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I-0-1-ln-lnx-1-x-2-dx-

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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 ))
∫0∞log(t)e2t+1etdt−logx=t⇒1=−xdtdx=∑∞n=1(−1)n+1∫0∞e−2nt+tlog(t)dt=∑∞n=1(−1)n+112n−1∫0∞e−ulog(u)−log(2n−1)2n−1∫0∞e−udu=−∑∞n=0(−1)nγ2n+1−∑∞n=0(−1)nlog(2n+1)2n+1=−πγ4+Φ′(1)Φ(α)=∑∞n=0(−1)n(2n+1)α⇒Φ(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,∞[,soyoucan′treplacexbyetinthepreviousequalityyoucouldthingaboutdividingbye2tthenumeratorandthedenominatorinthewaytoget∫0∞e−tlnt1+e−2tdtbeforereplacingxbye−2t
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) ))
e−tlog(t)1+e−2t=e−t.e2tlog(t)e2t+e2t−2t=etlog(t)1+e2t

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