西西爆难积分题求解 - Eufisky - The lost book
西西11年送的几个CMC培训题整理
一个与多项式分拆有关的级数题

西西爆难积分题求解

Eufisky posted @ 2014年8月27日 03:07 in 数学分析 with tags 积分计算 西西 , 1755 阅读

在这里,主要展示西西12年7月在百度贴吧数学吧中贴出的30个积分题的求解,本文中主要参考的是自己以前的摘录,不知何故,与西哥的版本有些出入,但基本包含了西哥的所有问题,来源于网友的解答均会注明出处。

1. \[\int_0^1 {\frac{{\ln \left( {1 + {x^{2 + \sqrt 3 }}} \right)}}{{1 + x}}dx}  = \frac{{{\pi ^2}}}{{12}}\left( {1 - \sqrt 3 } \right) + \ln 2 \cdot \ln \left( {1 + \sqrt 3 } \right)\]
2. \[\int_0^1 {\frac{{\ln \left( {1 + {x^{4 + \sqrt {15} }}} \right)}}{{1 + x}}dx}  = \frac{{{\pi ^2}}}{{12}}\left( {2 - \sqrt {15} } \right) + \ln \frac{{1 + \sqrt 5 }}{2} \cdot \ln \left( {2 + \sqrt 3 } \right) +  + \ln 2 \cdot \ln \left( {\sqrt 3  + \sqrt 5 } \right)\]
3. \[\int_0^1 {\frac{{\ln \left( {1 + {x^{6 + \sqrt {35} }}} \right)}}{{1 + x}}dx}  = \frac{{{\pi ^2}}}{{12}}\left( {3 - \sqrt {35} } \right) + \ln \frac{{1 + \sqrt 5 }}{2} \cdot \ln \left( {8 + 3\sqrt 7 } \right) +  + \ln 2 \cdot \ln \left( {\sqrt 5  + \sqrt 7 } \right)\]
4. \[\int_0^1 {\frac{{ar\tanh x\ln x}}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}dx} \]
5. \[\int_0^1 {\frac{{\arctan {x^{3 + \sqrt 8 }}}}{{1 + {x^2}}}dx} \]
6. \[\int_0^{\frac{\pi }{3}} {x{{\ln }^2}\left( {2\sin \frac{x}{2}} \right)dx} \]
7. \[\int_0^{\frac{\pi }{2}} {{{\ln }^n}\sin xdx} \]
8. \[\int_0^\infty  {\frac{{\sin x}}{{\cosh x - \cos x}} \cdot \frac{{{x^n}}}{{n!}}dx} \]
9. \[\int_0^\infty  {\frac{{\sin x}}{{\cosh x + \cos x}} \cdot \frac{{{x^n}}}{{n!}}dx} \]
10. \[\int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\ln \ln \left( {\tan x} \right)dx} \]
11. \[\int_0^\infty  {\frac{{x - \sin x}}{{\left( {{\pi ^2} + {x^2}} \right){x^3}}}dx}  = \frac{1}{{2{\pi ^3}}}\left( {1 + \frac{{{\pi ^2}}}{2} - \pi  - \frac{1}{{{e^\pi }}}} \right)\]
12. \[\int_0^\infty  {\frac{1}{{{a^2} + {x^2}}}\frac{{x - \sin x}}{{{x^3}}}dx}  = \frac{{{a^2} - 2a + 2 - 2{e^{ - a}}}}{{4{a^3}}},a > 0\]
13. \[\prod\limits_{k = 4}^\infty  {\left[ {1 - {{\left( {\frac{3}{k}} \right)}^3}} \right]}  = \frac{8}{{15561}} \cdot \frac{{\cos \left( {\frac{{3\sqrt 3 }}{2}\pi } \right)}}{\pi }\]
14. \[\sum\limits_{m = 1}^\infty  {\sum\limits_{n = {2^{m - 1}}}^{{2^m} - 1} {\frac{m}{{\left( {2n + 1} \right)\left( {2n + 2} \right)}}} }  = 1 - \gamma \]
15. \[\int_0^1 {\frac{{\left( {1 - x} \right)\ln x \cdot {e^{ - x}}}}{{\pi  - x}}dx} \]
16. \[\mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{1}{{2m}} + \ln \left( {\frac{e}{m}} \right) + \sum\limits_{n = 2}^m {\left( {\frac{1}{n} - \frac{{\varsigma \left( {1 - n} \right)}}{{{m^n}}}} \right)} } \right] = \gamma \]
17. \[\int_0^\infty  {\frac{1}{{x{e^x}\left( {{\pi ^2} + {{\ln }^2}x} \right)}}dx} \]
18. \[\int_0^\infty  {\frac{{\ln \left( {1 + {x^2}} \right)}}{{{e^{2\pi x}} - 1}}dx} \]
19. \[\sum\limits_{n = 1}^\infty  {\frac{{\sum\limits_{k = 1}^n {\frac{1}{{{k^4}}}} }}{{{n^2}}}} \]
20. \[\int_0^1 {\int_0^1 {\int_0^1 {\int_0^1 {\frac{{dwdxdydz}}{{\left( {wxy - 1} \right)\left( {zwx - 1} \right)\left( {yzw - 1} \right)\left( {xyz - 1} \right)}}} } } } \]
21. \[\int_0^\infty  {\frac{{{e^{ - {x^2}}}}}{{{\pi ^2} + {{\left( {\gamma  + x} \right)}^2}}}dx} \]
22. \[\sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 +  \cdots }}}}}}}}}} = \frac{2}{{\sqrt 3 }}\cos \left( {\frac{1}{3}\arccos \frac{{3\sqrt 3 }}{2}} \right)\]
23. \[\int_0^{2ar\cosh \pi } {\frac{{dx}}{{1 + \frac{{{{\sinh }^2}x}}{{{\pi ^4}}}}}} \]
24. \[\int_0^\infty  {\frac{{x{e^x}}}{{\sqrt {4{e^x} - 3} \left( {1 + 2{e^x} - \sqrt {4{e^x} - 3} } \right)}}dx} \]
25. \[\int_0^{\frac{\pi }{2}} {\frac{{{{\ln }^2}\left( {2\cos x} \right)}}{{{{\ln }^2}\left( {2\cos x} \right) + {x^2}}}dx} \]
26. \[\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{{n^n}}}\left( {\sum\limits_{k = 0}^n {\frac{{{n^k}}}{{k!}}}  - \sum\limits_{k = n + 1}^\infty {\frac{{{n^k}}}{{k!}}} } \right)\]
27. \[\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^2}}}\cos \left( {\frac{9}{{k\pi  + \sqrt {{k^2}{\pi ^2} - 9} }}} \right)} \]
28. \[\sum\limits_{n =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n\pi  + \phi } \right)}^2}}}\cos \left( {\sqrt {{n^2}{\pi ^2} + {a^2} - {\phi ^2}} } \right)}  = \frac{{a\cos a\cot \phi  + \phi \sin a}}{{a\sin \phi }}\]
29. \[\sum\limits_{n = 0}^\infty  {\frac{{{n^2}{\pi ^2} + {\phi ^2}}}{{{{\left( {{n^2}{\pi ^2} - {\phi ^2}} \right)}^2}}}{{\left( { - 1} \right)}^n}\cos \left( {\sqrt {{n^2}{\pi ^2} + {a^2} - {\phi ^2}} } \right)}  = \frac{{\cos \sqrt {{a^2} - {\phi ^2}} }}{{2{\phi ^2}}} + \frac{{a\cos a\cot \phi  + \phi \sin a}}{{2a\sin \phi }}\]
30. \[\sum\limits_{n = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^n}\left( {n + \frac{1}{2}} \right)}}{{{{\left( {n + \frac{1}{3}} \right)}^2}{{\left( {n + \frac{2}{3}} \right)}^2}}}} \cos \left[ {\pi \sqrt {\left( {n + \frac{1}{6}} \right)\left( {n + \frac{5}{6}} \right)} } \right] = {\pi ^2}{e^{\frac{{\pi \sqrt 3 }}{6}}}\]
31. \[\int_{ - 1}^1 {\frac{{\arctan x}}{{1 + x}}\ln \left( {\frac{{1 + {x^2}}}{2}} \right)dx} \]
32. \[\int_0^1 {\sin \left( {\pi x} \right){x^x}{{\left( {1 - x} \right)}^{1 - x}}dx} \]
33. \[\int_0^{\frac{\pi }{2}} {{x^2}{{\ln }^2}\left( {2\cos x} \right)dx} \]
34. \[\int_0^{\frac{\pi }{2}} {\frac{{{x^2}}}{{{x^2} + {{\ln }^2}\left( {2\cos x} \right)}}dx} \] 
35. \[\int_0^{\frac{\pi }{2}} {\frac{{{x^2}{{\ln }^2}\left( {2\cos x} \right)}}{{{x^2} + {{\ln }^2}\left( {2\cos x} \right)}}dx}\]
36. \[\mathop {\lim }\limits_{n \to \infty } \frac{n}{{\ln n}}\left[ {\frac{1}{\pi }{{\left( {\sum\limits_{k = 1}^n {\sin \frac{\pi }{{\sqrt {{n^2} + k} }}} } \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right]\]
37. \[\int_0^1 {{x^{ - x}}{{\left( {1 - x} \right)}^{ - 1 + x}}\sin \left( {\pi x} \right)dx = \frac{\pi }{e}}  = 1.15573\]
38. \[\int_0^{ + \infty } {\frac{{dx}}{{1 + x\left| {\sin x} \right|}}} \]
39. \[\int_0^{ + \infty } {\frac{{dx}}{{{x^2} + {{\left( {{n^2}{x^2} - 1} \right)}^2}}}}\]

参阅:[1]http://tieba.baidu.com/p/2114477017

[2]http://tieba.baidu.com/p/3148596990

[3]http://tieba.baidu.com/p/2121721064

[4]http://tieba.baidu.com/p/2908551228

[5]http://tieba.baidu.com/p/1700279486?qq-pf-to=pcqq.c2c


登录 *


loading captcha image...
(输入验证码)
or Ctrl+Enter