Question Number 138521 by mnjuly1970 last updated on 14/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….{Advanced}\:…\:…\:…\:{calculus}…….. \\ $$$$\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} {e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }\:{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:…..\ast\ast\ast\ast\ast\ast\ast\ast….. \\ $$
Answered by Dwaipayan Shikari last updated on 14/Apr/21
$$\frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{−{nx}} \\ $$$$\frac{{e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {ne}^{−{nx}} \\ $$$$\frac{{e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}\left({n}+\mathrm{1}\right){e}^{−\left({n}+\mathrm{1}\right){x}} }{\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} {e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {n}\left({n}+\mathrm{1}\right){e}^{−\left({n}+\mathrm{1}\right){x}} {x}^{\mathrm{2}} {dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−{log}\left(\mathrm{2}\right) \\ $$
Answered by mathmax by abdo last updated on 14/Apr/21
![Φ=∫_0 ^∞ ((x^2 e^x )/((1+e^x )^3 ))dx ⇒Φ=_(e^x =t) ∫_1 ^∞ ((ln^2 t .t)/((1+t)^3 ))(dt/t) =∫_1 ^∞ ((ln^2 t)/((1+t)^3 )) =_(t=(1/u)) −∫_0 ^1 ((ln^2 u)/((1+(1/u))^3 ))(−(du/u^2 )) =∫_0 ^1 ((ln^2 u)/(u^2 (1+u)^3 )).u^3 du =∫_0 ^1 ((uln^2 u)/((1+u)^3 ))du =∫_0 ^1 (((1+u−1)ln^2 u)/((1+u)^3 ))du =∫_0 ^1 ((ln^2 u)/((1+u)^2 ))du−∫_0 ^1 ((ln^2 u)/((1+u)^3 ))du by parts ∫_0 ^1 ((ln^2 u)/((1+u)^2 ))du =[(1−(1/(1+u)))ln^2 u]_0 ^1 −∫_0 ^(1 ) (1−(1/(1+u)))((2lnu)/u)du =−2∫_0 ^1 ((lnu)/(1+u))du =−2 ∫_0 ^1 lnuΣ_(n=0) ^∞ (−1)^n u^n du =−2Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 u^n lnu du U_n =∫_0 ^1 u^n lnu du =[(u^(n+1) /(n+1))lnu]_0 ^1 −∫_0 ^1 (u^n /(n+1))du =−(1/((n+1)^2 )) ⇒∫_0 ^1 ((log^2 u)/((1+u)^2 ))du =2Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =−2Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−2(2^(1−2) −1)ξ(2) =−2(−(1/2)).(π^2 /6)=(π^2 /6) ∫_0 ^1 ((log^2 x)/((1+x)^3 ))dx =∫_0 ^1 (1+x)^(−3) log^2 x dx by parts f^′ =(1+x)^(−3) and g=log^2 x ⇒ f=(1/(−2))(1+x)^(−2) ⇒∫_0 ^1 ((log^2 x)/((1+x)^3 ))dx={((1/2)−(1/(2(1+x)^2 )))log^2 x]_0 ^1 −∫_0 ^1 ((1/2)−(1/(2(1+x)^2 )))((2logx)/x)dx =−∫_0 ^1 (1−(1/((1+x)^2 )))((logx)/x)dx =−∫_0 ^1 (((1+2x+x^2 −1)/((1+x)^2 )))((logx)/x)dx =−∫_0 ^1 (((x+2)logx)/((1+x)^2 ))dx....be continued...](https://www.tinkutara.com/question/Q138557.png)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)^{\mathrm{3}} }\mathrm{dx}\:\Rightarrow\Phi=_{\mathrm{e}^{\mathrm{x}} =\mathrm{t}} \:\:\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{t}\:.\mathrm{t}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{3}} }\frac{\mathrm{dt}}{\mathrm{t}} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{t}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{3}} }\:=_{\mathrm{t}=\frac{\mathrm{1}}{\mathrm{u}}} \:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{u}}\right)^{\mathrm{3}} }\left(−\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\mathrm{u}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }.\mathrm{u}^{\mathrm{3}} \:\mathrm{du}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{uln}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\mathrm{du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{1}+\mathrm{u}−\mathrm{1}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\mathrm{du}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\mathrm{du}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{3}} }\mathrm{du}\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\mathrm{du}\:=\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{u}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}\:} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\right)\frac{\mathrm{2lnu}}{\mathrm{u}}\mathrm{du} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnu}}{\mathrm{1}+\mathrm{u}}\mathrm{du}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{lnu}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{u}^{\mathrm{n}} \:\mathrm{du} \\ $$$$=−\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{u}^{\mathrm{n}} \:\mathrm{lnu}\:\mathrm{du} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{u}^{\mathrm{n}} \:\mathrm{lnu}\:\mathrm{du}\:=\left[\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\mathrm{lnu}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}+\mathrm{1}}\mathrm{du} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{u}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\mathrm{du}\:=\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:=−\mathrm{2}\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right).\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{3}} \mathrm{log}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\:\mathrm{f}^{'} \:=\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{3}} \:\mathrm{and}\:\mathrm{g}=\mathrm{log}^{\mathrm{2}} \mathrm{x}\:\Rightarrow \\ $$$$\mathrm{f}=\frac{\mathrm{1}}{−\mathrm{2}}\left(\mathrm{1}+\mathrm{x}\right)^{−\mathrm{2}} \:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{3}} }\mathrm{dx}=\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\mathrm{log}^{\mathrm{2}} \mathrm{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\frac{\mathrm{2logx}}{\mathrm{x}}\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\frac{\mathrm{logx}}{\mathrm{x}}\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}+\mathrm{2x}+\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\right)\frac{\mathrm{logx}}{\mathrm{x}}\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{x}+\mathrm{2}\right)\mathrm{logx}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}….\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$