AOPS论坛上100+100个积分

100+10 rare and irresistible integrals

I bring you many beautiful integrals that I have collected over time, I hope you enjoy them as much as I do.
If you want to answer one of these integrals, please hide your answer.
#passion for this #Enjoy :showoff: :-D :weightlift: :stretcher:

1. Coxeter Integrals 0π2arccos(cosθ1+2cosθ)dθ=524π2

2. 0π2arccos(11+2cosθ)dθ=18π2

3. 0π2arccos(1cosθ2cosθ)dθ=1172π2

4. For any n natural number. Show that 02π(1+2cosx)ncosnx3+2cosxdx=2π5(35)n

5. Let 0<a<1 Prove that 02πcos23x1+a22acos2xdx=a2a+11aπ

6. For a>1 Prove that ππxsinx1+a22acosxdx=π2ln(1+1a)

7. 01lnln1x(1+x)2dx=12(lnπln2γ)

8. 0+sinhxcosh2xdxx=4Gπ where G is the Catalan's constant

9. Let z be a real number. Show that 12π02πlog|zeiθ|dθ={0 si |z|<1log|z| si |z|1

10. 0+exp(a2x(x6x2)2)dxx=πa

11. Let α>0 Prove that I(α)=0π2arctan(2αsin2xα21+cos2x)dx=πarctan(12α)

12. 01log(1x)x2zlog2x+(2πz)2dx=log(z!ezzz2πz),Re(z)>0

13.011xlogx(x+x2+x22+...)dx

14. Let ak>0 and a0>k=1nak. Show that 0+k=0nsin(akx)xdx=π2k=1nak

15. Let 0<z<1,α>0,βC
0+sin(αt1z+β)dt=Γ(z+1)αzsin(πz2+β)

16. Re(α)1
+|sinx|α1sinxxdx=2α1Γ2(α2)Γ(α)

17. 01sin(πx)xx(1x)1xdx=πe24

18. 0π4x3sin2xdx=3π4Gπ364+3π232log210564ς(3)

19. Let θ>0
+|cosθx|1+x2dx=4coshθarctaneθ

20. Let α0,θCπZ
+cosαx1+2cosθx+x2dx=πsinθcos(αcosθ)eαsinθ

21. Show that 0π2dθ1+sin2tanθ=π22(e2+322e23+22)

22. Let θ[0,π2) Prove that arctanxx22xsinθ+1dx

23. Given the function y(x):[0,1][0,1] continuous and decreasing such that xaxb=yayb. Compute 01ln(y(x))xdx

24. 01(1)[1994x]+[1995x](1993[1994x])(1994[1995x])dx

25. 01dx1+2F1(1n,x;1n;1n)=log(2n2n1)log(nn1)

26. 0+W(1x2)dx=2π

27. 0+W(x)xxdx=22π

28. Let α,β+. Integrate 0+(exp(θα)11+θβ)dθθ=1αγ where W is the Lambert W function

29. 0π2ln2sinxln2cosxsinxcosxdx=14(2ζ(5)ζ(2)ζ(3))

30. 0π24cos2x(lncosx)2dx=πln2+πln22π2+π312

31. 0101dxdy([xy]+1)2=12(ζ(3)+1ζ(2))

32. 0101ln(1xy)lnxlnydxdy=ζ(2)+ζ(3)+ζ(4)4

33. 0101...01ln(11inxi)1inlnxidx1dx2...dxn=(1)n1(2n+1k2nζ(k))

34. Prove that 0π2arctan(1(sinxcosx)2)dx=π(π4arctan212)

35. Let s>0 and α(0,1). Prove that 0+Lis(x)x1+αdx=παssin(πα)

36. limnππn!22ncos(ϕ)|k=1n(2neiϕk)|dϕ=2π

37. 062162+1lnxx22(15+83)x+1dxx1=23(23)G where G is the Catalan's constant

38. Let 0<r<1 and r<s Prove that 111x1+x1xlog|1+2rsx+(r2+s21)x212rsx+(r2+s21)x2|dx=4πarcsinr

39. 01cosh(αlnx)ln(1+x)dxx=12α(πcsc(πα)1α)

40. Let α0 be a real number. Prove that 0+lntan2(αx)1+x2dx=πlntanhα

41. Consider a>0, b. Prove that +a2(exaxb)2+(aπ)2dx=11+W(1aeba)

42. 0πsin(nα)arctan(tan(α2)tan(φ2))dα=π2n[(sec(φ)tan(φ))n(1)n)],|nZ+,0<φ<π2|

43. 0+cosαxcosβxsinθxdxx=log(coshβπ2θcoshαπ2θ)

44. 0πlog(1cosx)log(1+cosx)dx=πlog22π36

45. 0+arctanxsinh(πx2)dx=4logΓ(14)2logπ3log2

46. 0π2xcotxlogsinxdx=π348π4ln22

47. 01log(arcsechx)dx=γ2log22log(Γ(34)Γ(14))

48. 0118x2+16x41+7x28x4exp(4x1x21+8x2)dx=e1

49. Let |(n)|<1 Prove that 0+cos(nπx)cosh(πx)eiπx2dx=1+2sinn2π422coshnπ2+i12cosn2π422coshnπ2

50. Let f be a function of class C[0,a]. Prove that 02a02axx2x(x2+y2)4a2x2(x2+y2)2f(y)dydx=πa2(f(a)f(0))

51. 0+cosαxxsinhβxcoshγxdx=12log(coshαπ2γ+sinβπ2γcoshαπ2γsinβπ2γ)|Re(β)|<|Re(γ)|, |Re(β)|+|Im(α)|<|Re(γ)|

52. 0+sinαxxsinhβxsinhγxdx=arctan(tanβπ2γtanhαπ2γ)|Re(β)|<|Re(γ)|, |Re(β)|+|Im(α)|<|Re(γ)|

53. 01x1+x2arctanxln(1x2)dx=π348π8ln2+Gln2

54. 0π2arctan(αsinx)sinxdx=π2sinh1α

55. 0π2x2x2+log2(2cosx)dx=π8(1γ+log2π)

56. 0+sin(x2)ln2xdx=2π64(4ln2+2γπ)2

57. 0+eαxsin(βx)xs1dx=Γ(s)α2+β2sin(sarctanβα)

58. 0+eαxcos(βx)xs1dx=Γ(s)α2+β2cos(sarctanβα)

59. 0+11+eπxx1+x2dx=12(log2γ)

60. 0+cos(xp)exqx1+rdx=Γ(1rp)Γ(1+rp)Γ(1+r2p)Γ(1r2p)rΓ(1+rp)

61. π+sinxxdx+122π+sinxxdx+133π+sinxxdx+...=π2(1lnπ)

62. 0+sin(ωx2)x(ex1)dx=14ln(sinh(πω)πω)

63. 0+1cosxx2ekxdx=arctan1kkln(1+k2k)

64. 0+sinxsinxeαxdx=π2exp(α411+α2)(1+α2)34sin(32arctan1α1411+α2)

65. 01011x2(1+x2y2)ln2(xy)dxdy=2log(2Γ(34)Γ(14))

66. 0+sin(1x2)eαx2dx=12παe2αsin2α

67. 01ln(x2)(1+x2)(π2+ln2x)dx=ln212

68. 01ln(π2+ln2x)1+x2dx=πln(12π2Γ(14)Γ(34))

69. 0101x21(1+x2y2)ln2(xy)dxdx=122Cπ+ln(22πΓ2(14))

70. 0+dx(x2+π24)coshx=2ln2π

71. 01(tanh1x)zdx=ζ(z)22z1Γ(z+1)(2z2)zN,z2

72. 0+xex(0π2(1exxcsct)sec2tdt)2dx=13

73. 0+exlnln(ex+e2x1)dx=γ+4logΓ(14)3log22logπ

74. 010101{xy}{yz}{zx}dxdydz=1+ζ(2)ζ(3)63ζ(2)4

75. 0πxcot(x4)dx=2πlog2+8C

76. 0π2x2scossxsin(sx)dx=π4(γ+ψ(s+1))

77. 0+xs1(arctanx)2dx=π2ssinπs2(γ+ψ(1s2)+2log2)

78. 0π2xtansxdx=π4sinπs2(ψ(12)ψ(1s2))

79. 0+exp(x2)(x2+12)2dx=π

80. 0+1x(sinhαxsinhxαe2x)dx=log(πcosαπ2Γ2(α+12))|α|<1

81. 0+ln(x2+α2)coshx+costdx=2πsintlog(Γ(α2π+π+t2π)Γ(α2π+πt2π))+2tsintln2π

82. 0+(sinh(sx))2x(ex1)3dx=log(2πssin(2πs))0<s<12

83. 0+xs1sinh(πx)(cosh(πx)1)3dx=Γ(s)3πs(ζ(4s)ζ(2s))

84. 010101x2+y2+z2dxdydz=log(3+1)log22+34π24

85. 0101xα1yβ1(1+xy)log(xy)dxdy=1αβlog(Γ(α2)Γ(12+β2)Γ(β2)Γ(12+α2))

86. +11+x2α2k=1+1+x2(β+k)21+x2(α+k)2dx=πΓ(β+1)Γ(α)Γ(α+12)Γ(β+12)Γ(βα+12)Γ(βα+1)0<α<β+12

87. 0+1x(sinh(ax)sinhxae2x)dx=log(πcos(aπ2)Γ2(a+12))

88. 0+x2ex2erf(x)logxdx=2log216πγ+log216π(π+2)+G4π

89. 0π2sin(2nx)sinh(asinx)sin(acosx)dx=(1)n+1π4a2n(2n)!

90. Let β>0 and α(π2,π2). Prove that 0+etcosαtβ1cos(tsinα)dx=Γ(β)cos(βsinα)

91. 0+ln(1+x)ln(1+1x2)xdx=πG38ζ(3)

92. ++sign(x)sign(y)ex2+y22sin(xy)dxdy=22log(1+2)

93. 0+(xlog2(ex21)xex21log2(ex21)xex21log((ex21)2))dx=Gπ

94. 01B2n+1(x)cot(πx)dx=2(2n+1)!(1)n+1(2π)2n+1ζ(2n+1) where B2n+1(x) is the Bernoulli Polynomial

95. 0+x1+x4arctan(psinqx1+pcosqx)dx=π2arctan(psin(q2)eq2+pcos(q2))

96. Let m and a(1,1) Calculate
02πemcosθ(cos(msinθ)asin(θ+msinθ))12asinθ+a2dθ

97. Prove that 0101(xy)s1yn(1xy)log(xy)dxdy=Γ(s)Γ(s)log(n!)n

98. 0+sin(nx)(cotx+cothx)enxdx=π2sinh(nπ)cosh(nπ)cos(nπ)

99. 0π3log2(sinxsin(x+π3))dx=5π381

100. 02πx2log(1exp(ix))dx=2πζ(4)8iπ2ζ(3)

100+1. Let a(0,1) Prove that 01loglog1x1+2xcos(aπ)+x2dx=π2sin(aπ)(alog(2π)+logΓ(12+a2)Γ(12a2))

Bonus
1. 0+cosxx(0xsinttdt)2dx=76ζ(3)

2. 0+xa1sinxcosx+coshxdx=21a2Γ(a)sin(aπ4)(121a)ζ(a)

3. 0+cos(tx)(1+x2)cosh(πx2)dx=coshtlog(2cosht)tsinht

4. 0+ts1z1et1dt=Γ(s)Lis(z)

5. 0+exp(2u)(1usinhu1u2coshu)du=2log24Gπ

6. 0+sin(bt)teatlntdt=(γ+ln(a2+b2)2)arctanba

7. 01(at)ln(1t)12at+t2dt=π212(arccosaπ)28ln2(22a)8

8. 01{1x}2dx=ln(2π)1γ

9. 01{1x}2{11x}dx=2+γln(4π)

10. 01{xy}2dxdy=12ln(2π)13γ2

 

posted on   Eufisky  阅读(403)  评论(0编辑  收藏  举报

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