Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 122035 by mnjuly1970 last updated on 13/Nov/20

...nice calculus... prove that : ∫_0 ^( 1) ((ln^2 (1+x))/x)dx=^(??) ((ζ(3))/4) .m.n.

...nicecalculus...provethat:∫01ln2(1+x)xdx=??ζ(3)4.m.n.

Answered by mindispower last updated on 14/Nov/20

∫u^2 e^(−u) −u^2 e^(−u) −2ue^(−u) −2e^(−u) ∫_0 ^(ln(2)) (u^2 /(e^u −1))e^u du =∫_0 ^(ln(2)) Σ_(n≥0) u^2 e^(−nu) du=((ln^3 (2))/3) +Σ_(n≥1) ∫_0 ^(ln(2)) u^2 e^(−nu) du=Σ_(n≥1) (1/n^3 )∫_0 ^(nln(2)) x^2 e^(−x) dx =((ln^3 (2))/3)+Σ_(n≥1) (1/n^3 )[−x^2 e^(−x) −2xe^(−x) −2e^(−x) ]_0 ^(nln(2)) =((ln^3 (2))/3)+Σ_(n≥1) [−n^2 ln^2 (2).(1/2^n )−((2nln(2))/2^n )−2.(1/2^n )+2].(1/n^3 ) =((ln^3 (2))/3)−ln^2 (2)Σ_(n≥1) (1/(n.2^n ))−2ln(2)Σ_(n≥1) (1/(n^2 2^n ))−2Σ_(n≥1) (1/(n^3 2^n ))+2ζ(3) =−((2ln^3 (2))/3)−2ln(2)Li_2 ((1/2))−2Li_3 ((1/2))+2ζ(3) −2((ln^3 (2))/3)−2ln(2)((π^2 /(12))−(1/2)ln^2 (2))−2(((ln^3 (2))/6)−((π^2 ln(2))/(12))+(7/8)ζ(3))+2ζ(3) =−(7/4)ζ(3)+2ζ(3)=(1/4)ζ(3) Li_2 ((1/2))=(π^2 /(12))−((ln^2 (2))/2),Li_3 ((1/2))=((ln^3 (2))/6)−((π^2 ln(2))/(12))+((7ζ(3))/8)

∫u2e−u−u2e−u−2ue−u−2e−u∫0ln(2)u2eu−1eudu=∫0ln(2)∑n⩾0u2e−nudu=ln3(2)3+∑n⩾1∫0ln(2)u2e−nudu=∑n⩾11n3∫0nln(2)x2e−xdx=ln3(2)3+∑n⩾11n3[−x2e−x−2xe−x−2e−x]0nln(2)=ln3(2)3+∑n⩾1[−n2ln2(2).12n−2nln(2)2n−2.12n+2].1n3=ln3(2)3−ln2(2)∑n⩾11n.2n−2ln(2)∑n⩾11n22n−2∑n⩾11n32n+2ζ(3)=−2ln3(2)3−2ln(2)Li2(12)−2Li3(12)+2ζ(3)−2ln3(2)3−2ln(2)(π212−12ln2(2))−2(ln3(2)6−π2ln(2)12+78ζ(3))+2ζ(3)=−74ζ(3)+2ζ(3)=14ζ(3)Li2(12)=π212−ln2(2)2,Li3(12)=ln3(2)6−π2ln(2)12+7ζ(3)8

Commented by mnjuly1970 last updated on 14/Nov/20

Commented by mnjuly1970 last updated on 14/Nov/20

thank you mr power

thankyoumrpower

Commented by mindispower last updated on 14/Nov/20

hello sir realy nice please sir have you solution for Σ(1/(n^2 cos((π/2^n ))))

hellosirrealynicepleasesirhaveyousolutionforΣ1n2cos(π2n)

Terms of Service

Privacy Policy

Contact: info@tinkutara.com