西西爆难积分题求解
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. ∫10ln(1+x2+√3)1+xdx=π212(1−√3)+ln2⋅ln(1+√3)
2. ∫10ln(1+x4+√15)1+xdx=π212(2−√15)+ln1+√52⋅ln(2+√3)++ln2⋅ln(√3+√5)
3. ∫10ln(1+x6+√35)1+xdx=π212(3−√35)+ln1+√52⋅ln(8+3√7)++ln2⋅ln(√5+√7)
4. ∫10artanhxlnxx(1−x)(1+x)dx
5. ∫10arctanx3+√81+x2dx
6. ∫π30xln2(2sinx2)dx
7. ∫π20lnnsinxdx
8. ∫∞0sinxcoshx−cosx⋅xnn!dx
9. ∫∞0sinxcoshx+cosx⋅xnn!dx
10. ∫π2π4lnln(tanx)dx
11. ∫∞0x−sinx(π2+x2)x3dx=12π3(1+π22−π−1eπ)
12. ∫∞01a2+x2x−sinxx3dx=a2−2a+2−2e−a4a3,a>0
13. ∞∏k=4[1−(3k)3]=815561⋅cos(3√32π)π
14. ∞∑m=12m−1∑n=2m−1m(2n+1)(2n+2)=1−γ
15. ∫10(1−x)lnx⋅e−xπ−xdx
16. lim
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
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