Question Number 101277 by bobhans last updated on 01/Jul/20

∫_0 ^1  (((x−1) dx )/((x+1)ln (x)))

Answered by maths mind last updated on 01/Jul/20

=∫_0 ^1 ((x−1)/(x+1)).(x^a /(ln(x)))dx=f(a)  f′(a)=∫_0 ^1 ((x−1)/(x+1)).x^a   =∫_0 ^1 ((x^(a+1) /((x+1)))−(x^a /(x+1)))dx  =∫_0 ^1 (Σ_(k=0) ^(+∞) (−1)^k x^(k+a+1) −Σ_(k≥0) (−1)^k x^(k+a) )dx  −(1/(a+1))+2∫_0 ^1 Σ_(k≥0) ^(+∞) (((−1)^k )/(k+a+2))  =−(1/(a+1))+2∫_0 ^1 Σ_(k=0) ^(+∞) ((1/(2k+a+2))−(1/(2k+3+a)))  =−(1/(a+1))+2Σ_(k=0) ^(+∞) ((1/(2(k+1+(a/2))))−(1/(2k+2))−((1/(2(k+1+((a+1)/2))))−(1/(2k+2)))  =−(1/(a+1))+2{Σ_(k≥1) ((1/(2(k+(a/2))))−(1/(2k)))−(1/2)Σ_(k≥1) ((1/(k+((a+1)/2)))−(1/k))}  =−(1/(a+1))−H_(a/2) +H_((a+1)/2) =−(1/(a+1))−(Ψ((a/2)+1)+γ)+(Ψ(((a+1)/2)+1)+γ)  f′(a)=−(1/(a+1))+Ψ(((a+3)/2))−Ψ(((a+2)/2))  f(a)=−ln(a+1)+2(ln(Γ(((a+3)/2)))−ln(Γ(((a+2)/2))))  Γ(((a+2)/2)+(1/2))=2^(1−a−(3/2)) (√π).Γ(a+2)/(Γ(((a+2)/2)))  ⇒f(a)=−ln(a+1)+2ln(((Γ(((a+3)/2)))/(Γ(((a+2)/2)))))=−ln(a+1)+2ln(((2^(−1−a) (√π).Γ(a+2))/(Γ^2 (((a+2)/2)))))+c  lim_(a→∞) f(a)=0⇒  lim_(a→0)  −ln(a+1)+2ln(((2^(−(1/2)−a) (√(π.))Γ(a+2))/(Γ^2 (((a+2)/2)))))=2ln(((√π)/(√2)))=ln((π/2))  =∫_0 ^1 ((x−1)/((x+1)ln(x)))dx

Commented bymaths mind last updated on 01/Jul/20

somm justificTion  lim_(a→∞) f(a)=0  since ((x−1)/(ln(x))),is continus over ]0,1[ and  lim_(x→1) ((x−1)/(ln(x)))=1  with lim_(x→0) ((x−1)/(ln(x)))=0  ⇒((x−1)/(ln(x)))≤sup  {((x−1)/(ln(x))),x∈]0,1[ ,1}=c<∞  ⇒∫_0 ^1 ((x−1)/((x+1)ln(x)))x^a dx≤∫_0 ^1 cx^a .=(c/(a+1))→0,a→∞

Commented bybobhans last updated on 02/Jul/20

waw.....beautifull

Answered by abdomathmax last updated on 02/Jul/20

I =∫_0 ^1  (((x−1))/((x+1)lnx))dx  changement lnx =−t give  I =−∫_0 ^∞   ((e^(−t) −1)/((e^(−t) +1)(−t)))(−e^(−t) )dt  =−∫_0 ^∞  ((e^(−2t) −e^(−t) )/(t(e^(−t)  +1)))dt  =∫_0 ^∞  ((e^(−t) −e^(−2t) )/t)(Σ_(n=0) ^∞  (−1)^n  e^(−nt) )dt  =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^∞   ((e^(−(n+1)t) −e^(−(n+2)t) )/t) dt  let find u(a,b) =∫_0 ^∞  ((e^(−at) −e^(−bt) )/t) dt with a ,b>0  let I(ξ) =∫_ξ ^(+∞)  ((e^(−at) −e^(−bt) )/t) dt  u(a,b) =lim_(ξ →0^+ )    and  I(ξ) = ∫_ξ ^(+∞ )  (e^(−at) /t)dt(at =u) −∫_ξ ^(+∞)  (e^(−bt) /t)dt(bt =v)  =∫_(aξ) ^(+∞ )  (e^(−u) /(u/a))×(du/a) −∫_(bξ) ^(+∞)  (e^(−v) /(v/b))×(dv/b)  =∫_(aξ) ^∞  (e^(−u) /u)du −∫_(bξ) ^∞  (e^(−v) /v)dv  =∫_(aξ) ^(bξ)  (e^(−u) /u)du  but ∃ c ∈]aξ,bξ[ /I(ξ) =e^(−c) ∫_(aξ) ^(bξ)  (du/u)  =e^(−c) ln(((bξ)/(aξ))) =e^(−c) ln((b/a)) →ln((b/a)) (ξ→0) ⇒  u(a,b) =ln((b/a)) ⇒ I =Σ_(n=0) ^∞ (−1)^n ln(((n+2)/(n+1)))  ...be continued...