数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 置換積分法、累乗(平方)、累乗根(平方根)、三角関数(正弦と余弦)、対数関数

学習環境

解析入門(上) (松坂和夫数学入門シリーズ 4) (松坂和夫(著)、岩波書店)の第8章(積分の計算)、8.2(定積分の計算)、問題6の解答を求めてみる。

1. $t = x + \sqrt{x^{2} - 1}$
  
  とおくと、
  
  $\begin{array}{l} \frac{dt}{dx} = 1 + \frac{x}{\sqrt{x^{2} - 1}} = \frac{x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} \\ \int_{1}^{\infty} \frac{dx}{(x + a) \sqrt{x^{2} - 1}} \\ = \int_{1}^{\infty} \frac{1}{(x + a) \sqrt{x^{2} - 1}} \cdot \frac{\sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}} dt \\ = \int_{1}^{\infty} \frac{dt}{(x + a) (x + \sqrt{x^{2} - 1})} \\ = \int_{1}^{\infty} \frac{dt}{t (x + a)} \\ t - x = \sqrt{x^{2} - 1} \\ x^{2} - 2 t x + t^{2} = x^{2} - 1 \\ x = \frac{t^{2} + 1}{2 t} \\ \int_{1}^{\infty} \frac{dt}{t (x + a)} \\ = \int_{1}^{\infty} \frac{dt}{t (\frac{t^{2} + 1}{2 t} + a)} \\ = \int_{1}^{\infty} \frac{2}{t^{2} + 2 a t + 1} dt \\ = 2 \int_{1}^{\infty} \frac{dt}{{(t + a)}^{2} + (1 - a^{2})} \\ s = t + a \\ \frac{d s}{dt} = 1 \\ 2 \int_{1}^{\infty} \frac{dt}{{(t + a)}^{2} + (1 - a^{2})} \\ = 2 \int_{a}^{\infty} \frac{d s}{s^{2} + (1 - a^{2})} \\ = 2 \lim_{b \to \infty} \int_{a}^{b} \frac{d s}{s^{2} + (1 - a^{2})} \end{array}$
  
  場合分け。
  
  $|a| < 1$
  
  の場合、
  
  $\begin{array}{l} 2 \lim_{b \to \infty} \int_{a}^{b} \frac{d s}{s^{2} + (1 - a^{2})} \\ = 2 \lim_{n \to \infty} \int_{a}^{b} \frac{d s}{s^{2} + {(\sqrt{1 - a^{2}})}^{2}} \\ = 2 \lim_{b \to \infty} {[\frac{1}{\sqrt{1 - a^{2}}} \arctan \frac{s}{\sqrt{1 - a^{2}}}]}_{n} \\ = \frac{2}{\sqrt{1 - a^{2}}} \lim_{b \to \infty} (\arctan \frac{b}{\sqrt{1 - a^{2}}} - \arctan \frac{a}{\sqrt{1 - a^{2}}}) \\ = \frac{2}{\sqrt{1 - a^{2}}} (\frac{π}{2} - \arctan \frac{a}{\sqrt{1 - a^{2}}}) \end{array}$
  
  また、
  
  $|a| = 1$
  
  の場合、
  
  $\begin{array}{l} 2 \lim_{b \to \infty} \int_{a}^{b} \frac{d s}{s^{2}} \\ = 2 \lim_{b \to \infty} {[- \frac{1}{s}]}_{a}^{b} \\ = 2 \lim_{b \to \infty} (- \frac{1}{b} + \frac{1}{a}) \\ = \frac{2}{a} \end{array}$
  
  また、
  
  $|a| > 1$
  
  の場合、
  
  $\begin{array}{l} 2 \lim_{b \to \infty} \int_{a}^{b} \frac{d s}{s^{2} + (1 - a^{2})} \\ = 2 \lim_{b \to \infty} \int_{a}^{b} \frac{d s}{s^{2} - {(\sqrt{a^{2} - 1})}^{2}} \\ \frac{A}{s + \sqrt{a^{2} - 1}} + \frac{B}{s - \sqrt{a^{2} - 1}} \\ = \frac{(A + B) s + (B - A) \sqrt{a^{2} - 1}}{s^{2} - {(\sqrt{a^{2} - 1})}^{2}} \\ A + B = 0 \\ B - A = \frac{1}{\sqrt{a^{2} - 1}} \\ 2 B = \frac{1}{\sqrt{a^{2} - 1}} \\ B = \frac{1}{2 \sqrt{a^{2} - 1}} \\ A = - \frac{1}{2 \sqrt{a^{2} - 1}} \\ 2 \lim_{b \to \infty} \int_{a}^{b} \frac{d s}{s^{2} + (1 - a^{2})} \\ = \frac{1}{\sqrt{a^{2} - 1}} \lim_{b \to \infty} \int_{a}^{b} (\frac{1}{s - \sqrt{a^{2} - 1}} - \frac{1}{s + \sqrt{a^{2} - 1}}) d s \\ = \frac{1}{\sqrt{a^{2} - 1}} \lim_{b \to \infty} {[\log \frac{s - \sqrt{a^{2} - 1}}{s + \sqrt{a^{2} - 1}}]}_{a}^{b} \\ = \frac{1}{\sqrt{a^{2} - 1}} \lim_{b \to \infty} (\log \frac{b - \sqrt{a^{2} - 1}}{b + \sqrt{a^{2} - 1}} - \log \frac{a - \sqrt{a^{2} - 1}}{a + \sqrt{a^{2} - 1}}) \\ = \frac{1}{\sqrt{a^{2} - 1}} \log \frac{a + \sqrt{a^{2} - 1}}{a - \sqrt{a^{2} - 1}} \end{array}$
2. $\begin{array}{l} t = 1 + 2 a (\cos θ) + a^{2} \\ \frac{dt}{d θ} = - 2 a (\sin θ) \\ \int_{0}^{π} \frac{\sin θ}{\sqrt{1 + 2 a (\cos θ) + a^{2}}} d θ \\ = - \frac{1}{2 a} \int_{a + 2 a + 1}^{a^{2} - 2 a + 1} \frac{1}{\sqrt{t}} dt \\ = - \frac{1}{2 a} \cdot {[2 \sqrt{t}]}_{a^{2} + 2 a + 1}^{a^{2} - 2 a + 1} \\ = - \frac{1}{a} (\sqrt{a^{2} - 2 a + 1} - \sqrt{a^{2} + 2 a + 1}) \\ = - \frac{1}{a} (\sqrt{{(a - 1)}^{2}} + \sqrt{{(a + 1)}^{2}}) \\ a \geq 1 \\ - \frac{1}{a} (a - 1 + a + 1) \\ = - 2 \\ |a| \leq 1 \\ - \frac{1}{a} (1 - a + a + 1) = - \frac{2}{a} \\ a \leq - 1 \\ - \frac{1}{a} (1 - a - a - 1) \\ = 2 \end{array}$

#!/usr/bin/env python3 from sympy import pprint, symbols, Integral, sqrt, sin, cos, oo, pi, plot print('6.') x, a = symbols('x, a', real=True) f = 1 / ((x + a) * sqrt(x ** 2 - 1)) g = sin(x) / sqrt(1 + 2 * a * cos(x) + a ** 2) xs = [(1, oo), (0, pi)] for i, (h, (x1, x2)) in enumerate(zip([f, g], xs), 1): print(f'({i})') I = Integral(h, (x, x1, x2)) for o in [I, I.doit()]: pprint(o.simplify()) print() p = plot(*[h.subs({a: a0}) for h in [f, g] for a0 in range(-2, 3)], (x, -5, 5), ylim=(-5, 5), legend=False, show=False) colors = ['red', 'green', 'blue', 'brown', 'orange', 'purple', 'pink', 'gray', 'skyblue', 'yellow'] for o, color in zip(p, colors): o.line_color = color for o in zip(p, colors): pprint(o) print() p.show() p.save('sample6.png')

% ./sample6.py 6. (1) ∞ ⌠ ⎮ 1 ⎮ ─────────────────── dx ⎮ ________ ⎮ ╱ 2 ⎮ (a + x)⋅╲╱ x - 1 ⌡ 1 ⎧⎧a⋅(2⋅acosh(│a│) - ⅈ⋅π) + ⅈ⋅π⋅│a│ 2 ⎪⎪──────────────────────────────── for a > 1 ⎪⎪ ________ ⎪⎪ ╱ 2 ⎪⎪ 2⋅╲╱ a - 1 ⋅│a│ ⎪⎪ ⎪⎨ π⋅│a│ for 2⋅│arg(a)│ < 2⋅π ⎪⎪ -a⋅asin(│a│) + ───── ⎪⎪ 2 ⎪⎪ ──────────────────── otherwise ⎪⎪ ________ ⎨⎪ ╱ 2 ⎪⎩ ╲╱ 1 - a ⋅│a│ ⎪ ⎪ ∞ ⎪ ⌠ ⎪ ⎮ 1 ⎪ ⎮ ─────────────────── dx otherwise ⎪ ⎮ ________ ⎪ ⎮ ╱ 2 ⎪ ⎮ (a + x)⋅╲╱ x - 1 ⎪ ⌡ ⎩ 1 (2) π ⌠ ⎮ sin(x) ⎮ ──────────────────────── dx ⎮ _____________________ ⎮ ╱ 2 ⎮ ╲╱ a + 2⋅a⋅cos(x) + 1 ⌡ 0 ⎧ ______________ ______________ ⎪ ╱ 2 ╱ 2 ⎪- ╲╱ a - 2⋅a + 1 + ╲╱ a + 2⋅a + 1 ⎨─────────────────────────────────────── for a ≠ 0 ⎪ a ⎪ ⎩ 2 otherwise (cartesian line: 1/((x - 2)*sqrt(x**2 - 1)) for x over (-5.0, 5.0), red) (cartesian line: 1/((x - 1)*sqrt(x**2 - 1)) for x over (-5.0, 5.0), green) (cartesian line: 1/(x*sqrt(x**2 - 1)) for x over (-5.0, 5.0), blue) (cartesian line: 1/((x + 1)*sqrt(x**2 - 1)) for x over (-5.0, 5.0), brown) (cartesian line: 1/((x + 2)*sqrt(x**2 - 1)) for x over (-5.0, 5.0), orange) (cartesian line: sin(x)/sqrt(5 - 4*cos(x)) for x over (-5.0, 5.0), purple) (cartesian line: sin(x)/sqrt(2 - 2*cos(x)) for x over (-5.0, 5.0), pink) (cartesian line: sin(x) for x over (-5.0, 5.0), gray) (cartesian line: sin(x)/sqrt(2*cos(x) + 2) for x over (-5.0, 5.0), skyblue) (cartesian line: sin(x)/sqrt(4*cos(x) + 5) for x over (-5.0, 5.0), yellow) %

Kamimura's blog

2020年1月15日水曜日

数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 置換積分法、累乗(平方)、累乗根(平方根)、三角関数(正弦と余弦)、対数関数

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