calculate-0-x-n-sh-x-dx-with-n-integr-natural-

Question Number 85009 by mathmax by abdo last updated on 18/Mar/20

c a l c u l a t e \int_{0}^{\infty} \frac{x^{n}}{s h (x)} d x w i t h n i n t e g r n a t u r a l

Commented by mathmax by abdo last updated on 18/Mar/20

l e t A_{n} = \int_{0}^{\infty} \frac{x^{n}}{s h (x)} d x \Rightarrow A_{n} = 2 \int_{0}^{\infty} \frac{x^{n}}{e^{x} - e^{- x}} d x

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

= 2 \sum_{k = 0}^{\infty} \int_{0}^{\infty} x^{n} e^{- (2 k + 1) x} d x =_{(2 k + 1) x = t} 2 \sum_{k = 0}^{\infty} \int_{0}^{\infty} {(\frac{t}{2 k + 1})}^{n} e^{- t} \frac{d t}{(2 k + 1)}

= 2 \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{n + 1}} \int_{0}^{\infty} t^{n} e^{- t} d t = 2 Γ (n + 1) \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{n + 1}}

w e h a v e ξ (x) = \sum_{k = 1}^{\infty} \frac{1}{k^{x}} \Rightarrow ξ (n + 1) = \sum_{k = 1}^{\infty} \frac{1}{k^{n + 1}}

= \sum_{k = 1}^{\infty} \frac{1}{{(2 k)}^{n + 1}} + \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{n + 1}} = \frac{1}{2^{n + 1}} \sum_{k = 1}^{\infty} \frac{1}{k^{n + 1}} + \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{n + 1}}

\Rightarrow \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{n + 1}} = (1 - \frac{1}{2^{n + 1}}) ξ (n + 1) \Rightarrow

A_{n} = 2 Γ (n + 1) (1 - \frac{1}{2^{n + 1}}) ξ (n + 1)

= (2 - \frac{1}{2^{n}}) (n!) ξ (n + 1)

Answered by mind is power last updated on 18/Mar/20

\int_{0}^{+ \infty} \frac{2 x^{n} e^{x} d x}{(e^{x} - 1) (e^{x} + 1)}

= \int_{0}^{+ \infty} \frac{x^{n} d x}{e^{x} - 1} + \int_{0}^{+ \infty} \frac{x^{n}}{e^{x} + 1} d x

= \int_{0}^{+ \infty} \frac{x^{n} e^{- x}}{1 - e^{- x}} d x

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

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

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

= Γ (n + 1) ζ (n + 1)

\int_{0}^{+ \infty} x^{n} . \sum_{k ⩾ 0} {(- 1)}^{k} e^{- (1 + k) x} d x

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

= Γ (n + 1) \sum_{k ⩾ 0} {(- 1)}^{k} . \frac{1}{{(1 + k)}^{n + 1}}

\sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(1 + k)}^{n + 1}} = \sum_{k ⩾ 0} \frac{1}{{(1 + 2 k)}^{n + 1}} - \frac{1}{{(2 + 2 k)}^{n + 1}} = (1 - \frac{1}{2^{n}}) ζ (n + 1)

= Γ (n + 1) (1 - \frac{1}{2^{n}}) ζ (n + 1)

= Γ (n + 1) {2 - \frac{1}{2^{n}}} ζ (n + 1) = \int_{0}^{+ \infty} \frac{x^{n}}{s h (x)} d x

Commented by abdomathmax last updated on 18/Mar/20

t h a n k y o u s i r .

Commented by mind is power last updated on 18/Mar/20

w i t h e p l e a s u r