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Question Number 140885 by mnjuly1970 last updated on 13/May/21

                  ....... Advanced ::::::::::★★★:::::::::: Calculus.......        find the value of  the infinite series::                   Θ := Σ_(n=1) ^∞ (((−1)^(n−1) H_( 2n) )/(2n−1)) = ???            .......M.N.july.1970........

.......Advanced::::::::::::::::::::Calculus.......findthevalueoftheinfiniteseries::Θ:=n=1(1)n1H2n2n1=???.......M.N.july.1970........

Answered by mindispower last updated on 15/May/21

H_(2n) =H_(2n−1) +(1/(2n))  H_n =−n∫_0 ^1 x^(n−1) ln(1−x)dx...  ⇔Θ=Σ_(n≥1) (−1)^(n−1) .(−∫_0 ^1 x^(2n−2) ln(1−x)dx+(1/((2n)(2n−1))))  =−∫_0 ^1 Σ_(n≥1) (−x^2 )^(n−1) ln(1−x)dx+Σ_(n≥1) (((−1)^(n−1) )/(2n−1))+Σ_(n≥1) (((−1)^n )/(2n))  =∫_0 ^1 ((−ln(1−x))/(1+x^2 ))dx+Σ_(n≥0) (((−1)^n )/(2n+1))−Σ_(n≥0) (((−1)^n )/(n+1))  =(π/4)−ln(2)−∫_0 ^1 ((ln(1−x))/(1+x^2 ))dx_(=A)   x=tg(t)⇔A=−∫_0 ^(π/4) ln(((cos(t)−sin(t))/(cos(t))))dt  A=−∫_0 ^(π/4) ln(cos((π/4)+t))dt−∫_0 ^(π/4) ln(cos(t))dt  =−∫_0 ^(π/4) ln(sin(t))dt−∫_0 ^(π/4) ln(cos)t))dt  =−∫_0 ^(π/4) ln(((sin(2t))/2))dt =−(1/2)∫_0 ^(π/2) ln(sin(t))dt+(π/4)ln(2).

H2n=H2n1+12nHn=n01xn1ln(1x)dx...Θ=n1(1)n1.(01x2n2ln(1x)dx+1(2n)(2n1))=01n1(x2)n1ln(1x)dx+n1(1)n12n1+n1(1)n2n=01ln(1x)1+x2dx+n0(1)n2n+1n0(1)nn+1=π4ln(2)01ln(1x)1+x2dx=Ax=tg(t)A=0π4ln(cos(t)sin(t)cos(t))dtA=0π4ln(cos(π4+t))dt0π4ln(cos(t))dt=0π4ln(sin(t))dt0π4ln(cos)t))dt=0π4ln(sin(2t)2)dt=120π2ln(sin(t))dt+π4ln(2).

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