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Question Number 75773 by Ajao yinka last updated on 16/Dec/19

	Answered by mind is power last updated on 16/Dec/19
	$log(1+e^x )=x+log(1+e^(−x) ) =∫_0 ^(+∞) x^2 log(1+e^(−x) )dx log(1+e^(−x) )=Σ_(k≥0) (((−e^(−x) )^k )/k) ⇒Σ_(k≥1) ∫x^2 .(((−1)^k e^(−kx) )/k)dx =Σ_(k≥1) (((−1)^k )/k)∫_0 ^(+∞) x^2 e^(−kx) dx u=kx⇒dx=(du/k) =Σ_(k≥1) (((−1)^(k−1) )/k)∫_0 ^(+∞) (u^2 /k^3 ).e^(−u) du =Σ_(k≥1) (((−1)^(k−) )/k^4 )Γ(3) =−2Σ_(k≥1) (((−1)^k )/k^4 ) ζ(4)=Σ_(k≥1) (1/k^4 )⇒Σ_(k≥1) (((−1)^k )/k^4 )=((ζ(4))/2^4 )−(1−(1/2^4 ))ζ(4)=(−(1/2^3 )+1)ζ(4)=−((7/8).(π^4 /(90))) ⇒∫_0 ^(+∞) x^2 {log(1+e^x )−x)}dx=−2.−(7/8)((π^4 /(90)))=((7π^4 )/(360))$
	$\log (1 + e^{x}) = x + \log (1 + e^{- x})$ $= \int_{0}^{+ \infty} x^{2} \log (1 + e^{- x}) dx$ $\log (1 + e^{- x}) = \sum_{k ⩾ 0} \frac{{(- e^{- x})}^{k}}{k}$ $\Rightarrow \sum_{k ⩾ 1} \int x^{2} . \frac{{(- 1)}^{k} e^{- kx}}{k} dx$ $= \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} \int_{0}^{+ \infty} x^{2} e^{- kx} dx$ $u = kx \Rightarrow dx = \frac{du}{k}$ $= \sum_{k ⩾ 1} \frac{{(- 1)}^{k - 1}}{k} \int_{0}^{+ \infty} \frac{u^{2}}{k^{3}} . e^{- u} du$ $= \sum_{k ⩾ 1} \frac{{(- 1)}^{k -}}{k^{4}} Γ (3)$ $= - 2 \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k^{4}}$ $ζ (4) = \sum_{k ⩾ 1} \frac{1}{k^{4}} \Rightarrow \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k^{4}} = \frac{ζ (4)}{2^{4}} - (1 - \frac{1}{2^{4}}) ζ (4) = (- \frac{1}{2^{3}} + 1) ζ (4) = - (\frac{7}{8} . \frac{π^{4}}{90})$ $\Rightarrow \int_{0}^{+ \infty} x^{2} {\log (1 + e^{x}) - x)} dx = - 2 . - \frac{7}{8} (\frac{π^{4}}{90}) = \frac{7 π^{4}}{360}$
	Commented by Ajao yinka last updated on 16/Dec/19
	Nice
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