prove-that-0-t-x-1-e-t-1-dt-x-x-with-x-n-1-1-n-x-and-x-0-t-x-1-e-t-dt-x-gt-1-

Question Number 27345 by abdo imad last updated on 05/Jan/18

p r o v e t h a t \int_{0}^{\infty} \frac{t^{x - 1}}{e^{t} - 1} d t = ξ (x) Γ (x)

w i t h ξ (x) = \sum_{n = 1}^{\propto} \frac{1}{n^{x}} a n d Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t

(x > 1)

Commented by abdo imad last updated on 07/Jan/18

\int_{0}^{\propto} \frac{t^{x - 1}}{e^{t} - 1} d t = \int_{0}^{\propto} \frac{e^{- t} t^{x - 1}}{1 - e^{- t}} d t

= \int_{0}^{\propto} (\sum_{n = 0}^{\propto} e^{- n t}) e^{- t} t^{x - 1} d t

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

\int_{0}^{\propto} e^{- (n + 1) t} t^{x - 1} d t = \frac{1}{n + 1} \int_{0}^{\propto} e^{- u} {(\frac{u}{n + 1})}^{x - 1} d u

= \frac{1}{{(n + 1)}^{x}} \int_{0}^{\propto} e^{- u} u^{x - 1} d u = \frac{Γ (x)}{{(n + 1)}^{x}}

\Rightarrow \int_{0}^{\propto} \frac{t^{x - 1}}{e^{t} - 1} d t = \sum_{n = 0}^{\propto} (\frac{Γ (x)}{{(n + 1)}^{x}}) = (\sum_{n = 1}^{\propto} \frac{1}{n^{x}}) Γ (x)

= ξ (x) . Γ (x) .