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Advanced-Calculus-Q-Find-the-value-of-determinant-i-0-1-Ln-2-x-dx-ii-




Question Number 145602 by mnjuly1970 last updated on 06/Jul/21
                   .....Advanced .........Calculus.....     Q::      Find the value of ::                           determinant ((( i ::   𝛗 := ∫_0 ^( 1) Ln ( Γ ( 2 + x ) )dx = ?    )),(( ii ::   Ω := Σ_(n=1) ^∞ (( 1)/( n ( 2n + 3 ))) = ?)))                                               .....m.n.july.1970.....    ■
..AdvancedCalculus..Q::Findthevalueof::i::ϕ:=01Ln(Γ(2+x))dx=?ii::Ω:=n=11n(2n+3)=?..m.n.july.1970..◼
Answered by ajfour last updated on 06/Jul/21
(ii)  S=Σ_(n=1) ^∞ (1/(n(2n+3)))    ((3S)/2)=Σ_(n=1) ^∞ ((2n+3−2n)/(2n(2n+3)))          =Σ_(n=1) ^∞ ((1/(2n))−(1/(2n+3)))  ln (1−x)=−(x+(x^2 /2)+(x^3 /3)+..)  Σ_(n=1) ^∞ (1/(2n))=−(1/2)lim_(x→1)  ln (1−x)  ln (1+x)−ln (1−x)     =2(x+(x^3 /3)+(x^5 /5)+...)  lim_(x→1) [(1/2){ln (1+x)−ln (1−x)}    −x−(x^3 /3)] =Σ_(n=1) ^∞ (1/(2n+3))  ((3S)/2)=−(1/2)lim ln (1−x)−  lim_(x→1) [(1/2){ln (1+x)−ln (1−x)}                −x−(x^3 /3)]    ((3S)/2)= (4/3)−((ln 2)/2)    S=(8/9)−((ln 2)/3)
(ii)S=n=11n(2n+3)3S2=n=12n+32n2n(2n+3)=n=1(12n12n+3)ln(1x)=(x+x22+x33+..)n=112n=12limx1ln(1x)ln(1+x)ln(1x)=2(x+x33+x55+)limx1[12{ln(1+x)ln(1x)}xx33]=n=112n+33S2=12limln(1x)limx1[12{ln(1+x)ln(1x)}xx33]3S2=43ln22S=89ln23
Commented by mnjuly1970 last updated on 06/Jul/21
bravo mr ajfor your work    is admirable...
bravomrajforyourworkisadmirable
Answered by Olaf_Thorendsen last updated on 06/Jul/21
Ω = Σ_(n=1) ^∞ (1/(n(2n+3)))= (1/2)Σ_(n=0) ^∞ (1/((n+1)(n+(5/2))))  Ω = (1/2).((ψ((5/2))−ψ(1))/((5/2)−1)) = (1/3)(ψ((5/2))+γ)  ψ((5/2)) = ψ(1+(3/2)) = ψ((3/2))+(1/(3/2)) = ψ((3/2))+(2/3)  ψ((3/2)) = ψ(1+(1/2)) = ψ((1/2))+(1/(1/2)) = −2ln(2)−γ+2     Ω = (1/3)(−2ln(2)−γ+2+(2/3)+γ) = (2/3)((4/3)−ln(2))
Ω=n=11n(2n+3)=12n=01(n+1)(n+52)Ω=12.ψ(52)ψ(1)521=13(ψ(52)+γ)ψ(52)=ψ(1+32)=ψ(32)+13/2=ψ(32)+23ψ(32)=ψ(1+12)=ψ(12)+11/2=2ln(2)γ+2Ω=13(2ln(2)γ+2+23+γ)=23(43ln(2))
Commented by mnjuly1970 last updated on 06/Jul/21
grateful mr olaf   very nice as always..mercey
gratefulmrolafveryniceasalways..mercey
Answered by qaz last updated on 06/Jul/21
Γ(x)Γ(1−x)=(π/(sin πx))
Γ(x)Γ(1x)=πsinπx
Answered by Kamel last updated on 06/Jul/21
i) use t=1−x and: Γ(t)Γ(1−t)=(π/(sin(πt)))  ii.. Ω=Σ_(n=1) ^(+∞) (1/(n(2n+3)))=(2/3)Σ_(n=1) ^(+∞) ((1/(2n))−(1/(2n+3)))=(2/3)(Σ_(n=1) ^(+∞) ((1/(2n))−(1/(2n+1)))+Σ_(n=1) ^(+∞) ((1/(2n+1))−(1/(2(n+1)+1))))             =(2/3)(Σ_(n=1) ^(+∞) (((−1)^n )/n)+1+(1/3))=(2/3)((4/3)−Ln(2))=(8/9)−((2Ln(2))/3)
i)uset=1xand:Γ(t)Γ(1t)=πsin(πt)ii..Ω=+n=11n(2n+3)=23+n=1(12n12n+3)=23(+n=1(12n12n+1)++n=1(12n+112(n+1)+1))=23(+n=1(1)nn+1+13)=23(43Ln(2))=892Ln(2)3
Commented by mnjuly1970 last updated on 06/Jul/21
tashakor mr kamel
tashakormrkamel
Answered by ArielVyny last updated on 06/Jul/21
∅=∫_0 ^1 ln(Γ(2+x))dx  =∫_0 ^1 ln((x+1)!)dx  x+1=t→dt=dx  ∅=∫_1 ^2 ln(t!)dt=∫_1 ^2 ln[((t/e))^t (√(2πt))]  ∫_1 ^2 ln((t/e))^t +∫_1 ^2 ln((√(2πt)))dt  ∫_1 ^2 ln(e^(tln((t/e))) )+∫_1 ^2 ln((√(2π)))dt+(1/2)∫_1 ^2 ln(t)dt  ∫_1 ^2 tln((t/e))+ln((√(2π)))+(1/2)[xln(x)−x]_1 ^2   ∫_1 ^2 tln(t)−∫_1 ^2 tdt+ln((√(2π)))+(1/2)(2ln(2)+1)  ∫_1 ^2 tln(t)dt−(3/2)+(1/2)ln(2π)+((2ln2)/2)+(1/2)  ∫_1 ^2 tln(t)dt  du=ln(t)              v=t  u=tln(t)−t         dv=1  ∫_1 ^2 tln(t)=[t^2 ln(t)−t]−∫tlntdt+∫tdt  2∫_1 ^2 tln(t)dt=[t^2 ln(t)−t+(1/2)t^2 ]_1 ^2 =4ln(2)−2+(1/2)4+1−(1/2)  ∫_1 ^2 tln(t)dt=2ln(2)−1+1+(1/2)−(1/4)                         =2ln(2)+(1/4)  ∅=2ln(2)+(1/4)−(3/2)+((ln(2π))/2)+((2ln(2))/2)+(1/2)  ∅=2ln(2)+(1/4)−1+((ln(8π))/2)  ∅=−(3/4)+((2ln(4))/2)+((ln(8π))/2)  ∅=−(3/4)+(1/2)ln(8^3 π)
=01ln(Γ(2+x))dx=01ln((x+1)!)dxx+1=tdt=dx=12ln(t!)dt=12ln[(te)t2πt]12ln(te)t+12ln(2πt)dt12ln(etln(te))+12ln(2π)dt+1212ln(t)dt12tln(te)+ln(2π)+12[xln(x)x]1212tln(t)12tdt+ln(2π)+12(2ln(2)+1)12tln(t)dt32+12ln(2π)+2ln22+1212tln(t)dtdu=ln(t)v=tu=tln(t)tdv=112tln(t)=[t2ln(t)t]tlntdt+tdt212tln(t)dt=[t2ln(t)t+12t2]12=4ln(2)2+124+11212tln(t)dt=2ln(2)1+1+1214=2ln(2)+14=2ln(2)+1432+ln(2π)2+2ln(2)2+12=2ln(2)+141+ln(8π)2=34+2ln(4)2+ln(8π)2=34+12ln(83π)
Commented by mathmax by abdo last updated on 06/Jul/21
t!∼((t/e))^t (√(2πt))is valable pour t assez grand(t→∞)
t!(te)t2πtisvalablepourtassezgrand(t)

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