find-0-t-p-e-t-1-dt-with-p-N-

Question Number 40883 by prof Abdo imad last updated on 28/Jul/18

f i n d \int_{0}^{\infty} \frac{t^{p}}{e^{t} - 1} d t w i t h p \in N^{★}

Answered by maxmathsup by imad last updated on 31/Jul/18

l e t A_{p} = \int_{0}^{\infty} \frac{t^{p}}{e^{t} - 1} \Rightarrow A_{p} = \int_{0}^{\infty} \frac{e^{- t} t^{p}}{1 - e^{- t}}

= \int_{0}^{\infty} e^{- t} t^{p} (\sum_{n = 0}^{\infty} e^{- n t}) = \sum_{n = 0}^{\infty} \int_{0}^{\infty} t^{p} e^{- (n + 1) t} d t c h a n g e m e n t (n + 1) t = x

g i v e

A_{p} = \sum_{n = 0}^{\infty} {(\frac{x}{n + 1})}^{p} e^{- x} \frac{d x}{n + 1} = \sum_{n = 0}^{\infty} \int_{0}^{\infty} \frac{x^{p} e^{- x}}{{(n + 1)}^{p + 1}} d x

= \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{p + 1}} \int_{0}^{\infty} x^{p} e^{- x} d x b y p a r t s

w_{p} = \int_{0}^{\infty} x^{p} e^{- x} d x = {[- x^{p} e^{- x}]}_{0}^{+ \infty} + \int_{0}^{\infty} {p x}^{p - 1} e^{- p} d x

= {p w}_{p - 1} \Rightarrow w_{p} = p! w_{0} a n d w_{0} = \int_{0}^{\infty} e^{- x} d x = 1 \Rightarrow w_{p} = p! \Rightarrow

A_{p} = p! \sum_{n = 1}^{\infty} \frac{1}{n^{p + 1}} = p! ξ (p + 1) w i t h ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}}, x > 1