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Question Number 141082 by mnjuly1970 last updated on 15/May/21
........ nice ....... calculus ........ 𝛗:=Σ_(n=2) ^∞ ((ζ ( n ))/(n . 4^n ))=?
……..nice…….calculus……..ϕ:=∑∞n=2ζ(n)n.4n=?
Answered by mindispower last updated on 15/May/21
Γ(x+1)ζ(x+1)=∫_0 ^∞ (t^x /(e^t −1))dt φ=Σ_(n≥2) (1/(Γ(n))).(1/(n4^n ))∫_0 ^∞ (t^(n−1) /(e^t −1))dt =(1/4)Σ_(n≥2) .(1/(Γ(n+1)))∫_0 ^∞ ((t/4))^(n−1) .(1/(e^t −1))dt =(1/4)∫_0 ^∞ .(4/t).Σ_(n≥2) (((((t/4))^n )/(n!))).(1/(e^t −1))dt =∫_0 ^∞ (1/t)(((e^(t/4) −1−(t/4))/(e^t −1)))dt let f(α)=∫_0 ^∞ ((e^(αt) −1−αt)/(t(e^t −1)))dt φ=f((1/4)) f(0)=0 f′(α)=∫_0 ^∞ ((te^(αt) −t)/(t(e^t −1)))dt=∫_0 ^∞ ((e^(αt) −1)/(e^t −1))dt,0≤α<1 f′(a)=∫_1 ^∞ ((t^a −1)/(t−1)).(dt/t) =∫_0 ^1 (((1/t^a )−1)/(1−t)).dt =∫_0 ^1 ((t^(−a) −1)/(1−t))dt=−∫_0 ^1 ((1−t^(−a) )/(1−t))dt,Ψ(x+1)=−γ+∫_0 ^1 ((1−t^x )/(1−t)) f′(a)=−Ψ(1−a)−γ f(a)=log(Γ(1−a))−γa+c f(0)=c=0 φ=f((1/4))=log(Γ((3/4)))−(γ/4)
Γ(x+1)ζ(x+1)=∫0∞txet−1dtϕ=∑n⩾21Γ(n).1n4n∫0∞tn−1et−1dt=14∑n⩾2.1Γ(n+1)∫0∞(t4)n−1.1et−1dt=14∫0∞.4t.∑n⩾2((t4)nn!).1et−1dt=∫0∞1t(et4−1−t4et−1)dtletf(α)=∫0∞eαt−1−αtt(et−1)dtϕ=f(14)f(0)=0f′(α)=∫0∞teαt−tt(et−1)dt=∫0∞eαt−1et−1dt,0⩽α<1f′(a)=∫1∞ta−1t−1.dtt=∫011ta−11−t.dt=∫01t−a−11−tdt=−∫011−t−a1−tdt,Ψ(x+1)=−γ+∫011−tx1−tf′(a)=−Ψ(1−a)−γf(a)=log(Γ(1−a))−γa+cf(0)=c=0ϕ=f(14)=log(Γ(34))−γ4
Commented by mnjuly1970 last updated on 15/May/21
bravo bravo sir power...
bravobravosirpower…
Commented by mindispower last updated on 15/May/21
withe pleasur Sir
withepleasurSir
Answered by mnjuly1970 last updated on 15/May/21
Γ(x+1):= e^(−γx) Π_(n=1) ^∞ (e^(x/n) /(1+(x/n))) ln(Γ(x+1)):=−γx+Σ_(n≥1) ((x/n)−ln(1+(x/n))) ((ln(Γ(x+1)))/x):=−γ +Σ_(n≥1) ((1/n)−(1/x)Σ_(k=1) (((−1)^(k−1) x^k )/(k.n^k ))) :=−γ+Σ_(n≥1) (1/n)−Σ_(n≥1) Σ_(k≥1) (((−1)^(k−1) x^(k−1) )/(k.n^k )) :=−γ+Σ_(n≥1) (1/n)−Σ_(k≥1) ((((−1)^(k−1) x^(k−1) )/k)Σ_(n≥1) (1/n^k )) :=−γ+Σ_(n≥1) (1/n)−Σ_(k≥1) (((−1)^(k−1) x^(k−1) )/k)ζ(k) =−γ +ζ(1)−ζ(1)+((ζ(2)x)/2)−((ζ(3)x^2 )/3)+... =−γ+Σ_(k≥2) (((−1)^k x^(k−1) ζ(k))/k) ((ln(Γ(x+1)))/x)=−γ+(1/x)Σ_(k≥2) (((−1)^k x^k ζ(k))/k) ln(Γ(x+1))=−γx+Σ_(k≥2) (((−1)^k x^k ζ(k))/k) x:=−(1/4) ⇒ ln(Γ(1−(1/4)))=(γ/4)+Σ_(k≥2) (((ζ(k))/(k.4^k )) ) ln(Γ((3/4)))−(γ/4)=Σ_(k≥2) (((ζ (k ))/(k . 4^( k) )) )
Γ(x+1):=e−γx∏∞n=1exn1+xnln(Γ(x+1)):=−γx+∑n⩾1(xn−ln(1+xn))ln(Γ(x+1))x:=−γ+∑n⩾1(1n−1x∑k=1(−1)k−1xkk.nk):=−γ+∑n⩾11n−∑n⩾1∑k⩾1(−1)k−1xk−1k.nk:=−γ+∑n⩾11n−∑k⩾1((−1)k−1xk−1k∑n⩾11nk):=−γ+∑n⩾11n−∑k⩾1(−1)k−1xk−1kζ(k)=−γ+ζ(1)−ζ(1)+ζ(2)x2−ζ(3)x23+…=−γ+∑k⩾2(−1)kxk−1ζ(k)kln(Γ(x+1))x=−γ+1x∑k⩾2(−1)kxkζ(k)kln(Γ(x+1))=−γx+∑k⩾2(−1)kxkζ(k)kx:=−14⇒ln(Γ(1−14))=γ4+∑k⩾2(ζ(k)k.4k)ln(Γ(34))−γ4=∑k⩾2(ζ(k)k.4k)
Commented by mindispower last updated on 15/May/21
sir are you student ?
sirareyoustudent?
Answered by Dwaipayan Shikari last updated on 15/May/21
ψ(z+1)=−γ+Σ_(n=2) ^∞ ζ(n)z^(n−1) [log(Γ(z+1))]_0 ^(−(1/4)) =−∫_0 ^((−1)/4) γ+Σ_(n=2) ^∞ ∫_0 ^(−(1/4)) (−1)^n ζ(n)z^(n−1) dz log(Γ((3/4)))=(γ/4)+Σ((ζ(n))/(n4^n )) Σ((ζ(n))/(n4^n ))=log(Γ((3/4)))−(γ/4)
ψ(z+1)=−γ+∑∞n=2ζ(n)zn−1[log(Γ(z+1))]0−14=−∫0−14γ+∑∞n=2∫0−14(−1)nζ(n)zn−1dzlog(Γ(34))=γ4+Σζ(n)n4nΣζ(n)n4n=log(Γ(34))−γ4
Commented by mnjuly1970 last updated on 15/May/21
very very nice thank you mr payan... taylor expansion of digamma..
veryverynicethankyoumrpayan…taylorexpansionofdigamma..

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