cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A254133 Decimal expansion of Lamb's integral K_0.

Original entry on oeis.org

4, 9, 0, 7, 7, 2, 7, 2, 8, 9, 5, 8, 3, 4, 5, 1, 5, 9, 1, 6, 2, 7, 1, 7, 2, 5, 3, 2, 0, 3, 3, 8, 2, 6, 4, 0, 3, 8, 1, 9, 2, 3, 3, 4, 7, 7, 5, 8, 5, 8, 4, 6, 5, 6, 2, 4, 2, 6, 2, 6, 0, 1, 2, 0, 7, 8, 1, 3, 6, 3, 4, 1, 5, 5, 4, 8, 7, 8, 6, 9, 9, 9, 7, 1, 2, 5, 7, 2, 1, 8, 0, 3, 5, 7, 8, 9, 5, 5, 2, 3, 3, 4, 2
Offset: 0

Views

Author

Jean-François Alcover, Jan 26 2015

Keywords

Examples

			0.490772728958345159162717253203382640381923347758584656...
		

Crossrefs

Programs

  • Maple
    evalf(int(arctanh(1/sqrt(3 + x^2))/(1 + x^2), x=0..1), 120); # Vaclav Kotesovec, Jan 26 2015
  • Mathematica
    Ti2[x_] := (I/2)*(PolyLog[2, -I*x] - PolyLog[2, I*x]); K0 = (3/2)*Ti2[3 - 2 Sqrt[2]] + Pi/4*Log[1 + Sqrt[2]] - Catalan/2 // Re; RealDigits[K0, 10, 103] // First

Formula

K_0 = integral_[0..1] arctanh(1/sqrt(3 + x^2))/(1 + x^2) dx.
K_0 = 3/2*Ti_2(3 - 2*sqrt(2)) + Pi/4*log(1 + sqrt(2)) - Catalan/2, where Ti_2 is Lewin's arctan integral, Ti_2(x) = (i/2)*(Li_2(-i*x) - Li_2(i*x)).

A254134 Decimal expansion of Lamb's integral K_1.

Original entry on oeis.org

1, 6, 6, 1, 9, 0, 7, 8, 7, 4, 7, 3, 8, 1, 2, 3, 3, 7, 7, 4, 0, 6, 5, 8, 1, 6, 8, 6, 1, 6, 3, 0, 5, 9, 4, 9, 7, 3, 4, 8, 8, 6, 8, 6, 7, 3, 2, 5, 1, 2, 5, 8, 9, 1, 8, 3, 4, 1, 5, 0, 8, 1, 9, 4, 3, 4, 2, 3, 5, 4, 9, 3, 1, 0, 9, 3, 0, 4, 5, 2, 0, 6, 6, 9, 3, 8, 4, 8, 3, 8, 0, 5, 6, 8, 7, 2, 3, 4, 5, 1, 0, 3, 8
Offset: 1

Views

Author

Jean-François Alcover, Jan 26 2015

Keywords

Examples

			1.6619078747381233774065816861630594973488686732512589...
		

Crossrefs

Programs

  • Maple
    evalf(int(arcsec(x)/sqrt(x^2 - 4*x + 3), x=3..4), 120); # Vaclav Kotesovec, Jan 26 2015
  • Mathematica
    Cl2[x_] := (I/2)*(PolyLog[2, Exp[-I*x]] - PolyLog[2, Exp[I*x]]); th = (ArcTan[(16 - 3*Sqrt[15])/11] + Pi)/3; K1 = Cl2[th] - Cl2[th + Pi/3] - Cl2[th - Pi/2] + Cl2[th - Pi/6] - Cl2[3*th + Pi/3] + Cl2[3*th + 2*(Pi/3)] - Cl2[3*th - 5*(Pi/6)] + Cl2[3*th + 5*(Pi/6)] + (6*th - 5*(Pi/2))*Log[2 - Sqrt[3]] // Re; RealDigits[K1, 10, 103] // First

Formula

K_1 = integral_[3..4] arcsec(x)/sqrt(x^2 - 4*x + 3) dx.
K_1 = Cl_2(th) - Cl_2(th + Pi/3) - Cl_2(th - Pi/2) + Cl_2(th - Pi/6) - Cl_2(3*th + Pi/3) + Cl_2(3*th + 2*(Pi/3)) - Cl_2(3*th - 5*(Pi/6)) + Cl_2(3*th + 5*(Pi/6)) + (6*th - 5*(Pi/2))*log(2 - sqrt(3)), where Cl_2 is the Clausen function and th = (arctan((16 - 3*sqrt(15))/11) + Pi)/3.

A254135 Decimal expansion of Lamb's integral K_2.

Original entry on oeis.org

6, 9, 2, 6, 6, 0, 8, 1, 5, 1, 5, 2, 6, 4, 7, 5, 0, 6, 5, 0, 9, 4, 3, 1, 1, 8, 5, 8, 8, 4, 2, 7, 2, 4, 5, 8, 4, 6, 7, 1, 3, 4, 8, 3, 2, 8, 0, 7, 6, 6, 8, 8, 4, 2, 5, 8, 0, 7, 2, 0, 4, 5, 6, 9, 7, 1, 4, 9, 0, 6, 3, 0, 2, 1, 6, 3, 0, 0, 7, 0, 5, 2, 1, 4, 3, 3, 9, 1, 1, 7, 7, 2, 8, 2, 0, 4, 4, 2, 8, 6, 8, 3, 9
Offset: 0

Views

Author

Jean-François Alcover, Jan 26 2015

Keywords

Examples

			0.69266081515264750650943118588427245846713483280766884258...
		

Crossrefs

Programs

  • Maple
    evalf(int(sqrt(1 + sec(x)^2)*arctan(1/sqrt(1 + sec(x)^2)), x=0..Pi/4), 120); # Vaclav Kotesovec, Jan 26 2015
  • Mathematica
    Ti2[x_] := (I/2)* (PolyLog[2, -I *x] - PolyLog[2, I *x]); K2 = (1/2)*Ti2[-2 + Sqrt[3]] + (Pi/8)*Log[2 + Sqrt[3]] + Pi^2/32 // Re; RealDigits[K2, 10, 103] // First

Formula

K_2 = integral_[0..Pi/4] sqrt(1 + sec(x)^2)*arctan(1/sqrt(1 + sec(x)^2)) dx.
K_2 = (1/2)*Ti_2(-2 + sqrt(3)) + (Pi/8)*log(2 + sqrt(3)) + Pi^2/32, where Ti_2 is Lewin's arctan integral, Ti_2(x) = (i/2)*(Li_2(-i*x) - Li_2(i*x)).
Showing 1-3 of 3 results.