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

               .......Advanced ... ... ... calculus........           Ω=∫_0 ^( ∞) ((x^2 e^x )/((1+e^x )^3 )) dx =?            .....∗∗∗∗∗∗∗∗.....

.......Advanced.........calculus........Ω=0x2ex(1+ex)3dx=?..........

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

(1/(1+e^x ))=Σ_(n=1) ^∞ (−1)^(n+1) e^(−nx)   (e^x /((1+e^x )^2 ))=Σ_(n=0) ^∞  (−1)^(n+1) ne^(−nx)   (e^x /((1+e^x )^3 ))=(1/2)Σ_(n=0) ^∞ (((−1)^(n+1) n(n+1)e^(−(n+1)x) )/1)  ∫_0 ^∞ ((x^2 e^x )/((1+e^x )^3 ))dx  =(1/2)Σ_(n=0) ^∞ (−1)^(n+1) ∫_0 ^∞ n(n+1)e^(−(n+1)x) x^2 dx  =Σ_(n=0) ^∞ (((−1)^(n+1) n)/((n+1)^2 ))=(π^2 /(12))−log(2)

11+ex=n=1(1)n+1enxex(1+ex)2=n=0(1)n+1nenxex(1+ex)3=12n=0(1)n+1n(n+1)e(n+1)x10x2ex(1+ex)3dx=12n=0(1)n+10n(n+1)e(n+1)xx2dx=n=0(1)n+1n(n+1)2=π212log(2)

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...

Φ=0x2ex(1+ex)3dxΦ=ex=t1ln2t.t(1+t)3dtt=1ln2t(1+t)3=t=1u01ln2u(1+1u)3(duu2)=01ln2uu2(1+u)3.u3du=01uln2u(1+u)3du=01(1+u1)ln2u(1+u)3du=01ln2u(1+u)2du01ln2u(1+u)3dubyparts01ln2u(1+u)2du=[(111+u)ln2u]0101(111+u)2lnuudu=201lnu1+udu=201lnun=0(1)nundu=2n=0(1)n01unlnuduUn=01unlnudu=[un+1n+1lnu]0101unn+1du=1(n+1)201log2u(1+u)2du=2n=0(1)n(n+1)2=2n=1(1)nn2=2(2121)ξ(2)=2(12).π26=π2601log2x(1+x)3dx=01(1+x)3log2xdxbypartsf=(1+x)3andg=log2xf=12(1+x)201log2x(1+x)3dx={(1212(1+x)2)log2x]0101(1212(1+x)2)2logxxdx=01(11(1+x)2)logxxdx=01(1+2x+x21(1+x)2)logxxdx=01(x+2)logx(1+x)2dx....becontinued...

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