1. Feb 4, 2009

### project 33.1

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$$\pi^4/15$$

2. Feb 4, 2009