\mathrm{d} t}{t^3 + t^2}$$

#Unit Test 4-16
from sympy import *
init_printing()
x = Symbol('x')
f = 6 * x ** 5 / (x ** 2 + x ** 3)
integrate(f, x).subs(x, x ** (1 / 6))

$$6 x^{0.166666666666667} - 3 x^{0.333333333333333} + 2 x^{0.5} - 6 \log{\left (x^{0.166666666666667} + 1 \right )}$$


Unit Test 4-17
$$\int \sqrt {\frac{x+1}{x}} \mathrm{d} x = ?$$
解：先有理化
$$\int \sqrt {\frac{x+1}{x}} \mathrm{d} x = \int \sqrt {\frac{tan^2(t)+1}{tan^2(t)}} \mathrm{d} tan^2(t)$$

#Unit Test 4-17
from sympy import *
init_printing()
x = Symbol('x')
f = diff(tan(x) ** 2, x)
g = 2 / (cos(x) ** 3)
f, integrate(g, x).subs(x, atan(x))

$$\left ( \left(2 \tan^{2}{\left (x \right )} + 2\right) \tan{\left (x \right )}, \quad - \frac{2 x}{\sqrt{x^{2} + 1} \left(\frac{2 x^{2}}{x^{2} + 1} - 2\right)} - \frac{1}{2} \log{\left (\frac{x}{\sqrt{x^{2} + 1}} - 1 \right )} + \frac{1}{2} \log{\left (\frac{x}{\sqrt{x^{2} + 1}} + 1 \right )}\right )$$


Unit Test 4-18
$$\int \frac{\mathrm{d} x }{\sqrt{e^x+1}+\sqrt{e^x-1}} = ?$$
解：先有理化，再分部积分
$$\int \frac{\mathrm{d} x }{\sqrt{e^x+1}+\sqrt{e^x-1}} = \int \frac{\sqrt{e^x+1}-\sqrt{e^x-1}}{2} \mathrm{d} x \\ = \frac{1}{2}\int \sqrt{e^x+1} \mathrm{d} x - \frac{1}{2}\int \sqrt{e^x-1} \mathrm{d} x \\ = \frac{1}{2}\int t \mathrm{d} ln(t^2 -1) - \frac{1}{2}\int u \mathrm{d} ln(u^2 +1) \\ = \frac{1}{2}[t * ln(t^2 -1) - \int ln(t^2 -1) \mathrm{d} t - u * ln(u^2 +1) + \int ln(u^2 +1)\mathrm{d} u]$$

#Unit Test 4-18
from sympy import *
init_printing()
x = Symbol('x')
f = log(x ** 2 -1)
g = log(x ** 2 + 1)
integrate(f, x), integrate(g, x)

$$\left ( x \log{\left (x^{2} - 1 \right )} - 2 x - \log{\left (x - 1 \right )} + \log{\left (x + 1 \right )}, \quad x \log{\left (x^{2} + 1 \right )} - 2 x + 2 \operatorname{atan}{\left (x \right )}\right )$$

part1 = x * f - integrate(f, x)
part2 = x * g - integrate(g, x)
part1.subs(x, sqrt(E ** x + 1)) - part2.subs(x, sqrt(E ** x - 1))

$$- 2 \sqrt{e^{x} - 1} + 2 \sqrt{e^{x} + 1} + \log{\left (\sqrt{e^{x} + 1} - 1 \right )} - \log{\left (\sqrt{e^{x} + 1} + 1 \right )} + 2 \operatorname{atan}{\left (\sqrt{e^{x} - 1} \right )}$$


Unit Test 4-19
$$\int \frac{sin(x)}{1 + sin(x)} \mathrm{d} x = ?$$
解：简化分式
$$\int \frac{sin(x)}{1 + sin(x)} \mathrm{d} x = \int \mathrm{d} x - \int \frac{1}{1 + sin(x)} \mathrm{d} x$$

#Unit Test 4-19
from sympy import *
init_printing()
x = Symbol('x')
f = 1 / (sin(x) + 1)
integrate(f, x)

$$- \frac{2}{\tan{\left (\frac{x}{2} \right )} + 1}$$


Unit Test 4-20
$$\int \frac{x^2}{1 + x^2} atan(x)\mathrm{d} x = ?$$
解：简化分式
$$\int \frac{x^2}{1 + x^2} atan(x)\mathrm{d} x = \int x^2 atan(x)\mathrm{d} atan(x) \\ = \frac{1}{2} \int x^2(x^2 + 1) \mathrm{d} atan^2(x) \\ = \frac{1}{2}[x^2(x^2 + 1) * atan^2(x) - \int atan^2(x) \mathrm{d} x^2(x^2 + 1)] \\ = \frac{1}{2}[x^2(x^2 + 1) * atan^2(x) - 4\int atan^2(x) * x^3 \mathrm{d} x - 2\int atan^2(x) * x \mathrm{d} x]$$

#Unit Test 4-19
from sympy import *
init_printing()
x = Symbol('x')
f = atan(x) ** 2 * x
integrate(f, x)

$$\frac{x^{2}}{2} \operatorname{atan}^{2}{\left (x \right )} - x \operatorname{atan}{\left (x \right )} + \frac{1}{2} \log{\left (x^{2} + 1 \right )} + \frac{1}{2} \operatorname{atan}^{2}{\left (x \right )}$$

g = f * x ** 2
integrate(g, x)

$$\frac{x^{4}}{4} \operatorname{atan}^{2}{\left (x \right )} - \frac{x^{3}}{6} \operatorname{atan}{\left (x \right )} + \frac{x^{2}}{12} + \frac{x}{2} \operatorname{atan}{\left (x \right )} - \frac{1}{3} \log{\left (x^{2} + 1 \right )} - \frac{1}{4} \operatorname{atan}^{2}{\left (x \right )}$$

simplify(atan(x) ** 2 * (x ** 4 + x ** 2) - 4 * integrate(g, x) - 2 * integrate(f, x))

$$\frac{2 x^{3}}{3} \operatorname{atan}{\left (x \right )} - \frac{x^{2}}{3} + \frac{1}{3} \log{\left (x^{2} + 1 \right )}$$