let-y-gt-0-give-0-x-y-e-x-1-dx-at-form-of-series-

Question Number 48067 by maxmathsup by imad last updated on 18/Nov/18

l e t y > 0 g i v e \int_{0}^{\infty} \frac{x^{y}}{e^{x} - 1} d x a t f o r m o f s e r i e s .

Commented by maxmathsup by imad last updated on 19/Nov/18

l e t A (y) = \int_{0}^{\infty} \frac{x^{y}}{e^{x} - 1} d x \Rightarrow A (y) = \int_{0}^{\infty} \frac{e^{- x} x^{y}}{1 - e^{- x}} d x

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

=_{(n + 1) x = t} \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- t} {(\frac{t}{n + 1})}^{y} \frac{d t}{n + 1}

= \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{y + 1}} \int_{0}^{\infty} e^{- t} t^{y} d t b u t Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t (x > 0) \Rightarrow

\int_{0}^{\infty} t^{y} e^{- t} d t = Γ (y + 1) a n d \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{y + 1}} = \sum_{n = 1}^{\infty} \frac{1}{n^{y + 1}} = ξ (y + 1) \Rightarrow

A (y) = ξ (y + 1) Γ (y + 1) .