nice-calculus-find-0-x-2-ln-1-e-x-x-3-dx-

Question Number 130759 by mnjuly1970 last updated on 28/Jan/21

\dots n i c e c a l c u l u s \dots

f i n d : ϕ = \int_{0}^{\infty} (x^{2} l n (1 + e^{x}) - x^{3}) d x

Answered by mnjuly1970 last updated on 28/Jan/21

a n s :

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

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

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

\overset{n x = y}{=} \underset{n = 1}{\sum^{\infty}} [{(- 1)}^{(n - 1)} \frac{1}{n^{4}} \int_{0}^{\infty} y^{2} e^{- y} d y]

= Γ (3) \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{4}} = 2 [\frac{1}{1^{4}} - \frac{1}{2^{4}} + \frac{1}{3^{4}} - \frac{1}{4^{4}} + . .]

= 2 (ζ (4) - \frac{2}{2^{4}} (ζ (4))) = 2 (\frac{π^{4}}{90} - \frac{π^{4}}{8 * 90})

= \frac{π^{4}}{45} (1 - \frac{1}{8}) = \frac{7}{8} * \frac{π^{4}}{45} = \frac{7 π^{4}}{360} = \dots

c o r r e c t o r n o ? ? ?

Answered by Dwaipayan Shikari last updated on 28/Jan/21

\int_{0}^{\infty} x^{2} l o g (1 + e^{- x}) d x

= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n} \int_{0}^{\infty} e^{- n x} x^{2} d x = 2 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n^{4}} = \frac{π^{4}}{45} (1 - \frac{1}{2^{3}}) = \frac{7 π^{4}}{360}

Commented by mnjuly1970 last updated on 28/Jan/21

n i c e v e r y n i c e ::

η (s) = (1 - 2^{1 - s}) ζ (s)

η (4) = (1 - \frac{1}{8}) ζ (4) = \frac{7}{8} * \frac{π^{4}}{90}

= \frac{7 π^{4}}{720} \Rightarrow a n s := 2 (\frac{7 π^{4}}{720}) = \frac{7 π^{4}}{360}

Answered by Lordose last updated on 28/Jan/21

Ω = \int_{0}^{\infty} (x^{2} \ln (1 + e^{x}) - x^{3}) dx = \int_{0}^{\infty} x^{2} (\ln (1 + e^{- x})) dx

Ω \overset{u = e^{- x}}{=} \int_{0}^{1} \frac{1}{u} {(- lnu)}^{2} \ln (1 + u) du = \int_{0}^{1} u^{- 1} \ln (1 + u) \ln^{2} (u) du

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} u^{n - 1} \ln^{2} (u) du

Ω (n) = \int_{0}^{1} u^{n} du = \frac{u^{n + 1}}{n + 1} = \frac{1}{n + 1}

Ω^{'} (n) = \int_{0}^{1} u^{n} \ln (u) du = - \frac{1}{{(n + 1)}^{2}}

Ω^{″} (n) = \int_{0}^{1} u^{n} \ln^{2} (u) du = \frac{1}{{(n + 1)}^{3}}

Ω^{^{″}} (n - 1) = \frac{1}{n^{3}}

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{4}} = \frac{7 π^{4}}{720}

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