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Black body radiation integral

  1. Feb 4, 2009 #1
    Find[tex]\int\frac{x^3}{e^x-1}[/tex] evaluated between zero and infinitum. I got [tex]I=\displaystyle\int\displaystyle\frac{x^3}{e^x-1}dx=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t(t-1)}dt=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t-1}dt-\displaystyle\int\displaystyle\displaystyle\frac{Ln^3(t)}{t}dt=I_1-I_2[/tex]

    [tex]I_1=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t-1}dt=\displaystyle\int Ln^3(t)d(Ln(t-1))}[/tex]

    [tex]I_2=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t}dt=\displaystyle\frac{1}{4}Ln^4(t)+Cte=\displaystyle\frac{x^4}{4}+Cte[/tex]

    But I can not get near to the solution[tex]\pi^4/15[/tex]
     
  2. jcsd
  3. Feb 4, 2009 #2

    Dick

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