数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 部分積分法、累乗、指数関数、対数関数、積、漸化式、階乗

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7の解答を求めてみる。

1. $\begin{array}{l} \int x^{2 n + 1} e^{- x^{2}} dx \\ = \frac{x^{2 (n + 1)}}{2 (n + 1)} e^{- x^{2}} - \frac{1}{2 (n + 1)} \int x^{2 (n + 1)} e^{- x^{2}} (- 2 x) dx \\ = \frac{x^{2 (n + 1)}}{2 (n + 1)} e^{- x^{2}} + \frac{1}{n + 1} \int x^{2 (n + 1) + 1} e^{- x^{2}} dx \\ \int_{0}^{\infty} x^{2 n + 1} e^{- x^{2}} dx = \frac{1}{n + 1} \int_{0}^{\infty} x^{2 (n + 1) + 1} e^{- x^{2}} dx \\ \int_{0}^{\infty} x^{2 (n + 1) + 1} e^{- x^{2}} dx = (n + 1) \int_{0}^{\infty} x^{2 n + 1} e^{- x^{2}} dx \\ \int_{0}^{\infty} x e^{- x^{2}} dx \\ = \lim_{b \to \infty} \int_{0}^{b} x e^{- x^{2}} dx \\ = - \frac{1}{2} \lim_{b \to \infty} {[e^{- x^{2}}]}_{0}^{b} \\ = - \frac{1}{2} \lim_{b \to \infty} (e^{- b^{2}} - 1) \\ = \frac{1}{2} \end{array}$
  
  よって、
  
  $\int_{0}^{n} x^{2 n + 1} e^{- x^{2}} dx = \frac{n!}{2}$
2. $\begin{array}{l} \int x^{α} {(\log x)}^{n} dx \\ = \frac{x^{α + 1}}{α + 1} {(\log x)}^{n} - \frac{1}{α + 1} \int x^{α + 1} n {(\log x)}^{n - 1} \cdot \frac{1}{x} dx \\ = \frac{x^{α + 1}}{α + 1} {(\log x)}^{n} - \frac{n}{α + 1} \int x^{α} {(\log x)}^{n - 1} dx \\ \int_{0}^{1} x^{α} {(\log x)}^{n} dx = - \frac{n}{α + 1} \int_{0}^{1} x^{α} {(\log x)}^{n - 1} dx \\ \int_{0}^{1} x^{α} \log x dx \\ = - \frac{1}{α + 1} \int_{0}^{1} x^{α} dx \\ = - \frac{1}{{(α + 1)}^{2}} {[x^{α + 1}]}_{0}^{1} \\ = - \frac{1}{{(α + 1)}^{2}} \end{array}$
  
  よって、
  
  $\begin{array}{l} \int_{0}^{1} x^{α} {(\log x)}^{n} dx \\ = {(- 1)}^{n} \frac{n!}{{(x + 1)}^{2 + n - 1}} \\ = {(- 1)}^{n} \frac{n!}{{(α + 1)}^{n + 1}} \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import pprint, symbols, Integral, log, exp, oo, Rational, plot print('7.') x = symbols('x', real=True) n = symbols('n', integer=True, nonnegative=True) f = x ** (2 * n + 1) * exp(-x ** 2) a = 2 g1 = x ** -Rational(1, 2) * log(x) ** n g2 = x ** 2 * log(x) ** n xs = [(0, oo), (0, 1), (0, 1)] for i, (h, (x1, x2)) in enumerate(zip([f, g1, g2], 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({n: n0}) for h in [f, g1, g2] for n0 in range(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) p.show() p.save('sample7.png')

% ./sample7.py 7. (1) ∞ ⌠ ⎮ 2 ⎮ 2⋅n + 1 -x ⎮ x ⋅ℯ dx ⌡ 0 n! ── 2 (2) 1 ⌠ ⎮ n ⎮ log (x) ⎮ ─────── dx ⎮ √x ⌡ 0 1 ⌠ ⎮ n ⎮ log (x) ⎮ ─────── dx ⎮ √x ⌡ 0 (3) 1 ⌠ ⎮ 2 n ⎮ x ⋅log (x) dx ⌡ 0 1 ⌠ ⎮ 2 n ⎮ x ⋅log (x) dx ⌡ 0 (cartesian line: x*exp(-x**2) for x over (-5.0, 5.0), red) (cartesian line: x**3*exp(-x**2) for x over (-5.0, 5.0), green) (cartesian line: x**5*exp(-x**2) for x over (-5.0, 5.0), blue) (cartesian line: 1/sqrt(x) for x over (-5.0, 5.0), brown) (cartesian line: log(x)/sqrt(x) for x over (-5.0, 5.0), orange) (cartesian line: log(x)**2/sqrt(x) for x over (-5.0, 5.0), purple) (cartesian line: x**2 for x over (-5.0, 5.0), pink) (cartesian line: x**2*log(x) for x over (-5.0, 5.0), gray) (cartesian line: x**2*log(x)**2 for x over (-5.0, 5.0), skyblue) %

Kamimura's blog

2020年1月16日木曜日

数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 部分積分法、累乗、指数関数、対数関数、積、漸化式、階乗

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