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Question Number 133530 by snipers237 last updated on 22/Feb/21

 I=∫_0 ^1   ((ln(−lnx))/(1+x^2 )) dx

I=01ln(lnx)1+x2dx

Commented by Dwaipayan Shikari last updated on 22/Feb/21

∫_0 ^∞ ((log(t))/(e^(2t) +1))e^t dt              −logx=t⇒1=−x(dt/dx)  =Σ_(n=1) ^∞ (−1)^(n+1) ∫_0 ^∞ e^(−2nt+t) log(t)dt  =Σ_(n=1) ^∞ (−1)^(n+1) (1/(2n−1))∫_0 ^∞ e^(−u) log(u)−((log(2n−1))/(2n−1)) ∫_0 ^∞ e^(−u) du  =−Σ_(n=0) ^∞ (((−1)^n γ)/(2n+1))−Σ_(n=0) ^∞ (−1)^n ((log(2n+1))/(2n+1))  =−((πγ)/4)+Φ′(1)  Φ(α)=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^α ))⇒Φ(α)=−Σ_(n=0) ^∞ (((−1)^n log(2n+1))/((2n+1)^α ))

0log(t)e2t+1etdtlogx=t1=xdtdx=n=1(1)n+10e2nt+tlog(t)dt=n=1(1)n+112n10eulog(u)log(2n1)2n10eudu=n=0(1)nγ2n+1n=0(1)nlog(2n+1)2n+1=πγ4+Φ(1)Φ(α)=n=0(1)n(2n+1)αΦ(α)=n=0(1)nlog(2n+1)(2n+1)α

Commented by snipers237 last updated on 23/Feb/21

thanks for crying sir , but i think  if x∈[0,1] then i can write (1/(1+x^2 ))=Σ_(n=0) ^∞ (−x^2 )^n    but when t∈[0,∞[ , we have e^t ∈[1,∞[ , so you can′t replace x by e^t  in the previous equality   you could thing about dividing by  e^(2t)    the numerator and the denominator in the way to get   ∫_0 ^∞ ((e^(−t) lnt)/(1+e^(−2t) ))dt    before replacing x by e^(−2t)

thanksforcryingsir,butithinkifx[0,1]thenicanwrite11+x2=n=0(x2)nbutwhent[0,[,wehaveet[1,[,soyoucantreplacexbyetinthepreviousequalityyoucouldthingaboutdividingbye2tthenumeratorandthedenominatorinthewaytoget0etlnt1+e2tdtbeforereplacingxbye2t

Commented by Dwaipayan Shikari last updated on 23/Feb/21

((e^(−t) log(t))/(1+e^(−2t) ))=((e^(−t) .e^(2t) log(t))/(e^(2t) +e^(2t−2t) ))=((e^t log(t))/(1+e^(2t) ))

etlog(t)1+e2t=et.e2tlog(t)e2t+e2t2t=etlog(t)1+e2t

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