1-calculate-A-n-0-x-n-1-e-x-1-dx-with-n-integr-natural-n-2-2-find-the-value-of-0-x-e-x-1-dx-

Question Number 53599 by maxmathsup by imad last updated on 23/Jan/19

1) c a l c u l a t e A_{n} = \int_{0}^{\infty} \frac{x^{n - 1}}{e^{x} + 1} d x w i t h n i n t e g r n a t u r a l (n ⩾ 2)

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x}{e^{x} + 1} d x

Commented by maxmathsup by imad last updated on 24/Jan/19

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

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

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

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

= Γ (n) . \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p^{n}} l e t r e m e m b e r ξ (x) = \sum_{p = 1}^{\infty} \frac{1}{p^{x}} w i t h x > 1

w e h a v e \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p^{n}} = - \sum_{k = 1}^{\infty} \frac{1}{{(2 k)}^{n}} + \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{n}}

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

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

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

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

Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t \Rightarrow Γ (n) = \int_{0}^{\infty} t^{n - 1} e^{- t} d t = {[\frac{1}{n} t^{n} e^{- t}]}_{0}^{+ \infty} + \int_{0}^{\infty} \frac{t^{n}}{n} e^{- t} d t

= \frac{1}{n} Γ (n + 1) \Rightarrow Γ (n + 1) = n Γ (n) \Rightarrow Γ (n + 1) = n! \Rightarrow

A_{n} = (1 - 2^{1 - n}) ξ (n) (n - 1)!

2) w e s e e t h a t \int_{0}^{\infty} \frac{x}{e^{x} + 1} d x = A_{2} = (1 - \frac{1}{2}) ξ (2) = \frac{1}{2} . \frac{π^{2}}{6} = \frac{π^{2}}{12} .