project 33.1
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Find\int\frac{x^3}{e^x-1} evaluated between zero and infinitum. I got 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
I_1=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t-1}dt=\displaystyle\int Ln^3(t)d(Ln(t-1))}
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
But I can not get near to the solution\pi^4/15
I_1=\displaystyle\int\displaystyle\frac{Ln^3(t)}{t-1}dt=\displaystyle\int Ln^3(t)d(Ln(t-1))}
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
But I can not get near to the solution\pi^4/15