Clausen Function

Author

Eric W. Weisstein
October 3, 2007

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/ClausenFunction.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/ClausenFunction.html.

©2007 Wolfram Research, Inc. except for portions noted otherwise

Plot

Plot[Evaluate[Table[ClausenCl[n,x],{n,5}]],{x,0,2Pi},PlotStyle->{Red,Yellow,Green,Blue,Purple},
AxesLabel->{x,(Subscript[Cl, n])[x]}]

Sums

v4.2

Sum[Cos[k x]/k^n,{k,Infinity}]

Sum[Sin[k x]/k^n,{k,Infinity}]

v5.0

Sum[Cos[k x]/k^n,{k,Infinity}]

Sum[Sin[k x]/k^n,{k,Infinity}]

v5.1

Sum[Cos[k x]/k^n,{k,Infinity}]

Sum[Sin[k x]/k^n,{k,Infinity}]

v5.2

Sum[Cos[k x]/k^n,{k,Infinity}]//Timing

Sum[Sin[k x]/k^n,{k,Infinity}]//Timing

v6

Sum[Cos[k x]/k^n,{k,Infinity}]//Timing

Sum[Sin[k x]/k^n,{k,Infinity}]//Timing

V7

Sum[Cos[k x]/k^n,{k,Infinity}]//Timing

Sum[Sin[k x]/k^n,{k,Infinity}]//Timing

Special Values

Table[ClausenCl[i,x],{i,5}]//TableForm

(Subscript[Cl, 1])(x)

ClausenCl[1,x]

Plot[Subtract@@{(1/2) (-Log[1-(E^(-I x))]-Log[1-(E^(I x))]),-Log[Abs[2Sin[x/2]]]},{x,-20,20},PlotStyle->Red]

FullSimplify[Subtract@@{(1/2) (-Log[1-(E^(-I x))]-Log[1-(E^(I x))]),-Log[Abs[2Sin[x/2]]]},x∈Reals]

FullSimplify[Subtract@@{(1/2) (-Log[1-(E^(-I x))]-Log[1-(E^(I x))]),-Log[Abs[2Sin[x/2]]]},x>0]

Subtract@@@Table[{(1/2) (-Log[1-(E^(-I x))]-Log[1-(E^(I x))]),-Log[Abs[2Sin[x/2]]]}/.x->Random[Complex,5(I+1){-1,1},50],{10}]//Chop//Union

Subtract@@@Table[{(1/2) (-Log[1-(E^(-I x))]-Log[1-(E^(I x))]),-Log[Abs[2Sin[x/2]]]}/.x->Random[Real,10{-1,1},50],{10}]//Chop//Union

(Subscript[Cl, 2])(x)

ClausenCl[2,x]

-Integrate[Log[2 Sin[t/2]],{t,0,x}]

Plot[Subtract@@{(1/2) (-Log[1-(E^(-I x))]-Log[1-(E^(I x))]),-Log[Abs[2Sin[x/2]]]},{x,-20,20},PlotStyle->Red]

Opposite Parity

FullSimplify[Table[Sum[((Sin[k x])/(k^n)),{k,∞}],{n,1,5,2}],0<x<2π]

FullSimplify[Subtract@@{(1/2) I (-Log[1-(E^(-I x))]+Log[1-(E^(I x))]),(π-x)/2},0<x<2π]

FunctionExpand[%216,0<x<2π]

%216/.x->1.5

FullSimplify[Table[Sum[((Cos[k x])/(k^n)),{k,∞}],{n,2,4,2}],0<x<2π]