Advanced-calculus-0-x-2-e-x-1-e-x-3-dx-

Question Number 138521 by mnjuly1970 last updated on 14/Apr/21

\dots \dots . A d v a n c e d \dots \dots \dots c a l c u l u s \dots \dots . .

Ω = \int_{0}^{\infty} \frac{x^{2} e^{x}}{{(1 + e^{x})}^{3}} d x = ?

\dots . . * * * * * * * * \dots . .

Answered by Dwaipayan Shikari last updated on 14/Apr/21

\frac{1}{1 + e^{x}} = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} e^{- n x}

\frac{e^{x}}{{(1 + e^{x})}^{2}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n + 1} {n e}^{- n x}

\frac{e^{x}}{{(1 + e^{x})}^{3}} = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} n (n + 1) e^{- (n + 1) x}}{1}

\int_{0}^{\infty} \frac{x^{2} e^{x}}{{(1 + e^{x})}^{3}} d x

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

= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} n}{{(n + 1)}^{2}} = \frac{π^{2}}{12} - l o g (2)

Answered by mathmax by abdo last updated on 14/Apr/21

Φ = \int_{0}^{\infty} \frac{x^{2} e^{x}}{{(1 + e^{x})}^{3}} dx \Rightarrow Φ =_{e^{x} = t} \int_{1}^{\infty} \frac{\ln^{2} t . t}{{(1 + t)}^{3}} \frac{dt}{t}

= \int_{1}^{\infty} \frac{\ln^{2} t}{{(1 + t)}^{3}} =_{t = \frac{1}{u}} - \int_{0}^{1} \frac{\ln^{2} u}{{(1 + \frac{1}{u})}^{3}} (- \frac{du}{u^{2}})

= \int_{0}^{1} \frac{\ln^{2} u}{u^{2} {(1 + u)}^{3}} . u^{3} du = \int_{0}^{1} \frac{{uln}^{2} u}{{(1 + u)}^{3}} du

= \int_{0}^{1} \frac{(1 + u - 1) \ln^{2} u}{{(1 + u)}^{3}} du = \int_{0}^{1} \frac{\ln^{2} u}{{(1 + u)}^{2}} du - \int_{0}^{1} \frac{\ln^{2} u}{{(1 + u)}^{3}} du by parts

\int_{0}^{1} \frac{\ln^{2} u}{{(1 + u)}^{2}} du = {[(1 - \frac{1}{1 + u}) \ln^{2} u]}_{0}^{1} - \int_{0}^{1} (1 - \frac{1}{1 + u}) \frac{2 lnu}{u} du

= - 2 \int_{0}^{1} \frac{lnu}{1 + u} du = - 2 \int_{0}^{1} lnu \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} du

= - 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} u^{n} lnu du

U_{n} = \int_{0}^{1} u^{n} lnu du = {[\frac{u^{n + 1}}{n + 1} lnu]}_{0}^{1} - \int_{0}^{1} \frac{u^{n}}{n + 1} du

= - \frac{1}{{(n + 1)}^{2}} \Rightarrow \int_{0}^{1} \frac{\log^{2} u}{{(1 + u)}^{2}} du = 2 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}}

= - 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = - 2 (2^{1 - 2} - 1) ξ (2) = - 2 (- \frac{1}{2}) . \frac{π^{2}}{6} = \frac{π^{2}}{6}

\int_{0}^{1} \frac{\log^{2} x}{{(1 + x)}^{3}} dx = \int_{0}^{1} {(1 + x)}^{- 3} \log^{2} x dx

by parts f^{^{'}} = {(1 + x)}^{- 3} and g = \log^{2} x \Rightarrow

f = \frac{1}{- 2} {(1 + x)}^{- 2} \Rightarrow \int_{0}^{1} \frac{\log^{2} x}{{(1 + x)}^{3}} dx = {{(\frac{1}{2} - \frac{1}{2 {(1 + x)}^{2}}) \log^{2} x]}_{0}^{1}

- \int_{0}^{1} (\frac{1}{2} - \frac{1}{2 {(1 + x)}^{2}}) \frac{2 logx}{x} dx

= - \int_{0}^{1} (1 - \frac{1}{{(1 + x)}^{2}}) \frac{logx}{x} dx

= - \int_{0}^{1} (\frac{1 + 2 x + x^{2} - 1}{{(1 + x)}^{2}}) \frac{logx}{x} dx

= - \int_{0}^{1} \frac{(x + 2) logx}{{(1 + x)}^{2}} dx \dots . be continued \dots