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Question Number 60727 by Mr X pcx last updated on 25/May/19
calculate ∫_1 ^(+∞)  ((ln(lnx))/(1+x^2 ))dx
calculate1+ln(lnx)1+x2dx
Commented by maxmathsup by imad last updated on 26/May/19
let A =∫_1 ^(+∞)  ((ln(lnx))/(1+x^2 )) dx  changement ln(x)=t give  A =∫_0 ^(+∞)  ((ln(t))/(1+e^(2t) )) e^t  dt =∫_0 ^∞    ((e^(−t) ln(t))/(1+e^(−2t) )) dt  =∫_0 ^∞    e^(−t) ln(t)(Σ_(n=0) ^∞  (−1)^n  e^(−2nt) )dt  =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^∞    e^(−(2n+1)t)  ln(t)dt =Σ_(n=0) ^∞ (−1)^n  A_n   A_n =∫_0 ^∞   e^(−(2n+1)t) ln(t)dt =_((2n+1)t =u)    ∫_0 ^∞    e^(−u) ln((u/(2n+1)))(du/(2n+1))  =(1/(2n+1)) ∫_0 ^∞   e^(−u) {ln(u)−ln(2n+1))du  =(1/(2n+1)){   ∫_0 ^∞  e^(−u) ln(u)du −ln(2n+1)∫_0 ^∞  e^(−u)  du}  ∫_0 ^∞  e^(−u) ln(u)du =−γ  (result proved)  ∫_0 ^∞  e^(−u)  du =[−e^(−u) ]_0 ^∞  =1 ⇒A_n =((−γ)/(2n+1)) −((ln(2n+1))/(2n+1)) ⇒  A =Σ_(n=0) ^∞ (−1)^n {−(γ/(2n+1)) −((ln(2n+1))/(2n+1))}  =−γ Σ_(n=0) ^∞  (((−1)^n )/(2n+1)) −Σ_(n=0) ^∞  (((−1)^n )/(2n+1))ln(2n+1)  =−((γπ)/4) −Σ_(n=0) ^∞  (−1)^n  ((ln(2n+1))/(2n+1))    with γ constant of Euler ...be continued...
letA=1+ln(lnx)1+x2dxchangementln(x)=tgiveA=0+ln(t)1+e2tetdt=0etln(t)1+e2tdt=0etln(t)(n=0(1)ne2nt)dt=n=0(1)n0e(2n+1)tln(t)dt=n=0(1)nAnAn=0e(2n+1)tln(t)dt=(2n+1)t=u0euln(u2n+1)du2n+1=12n+10eu{ln(u)ln(2n+1))du=12n+1{0euln(u)duln(2n+1)0eudu}0euln(u)du=γ(resultproved)0eudu=[eu]0=1An=γ2n+1ln(2n+1)2n+1A=n=0(1)n{γ2n+1ln(2n+1)2n+1}=γn=0(1)n2n+1n=0(1)n2n+1ln(2n+1)=γπ4n=0(1)nln(2n+1)2n+1withγconstantofEulerbecontinued
Commented by maxmathsup by imad last updated on 26/May/19
let try another way   by use of chang. x =(1/t) ⇒  I = ∫_1 ^(+∞)   ((ln(lnx))/(1+x^2 )) dx =−∫_0 ^1  ((ln(ln((1/t))))/(1+(1/t^2 ))) (−(dt/t^2 ))  = ∫_0 ^1    ((ln(ln((1/t))))/(1+t^2 )) dt =∫_0 ^1   ((ln(ln((1/t))))/((1+it)(1−it)))dt  = (1/2)∫_0 ^1   ((1/(1+it)) +(1/(1−it)))ln(ln((1/t)))dt  =(1/2) ∫_0 ^1   ((ln{ln((1/t))})/(1+it))dt +(1/2) ∫_0 ^1  ((ln{ln((1/t))})/(1−it)) dt  let W =∫_0 ^1  ((ln{ln((1/t))})/(1+it)) dt  chang . ln((1/t)) =u give (1/t) =e^u  ⇒t =e^(−u)  ⇒  W =−∫_0 ^∞    ((lnu)/(1+ie^(−u) )) (−e^(−u) )du = ∫_0 ^∞   ((e^(−u) ln(u))/(1+ie^(−u) )) du  =∫_0 ^∞  e^(−u) ln(u)(Σ_(n=0) ^∞  (−ie^(−u) )^n )du  =Σ_(n=0) ^∞  (−i)^n  ∫_0 ^∞    e^(−(n+1)u) ln(u)du  ∫_0 ^∞ e^(−(n+1)u) ln(u)du =_((n+1)u=t)    ∫_0 ^∞   e^(−t) ln((t/(n+1)))(dt/(n+1))  =(1/(n+1)) ∫_0 ^∞   { e^(−t) ln(t)−e^(−t) ln(n+1)}dt  =(1/(n+1)) ∫_0 ^∞    e^(−t) ln(t) −((ln(n+1))/(n+1)) ∫_0 ^∞    e^(−t)  dt  =((−γ)/(n+1)) −((ln(n+1))/(n+1)) ⇒ W =Σ_(n=0) ^∞ ((−γ(−i)^n )/(n+1)) −Σ_(n=0) ^∞  (((−i)^n ln(n+1))/(n+1))  H =∫_0 ^1   ((ln{ln((1/t))})/(1−it ))dt =_(ln((1/t))=u)     −∫_0 ^∞   ((ln(u))/(1−i e^(−u) ))(−e^(−u) )du  =∫_0 ^∞    ((e^(−u) ln(u))/(1−ie^(−u) )) du =∫_0 ^∞     e^(−u) lnu {Σ_(n=0) ^∞  i^n  e^(−nu) }du  =Σ_(n=0) ^∞   i^n  ∫_0 ^∞   e^(−(n+1)u)  ln(u)du =....be continued....
lettryanotherwaybyuseofchang.x=1tI=1+ln(lnx)1+x2dx=01ln(ln(1t))1+1t2(dtt2)=01ln(ln(1t))1+t2dt=01ln(ln(1t))(1+it)(1it)dt=1201(11+it+11it)ln(ln(1t))dt=1201ln{ln(1t)}1+itdt+1201ln{ln(1t)}1itdtletW=01ln{ln(1t)}1+itdtchang.ln(1t)=ugive1t=eut=euW=0lnu1+ieu(eu)du=0euln(u)1+ieudu=0euln(u)(n=0(ieu)n)du=n=0(i)n0e(n+1)uln(u)du0e(n+1)uln(u)du=(n+1)u=t0etln(tn+1)dtn+1=1n+10{etln(t)etln(n+1)}dt=1n+10etln(t)ln(n+1)n+10etdt=γn+1ln(n+1)n+1W=n=0γ(i)nn+1n=0(i)nln(n+1)n+1H=01ln{ln(1t)}1itdt=ln(1t)=u0ln(u)1ieu(eu)du=0euln(u)1ieudu=0eulnu{n=0inenu}du=n=0in0e(n+1)uln(u)du=.becontinued.

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