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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))

$$\mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)=\mathrm{x}+\mathrm{log}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{x}} \right) \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \mathrm{x}^{\mathrm{2}} \mathrm{log}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{x}} \right)\mathrm{dx} \\ $$$$\mathrm{log}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{x}} \right)=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{e}^{−\mathrm{x}} \right)^{\mathrm{k}} }{\mathrm{k}} \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\int\mathrm{x}^{\mathrm{2}} .\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{e}^{−\mathrm{kx}} \:}{\mathrm{k}}\mathrm{dx} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\int_{\mathrm{0}} ^{+\infty} \mathrm{x}^{\mathrm{2}} \mathrm{e}^{−\mathrm{kx}} \mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{kx}\Rightarrow\mathrm{dx}=\frac{\mathrm{du}}{\mathrm{k}} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} }{\mathrm{k}}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{k}^{\mathrm{3}} }.\mathrm{e}^{−\mathrm{u}} \mathrm{du} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}−} }{\mathrm{k}^{\mathrm{4}} }\Gamma\left(\mathrm{3}\right) \\ $$$$=−\mathrm{2}\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}^{\mathrm{4}} } \\ $$$$\zeta\left(\mathrm{4}\right)=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{4}} }\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}^{\mathrm{4}} }=\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{2}^{\mathrm{4}} }−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\right)\zeta\left(\mathrm{4}\right)=\left(−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\mathrm{1}\right)\zeta\left(\mathrm{4}\right)=−\left(\frac{\mathrm{7}}{\mathrm{8}}.\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\right) \\ $$$$\left.\Rightarrow\int_{\mathrm{0}} ^{+\infty} \mathrm{x}^{\mathrm{2}} \left\{\mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)−\mathrm{x}\right)\right\}\mathrm{dx}=−\mathrm{2}.−\frac{\mathrm{7}}{\mathrm{8}}\left(\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\right)=\frac{\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{360}} \\ $$$$ \\ $$

Commented by Ajao yinka last updated on 16/Dec/19

Nice

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