数学 - Python - 解析学 - 積分法 - 不定積分、広義積分(arcsine、arctan、数列の極限、級数)

解析入門〈2〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

解析入門〈2〉(松坂 和夫(著)、岩波書店)の第7章(積分法)、7.3(不定積分、広義積分)、問題1、2.を取り組んでみる。

1. 1. 
      $\begin{array}{l}{\left[\arcsin \frac{x}{a}\right]}_0^a\\ =\arcsin 1-\arcsin 0\\ =\frac{\pi }{2}-0\\ =\frac{\pi }{2}\end{array}$
   2. 
      $\begin{array}{l}{\left[\frac{1}{a}\arctan \frac{x}{a}\right]}_{-\infty }^{\infty }\\ =\frac{1}{a}\left(\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)\right)\\ =\frac{\pi }{a}\end{array}$
2. 1. 
      $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{\left(0+\frac{k\left(1-0\right)}{n}\right)}^{\alpha }\frac{1-0}{n}}\\ ={\displaystyle {\int }_0^1{x}^{\alpha }dx}\\ ={\left[\frac{x^{\alpha +1}}{\alpha +1}\right]}_0^1\\ =\frac{1}{\alpha +1}\end{array}$
   2. 
      $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\sin \pi \left(0+\frac{k\left(1-0\right)}{n}\right)\frac{1-0}{n}}\\ ={\displaystyle {\int }_0^1\sin \pi xdx}\\ ={\left[-\frac{1}{\pi }\cos \pi x\right]}_0^1\\ =-\frac{1}{\pi }\left(\cos \pi -\cos \pi \right)\\ =-\frac{1}{\pi }\left(-1-1\right)\\ =\frac{2}{\pi }\end{array}$
   3. 
      $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{n+i}}\\ =\underset{n\to \infty }{\lim }\frac{2-1}{n}{\displaystyle \sum _{i=1}^n\frac{n}{n+i\left(2-1\right)}}\\ =\underset{n\to \infty }{\lim }\frac{2-1}{n}{\displaystyle \sum _{i=1}^n\frac{1}{1+\frac{i\left(2-1\right)}{n}}}\\ ={\displaystyle {\int }_1^2\frac{1}{x}dx}\\ ={\left[\log x\right]}_1^2\\ =\log 2-\log 1\\ =\log 2\end{array}$
   4. 
      $\begin{array}{l}\underset{n\to \infty }{\lim }\underset{n}{\overset{i=1}{{{\displaystyle \sum }}^{\text{}}}}\frac{\sqrt{i}}{n\sqrt{n}}\hfill \\ =\underset{n\to \infty }{\lim }\frac{1-0}{n}\underset{n}{\overset{i=1}{{{\displaystyle \sum }}^{\text{}}}}\sqrt{0+\frac{i(1-0)}{n}}\hfill \\ =\underset{1}{\overset{0}{{{\displaystyle \int }}^{\text{}}}}\sqrt{x}dx\hfill \\ ={[\frac{2}{3}{x}^{\frac{3}{2}}]}_0^1\hfill \\ =\frac{2}{3}\hfill \end{array}$
   5. 
      $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{1}{\sqrt{n^2+in}}}\\ =\underset{n\to \infty }{\lim }\frac{2-1}{n}{\displaystyle \sum _{i=1}^n\frac{1}{\sqrt{1+\frac{i\left(2-1\right)}{n}}}}\\ ={\displaystyle {\int }_1^2\frac{1}{\sqrt{x}}dx}\\ ={\left[2x^{\frac{1}{2}}\right]}_1^2\\ =2\left(\sqrt{2}-1\right)\end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Integral, pi, sin, sqrt

x = symbols('x')
a = symbols('a', positive=True)
fs = [(x ** a, (0, 1)),
      (sin(pi * x), (0, 1)),
      (1 / x, (1, 2)),
      (sqrt(x), (0, 1)),
      (1 / sqrt(x), (1, 2))]
for i, (f, (x1, x2)) in enumerate(fs, 1):
    g = Integral(f, (x, x1, x2))
    pprint(g)
    result = g.doit()
    pprint(result.expand())
    pprint(result.factor())
    print()

入出力結果(Terminal, IPython)

$ ./sample1.py
1      
⌠      
⎮  a   
⎮ x  dx
⌡      
0      
  1  
─────
a + 1
  1  
─────
a + 1

1            
⌠            
⎮ sin(π⋅x) dx
⌡            
0            
2
─
π
2
─
π

2     
⌠     
⎮ 1   
⎮ ─ dx
⎮ x   
⌡     
1     
log(2)
log(2)

1      
⌠      
⎮ √x dx
⌡      
0      
2/3
2/3

2      
⌠      
⎮ 1    
⎮ ── dx
⎮ √x   
⌡      
1      
-2 + 2⋅√2
2⋅(-1 + √2)

$