find-the-value-of-0-x-e-x-1-dx-

Question Number 27481 by abdo imad last updated on 07/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\propto} \frac{\sqrt{x}}{e^{x} - 1} d x .

Commented by abdo imad last updated on 10/Jan/18

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

s o \int_{0}^{\infty} \frac{\sqrt{t}}{e^{t} - 1} d t = \int_{0}^{\infty} \frac{t^{\frac{3}{2} - 1}}{e^{t} - 1} = ξ (\frac{3}{2}) Γ (\frac{3}{2})

t h e r e l a t i o n Γ (x + 1) = x Γ (x) g i v e Γ (\frac{3}{2}) = \frac{1}{2} Γ (\frac{1}{2})

b u t Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t \Rightarrow Γ (\frac{1}{2}) = \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t t h e c h . \sqrt{t} = u

g i v e \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t == \int_{0}^{\infty} \frac{e^{- u^{2}}}{u} (2 u) d u = 2 \int_{0}^{\infty} e^{- u^{2}} d u = \sqrt{π}

\Rightarrow Γ (\frac{1}{2}) = \sqrt{π}

\Rightarrow \int_{0}^{\infty} \frac{\sqrt{t}}{e^{t} - 1} = \sqrt{π} ξ (\frac{3}{2}) a n d ξ (\frac{3}{2}) = \sum_{n = 1}^{\propto} \frac{1}{n \sqrt{n}} .

S_{n} = \sum_{k = 1}^{n} \frac{1}{k \sqrt{k}} = \sum_{k = 1}^{n} \int_{k}^{k + 1} \frac{d t}{k \sqrt{k}}

b u t k ⩽ t ⩽ k + 1 \Rightarrow \sqrt{k} ⩽ \sqrt{t} ⩽ \sqrt{k + 1} \Rightarrow k \sqrt{k} ⩽ t \sqrt{t} ⩽ (k + 1) \sqrt{k + 1}

\Rightarrow \frac{1}{(k + 1) \sqrt{k + 1}} ⩽ \frac{1}{t \sqrt{t}} ⩽ \frac{1}{k \sqrt{k}}

\Rightarrow \sum_{k = 1}^{n}_{(} \frac{1}{k + 1) \sqrt{k + 1}} ⩽ \sum_{k = 1}^{n} \int_{k}^{k + 1} t^{- \frac{3}{2}} d t ⩽ S_{n}

\Rightarrow \sum_{k = 2}^{n + 1} \frac{1}{k \sqrt{k}} ⩽ \sum_{k = 1}^{n} {[\frac{1}{- \frac{3}{2} + 1} t^{- \frac{3}{2} + 1}]}_{k}^{k + 1} ⩽ S_{n}

S_{n} + \frac{1}{n + 1} - 1 ⩽ - 2 \sum_{k = 1}^{n} ({(k + 1)}^{- \frac{1}{2}} - k^{- \frac{1}{2}}) ⩽ S_{n}

S_{n} - \frac{n}{n + 1} ⩽ - 2 \sum_{k = 1}^{n} (\frac{1}{\sqrt{k + 1}} - \frac{1}{\sqrt{k}}) ⩽ S_{n}

S_{n} - \frac{n}{n + 1} ⩽ - 2 (\frac{1}{\sqrt{2}} - 1 + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{2}} + \dots \frac{1}{\sqrt{n + 1}} - \frac{1}{\sqrt{n}}) ⩽ S_{n}

S_{n} - \frac{n}{n + 1} ⩽ - 2 (\frac{1}{\sqrt{n + 1}} - 1) ⩽ S_{n}

S_{n} - \frac{n}{n + 1} ⩽ 2 - \frac{2}{\sqrt{n + 1}} ⩽ S_{n} \dots . .