find-0-t-n-e-t-1-dt-by-using-x-for-n-integr-x-n-1-1-n-x-with-x-gt-1-

Question Number 33119 by abdo imad last updated on 10/Apr/18

f i n d \int_{0}^{\infty} \frac{t^{n}}{e^{t} - 1} d t b y u s i n g ξ (x) f o r n i n t e g r

ξ (x) = \sum_{n = 1}^{\infty} \frac{1}{n^{x}} w i t h x > 1 .

Commented by prof Abdo imad last updated on 15/Apr/18

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

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

= \sum_{p = 0}^{\infty} \int_{0}^{\infty} t^{n} e^{- (p + 1) t} d t t h e c h (p + 1) t = u g i v e

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

= \sum_{p = 0}^{\infty} \frac{1}{{(p + 1)}^{n + 1}} \int_{0}^{\infty} u^{n} e^{- u} d u b u t w e k n o w t h a t

Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t f o r x > 0 \Rightarrow

\int_{0}^{\infty} u^{n} e^{- u} d u = Γ (n + 1) a n d \sum_{p = 0}^{\infty} \frac{1}{{(p + 1)}^{n + 1}}

= \sum_{p = 1}^{\infty} \frac{1}{p^{n + 1}} = ξ (n + 1) \Rightarrow

A_{n} = Γ (n + 1) . ξ (n + 1) .