Debye Functions

Author

Eric W. Weisstein
February 17, 2008

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

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

©2008 Wolfram Research, Inc. except for portions noted otherwise

(∫_0^x)((t^n)/(-1+(e^t)))dt

Integral

Integrate[t^n/(E^t-1),{t,0,x}]

d1=Table[Integrate[t^n/(E^t-1),{t,0,x},Assumptions->-2π<x<2π],{n,5}]

d1//Column//TraditionalForm

MakeBoxes[Debye[n_,x_,i_],TraditionalForm]:=RowBox[{SubsuperscriptBox["D",MakeBoxes[n,TraditionalForm],RowBox[{"(",MakeBoxes[i,TraditionalForm],")"}]],"(",MakeBoxes[x,TraditionalForm],")"}]

Plot[d1,{x,-2π,2π},PlotStyle->{Red,Yellow,Green,Blue,Purple},WorkingPrecision->20,AxesLabel->{x,Debye[n,x,1]}]

A&S Table

Chop[Table[MapIndexed[((#2[[1]])/(x^#2[[1]]))#1&,d1],{x,0,1,1/10}]//N[#,10]&]//TableForm//Quiet

In Terms of Sum

Integrate[t^n/(E^t-1),{t,0,x}]==x^n*(1/n-x/(2*(n+1))+Sum[(BernoulliB[2*k]*x^(2*k))/((2*k+n)*(2*k)!),{k,1,Infinity}])

(∑_(k=1)^∞)(((x^(2 k)) BernoulliB[2 k])/((2 k+n) (2 k)!))

Integrate[Normal[Series[t^n/(E^t-1),{t,0,10}]],{t,0,x},GenerateConditions->False]

Simplify[Series[%-x^n*(1/n-x/(2*(n+1))+Sum[(BernoulliB[2*k]*x^(2*k))/((2*k+n)*(2*k)!),{k,1,20}]),{x,0,8}]]

(∫_x^∞)((t^n)/((e^t)-1))dt

Integral

Integrate[t^n/(E^t-1),{t,x,Infinity}]

d2=Table[Integrate[t^n/(E^t-1),{t,x,Infinity},Assumptions->x>0],{n,4}]

d2//Column//TraditionalForm

d2[[1]]/.x->1/10

NIntegrate[(t/(-1+(E^t))),{t,(1/10),∞}]

1!Zeta[1+1]-d1[[1]]/.x->.1

In Terms of Sum

Integrate[t^n/(E^t-1),{t,x,Infinity}]==Sum[(1+k)^(-1-n)*Gamma[1+n,(k+1)*x],{k,0,Infinity}]==Sum[(x^n/(k*E^(k*x)))*Sum[((-1)^j*Pochhammer[-n,j])/(k^j*x^j),{j,0,n}],{k,1,Infinity}]

Sum[(1+k)^(-1-n)*Gamma[1+n,(k+1)*x],{k,0,Infinity}]

Sum[(x^n/(k*E^(k*x)))*Sum[((-1)^j*Pochhammer[-n,j])/(k^j*x^j),{j,0,n}],{k,1,Infinity}]

FullSimplify[-(((x^n) (((k x))^(-n)) (-π Gamma[1+n,k x]+(((-(1/(k x))))^n) (((k x))^n) Gamma[1+n] Gamma[0,k x] Sin[n π]))/(k π)),n≥0&&n∈Integers]

Sum[(x^(1+n)) ExpIntegralE[-n,k x],{k,∞}]

Table[(∑_k^∞)(x^(1+n)) ExpIntegralE[-n,k x],{n,2}]

(∫_0^∞)((t^n)/((e^t)-1))dt==n! ζ(n+1)

Assuming[Re[n]>0,Integrate[t^n/(E^t-1),{t,0,Infinity}]//FullSimplify]