0-x-3-e-2x-e-x-

Question Number 80160 by peter frank last updated on 31/Jan/20

\int_{0}^{\infty} \frac{x^{3}}{e^{2 x} - e^{x}}

Commented by mathmax by abdo last updated on 31/Jan/20

I = \int_{0}^{\infty} \frac{x^{3}}{e^{2 x} - e^{x}} d x \Rightarrow I = \int_{0}^{\infty} \frac{e^{- 2 x} x^{3}}{1 - e^{- x}} d x = \int_{0}^{\infty} e^{- 2 x} x^{3} (\sum_{n = 0}^{\infty} e^{- n x})

= \sum_{n = 0}^{\infty} \int_{0}^{\infty} x^{3} e^{- (n + 2) x} d x =_{(n + 2) x = t} \sum_{n = 0}^{\infty} \int_{0}^{\infty} \frac{t^{3}}{{(n + 2)}^{3}} e^{- t} \frac{d t}{(n + 2)}

= \sum_{n = 0}^{\infty} \frac{1}{{(n + 2)}^{4}} \int_{0}^{\infty} t^{3} e^{- t} d t l e t r e m e m e r Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t

\Rightarrow I = Γ (4) \sum_{n = 0}^{\infty} \frac{1}{{(n + 2)}^{4}} = Γ (4) \sum_{n = 2}^{\infty} \frac{1}{n^{4}}

Γ (4) = 3! = 6 a n d \sum_{n = 2}^{\infty} \frac{1}{n^{4}} = \sum_{n = 1}^{\infty} \frac{1}{n^{4}} - 1 = \frac{π^{4}}{90} - 1 \Rightarrow

I = 6 (\frac{π^{4}}{90} - 1) = \frac{3.2}{3.30} π^{4} - 6 = \frac{π^{4}}{15} - 6