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Question Number 69233 by ~ À ® @ 237 ~ last updated on 21/Sep/19
Prove that B=∫_0 ^1 [ln(−lnu)]^2 du = γ^2 + ζ(2)
ProvethatB=∫01[ln(−lnu)]2du=γ2+ζ(2)
Commented by mathmax by abdo last updated on 22/Sep/19
B=∫_0 ^1 {ln(−lnx)}^2 dx changement −lnx =t give x =e^(−t) B =−∫_0 ^∞ (ln(t)^2 (−e^(−t) )dt =∫_0 ^∞ (ln(t))^2 e^(−t) dt we have Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt with x>0 ⇒Γ(x)=∫_0 ^∞ e^((x−1)lnt) e^(−t) dt ⇒Γ^′ (x) =∫_0 ^∞ ln(t) t^(x−1) e^(−t) dt ⇒Γ^((2)) (x)=∫_0 ^∞ (ln(t))^2 t^(x−1) e^(−t) dt ⇒ Γ^((2)) (1) =∫_0 ^∞ (ln(t))^(2 ) e^(−t) dt but we have the formula ((Γ^′ (x))/(Γ(x))) =−γ−(1/x) +Σ_(n=1) ^∞ ((1/n)−(1/(n+x))) by derivation we get ((Γ^((2)) (x)Γ(x)−(Γ^′ (x))^2 )/((Γ(x))^2 )) =(1/x^2 ) +Σ_(n=1) ^∞ (1/((n+x)^2 )) ⇒ ((Γ^((2)) (1)Γ(1)−(Γ^′ (1))^2 )/((Γ(1))^2 )) =1+Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=0) ^∞ (1/((n+1)^2 )) =(π^2 /6) Γ(1)=1 and Γ^′ (1) =∫_0 ^∞ e^(−t) ln(t)dt =−γ (this result is proved) ⇒ Γ^((2)) (1) −γ^2 =(π^2 /6) ⇒Γ^((2)) (1) =γ^2 +(π^2 /6) =B
B=∫01{ln(−lnx)}2dxchangement−lnx=tgivex=e−tB=−∫0∞(ln(t)2(−e−t)dt=∫0∞(ln(t))2e−tdtwehaveΓ(x)=∫0∞tx−1e−tdtwithx>0⇒Γ(x)=∫0∞e(x−1)lnte−tdt⇒Γ′(x)=∫0∞ln(t)tx−1e−tdt⇒Γ(2)(x)=∫0∞(ln(t))2tx−1e−tdt⇒Γ(2)(1)=∫0∞(ln(t))2e−tdtbutwehavetheformulaΓ′(x)Γ(x)=−γ−1x+∑n=1∞(1n−1n+x)byderivationwegetΓ(2)(x)Γ(x)−(Γ′(x))2(Γ(x))2=1x2+∑n=1∞1(n+x)2⇒Γ(2)(1)Γ(1)−(Γ′(1))2(Γ(1))2=1+∑n=1∞1(n+1)2=∑n=0∞1(n+1)2=π26Γ(1)=1andΓ′(1)=∫0∞e−tln(t)dt=−γ(thisresultisproved)⇒Γ(2)(1)−γ2=π26⇒Γ(2)(1)=γ2+π26=B
Answered by mind is power last updated on 21/Sep/19
−ln(u)=t ⇒∫_0 ^(+∞) [ln(t)]^2 e^(−t) dt Γ(z)=∫_0 ^(+∞) t^(z−1) e^(−t) dt ⇒Γ′(z)=∫_0 ^(+∞) (d/dz)e^((z−1)ln(t)) e^(−t) dt ⇒Γ′(z)=∫_0 ^(+∞) ln(t)t^(z−1) e^(−t) dt ⇒Γ′′(z)=∫_0 ^(+∞) (ln(t))^2 t^(z−1) e^(−t) dt ⇒Γ′′(1)=∫_0 ^(+∞) (ln(t))^2 e^(−t) dt we have ψ(z)=((Γ′(z))/(Γ(z)))=−γ−(1/z)+Σ_(n≥1) (1/n)−(1/(n+z))...psi function to prove it tack ln(Γ) Γ′(1)=−γ (((Γ′(z))/(Γ(z))))^′ =((Γ′′(z)Γ(z)−(Γ′(z))^2 )/(Γ(z)^2 ))=(1/z^2 )+Σ_(n≥1) (1/((n+z)^2 )) ⇒((Γ′′(1)Γ(1)−(Γ′(1))^2 )/(Γ(1)^2 ))=1+Σ_(n≥1) (1/((1+n)^2 ))=Σ_(n≥1) (1/n^2 )=ζ(2) Γ(1)=1 .Γ′(1)=−γ ⇒Γ′′(1)−γ^2 =ζ(2)⇒Γ′′(1)=γ^2 +ζ(2)
−ln(u)=t⇒∫0+∞[ln(t)]2e−tdtΓ(z)=∫0+∞tz−1e−tdt⇒Γ′(z)=∫0+∞ddze(z−1)ln(t)e−tdt⇒Γ′(z)=∫0+∞ln(t)tz−1e−tdt⇒Γ″(z)=∫0+∞(ln(t))2tz−1e−tdt⇒Γ″(1)=∫0+∞(ln(t))2e−tdtwehaveψ(z)=Γ′(z)Γ(z)=−γ−1z+∑n⩾11n−1n+z…psifunctiontoproveittackln(Γ)Γ′(1)=−γMissing \left or extra \right⇒Γ″(1)Γ(1)−(Γ′(1))2Γ(1)2=1+∑n⩾11(1+n)2=∑n⩾11n2=ζ(2)Γ(1)=1.Γ′(1)=−γ⇒Γ″(1)−γ2=ζ(2)⇒Γ″(1)=γ2+ζ(2)
Commented by ~ À ® @ 237 ~ last updated on 21/Sep/19
Thanks you sir ! have you prove that Γ′(1)=−γ ?
Thanksyousir!haveyouprovethatΓ′(1)=−γ?
Commented by mind is power last updated on 21/Sep/19
ψ(1)=((Γ′(1))/(Γ(1)))=−γ−(1/1)+Σ_(n≥1) (1/n)−(1/(n+1))=−γ ⇒Γ′(1)=−γ
ψ(1)=Γ′(1)Γ(1)=−γ−11+∑n⩾11n−1n+1=−γ⇒Γ′(1)=−γ
Commented by ~ À ® @ 237 ~ last updated on 21/Sep/19
you′re solving a question by another : else thanks!
you′resolvingaquestionbyanother:elsethanks!
Commented by mind is power last updated on 21/Sep/19
Γ(z)=(e^(−γz) /z).∐_(k≥1) (e^(z/k) /(1+(z/k))) ⇒ln(Γ(z))=−γz−ln(z)+Σ_(k≥1) (z/k)−ln(1+(z/k)) ⇒((Γ′(z))/(Γ(z)))=−γ−(1/z)+Σ_(k≥1) (1/k)−(1/k).(k/(k+z)) ⇒((Γ′(z))/(Γ(z)))=−γ−(1/z)+Σ_(k≥1) (1/k)−(1/(k+z))
Γ(z)=e−γzz.∐k⩾1ezk1+zk⇒ln(Γ(z))=−γz−ln(z)+∑k⩾1zk−ln(1+zk)⇒Γ′(z)Γ(z)=−γ−1z+∑k⩾11k−1k.kk+z⇒Γ′(z)Γ(z)=−γ−1z+∑k⩾11k−1k+z

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