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Question Number 141082 by mnjuly1970 last updated on 15/May/21
........ nice ....... calculus ........ 𝛗:=Ξ£_(n=2) ^∞ ((ΞΆ ( n ))/(n . 4^n ))=?
$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:……..\:{nice}\:\:…….\:\:{calculus}\:…….. \\ $$$$\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\zeta\:\left(\:{n}\:\right)}{{n}\:.\:\mathrm{4}^{{n}} }=? \\ $$$$ \\ $$
Answered by mindispower last updated on 15/May/21
Ξ“(x+1)ΞΆ(x+1)=∫_0 ^∞ (t^x /(e^t βˆ’1))dt Ο†=Ξ£_(nβ‰₯2) (1/(Ξ“(n))).(1/(n4^n ))∫_0 ^∞ (t^(nβˆ’1) /(e^t βˆ’1))dt =(1/4)Ξ£_(nβ‰₯2) .(1/(Ξ“(n+1)))∫_0 ^∞ ((t/4))^(nβˆ’1) .(1/(e^t βˆ’1))dt =(1/4)∫_0 ^∞ .(4/t).Ξ£_(nβ‰₯2) (((((t/4))^n )/(n!))).(1/(e^t βˆ’1))dt =∫_0 ^∞ (1/t)(((e^(t/4) βˆ’1βˆ’(t/4))/(e^t βˆ’1)))dt let f(Ξ±)=∫_0 ^∞ ((e^(Ξ±t) βˆ’1βˆ’Ξ±t)/(t(e^t βˆ’1)))dt Ο†=f((1/4)) f(0)=0 fβ€²(Ξ±)=∫_0 ^∞ ((te^(Ξ±t) βˆ’t)/(t(e^t βˆ’1)))dt=∫_0 ^∞ ((e^(Ξ±t) βˆ’1)/(e^t βˆ’1))dt,0≀α<1 fβ€²(a)=∫_1 ^∞ ((t^a βˆ’1)/(tβˆ’1)).(dt/t) =∫_0 ^1 (((1/t^a )βˆ’1)/(1βˆ’t)).dt =∫_0 ^1 ((t^(βˆ’a) βˆ’1)/(1βˆ’t))dt=βˆ’βˆ«_0 ^1 ((1βˆ’t^(βˆ’a) )/(1βˆ’t))dt,Ξ¨(x+1)=βˆ’Ξ³+∫_0 ^1 ((1βˆ’t^x )/(1βˆ’t)) fβ€²(a)=βˆ’Ξ¨(1βˆ’a)βˆ’Ξ³ f(a)=log(Ξ“(1βˆ’a))βˆ’Ξ³a+c f(0)=c=0 Ο†=f((1/4))=log(Ξ“((3/4)))βˆ’(Ξ³/4)
$$\Gamma\left({x}+\mathrm{1}\right)\zeta\left({x}+\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}} }{{e}^{{t}} βˆ’\mathrm{1}}{dt} \\ $$$$\phi=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\Gamma\left({n}\right)}.\frac{\mathrm{1}}{{n}\mathrm{4}^{{n}} }\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{n}βˆ’\mathrm{1}} }{{e}^{{t}} βˆ’\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{2}} {\sum}.\frac{\mathrm{1}}{\Gamma\left({n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \left(\frac{{t}}{\mathrm{4}}\right)^{{n}βˆ’\mathrm{1}} .\frac{\mathrm{1}}{\boldsymbol{{e}}^{\boldsymbol{{t}}} βˆ’\mathrm{1}}\boldsymbol{{dt}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} .\frac{\mathrm{4}}{{t}}.\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{\left(\frac{{t}}{\mathrm{4}}\right)^{{n}} }{{n}!}\right).\frac{\mathrm{1}}{{e}^{{t}} βˆ’\mathrm{1}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}}\left(\frac{{e}^{\frac{{t}}{\mathrm{4}}} βˆ’\mathrm{1}βˆ’\frac{{t}}{\mathrm{4}}}{{e}^{{t}} βˆ’\mathrm{1}}\right){dt} \\ $$$${let}\:{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{\alpha{t}} βˆ’\mathrm{1}βˆ’\alpha{t}}{{t}\left({e}^{{t}} βˆ’\mathrm{1}\right)}{dt} \\ $$$$\phi={f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}'\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{te}^{\alpha{t}} βˆ’{t}}{{t}\left({e}^{{t}} βˆ’\mathrm{1}\right)}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{\alpha{t}} βˆ’\mathrm{1}}{{e}^{{t}} βˆ’\mathrm{1}}{dt},\mathrm{0}\leqslant\alpha<\mathrm{1} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{1}} ^{\infty} \frac{{t}^{{a}} βˆ’\mathrm{1}}{{t}βˆ’\mathrm{1}}.\frac{{dt}}{{t}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}}{{t}^{{a}} }βˆ’\mathrm{1}}{\mathrm{1}βˆ’{t}}.{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{βˆ’{a}} βˆ’\mathrm{1}}{\mathrm{1}βˆ’{t}}{dt}=βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{t}^{βˆ’{a}} }{\mathrm{1}βˆ’{t}}{dt},\Psi\left({x}+\mathrm{1}\right)=βˆ’\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{t}^{{x}} }{\mathrm{1}βˆ’{t}} \\ $$$${f}'\left({a}\right)=βˆ’\Psi\left(\mathrm{1}βˆ’{a}\right)βˆ’\gamma \\ $$$${f}\left({a}\right)={log}\left(\Gamma\left(\mathrm{1}βˆ’{a}\right)\right)βˆ’\gamma{a}+{c} \\ $$$${f}\left(\mathrm{0}\right)={c}=\mathrm{0} \\ $$$$\phi={f}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)={log}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)βˆ’\frac{\gamma}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 15/May/21
bravo bravo sir power...
$${bravo}\:{bravo}\:{sir}\:{power}… \\ $$
Commented by mindispower last updated on 15/May/21
withe pleasur Sir
$${withe}\:{pleasur}\:{Sir} \\ $$
Answered by mnjuly1970 last updated on 15/May/21
Ξ“(x+1):= e^(βˆ’Ξ³x) Ξ _(n=1) ^∞ (e^(x/n) /(1+(x/n))) ln(Ξ“(x+1)):=βˆ’Ξ³x+Ξ£_(nβ‰₯1) ((x/n)βˆ’ln(1+(x/n))) ((ln(Ξ“(x+1)))/x):=βˆ’Ξ³ +Ξ£_(nβ‰₯1) ((1/n)βˆ’(1/x)Ξ£_(k=1) (((βˆ’1)^(kβˆ’1) x^k )/(k.n^k ))) :=βˆ’Ξ³+Ξ£_(nβ‰₯1) (1/n)βˆ’Ξ£_(nβ‰₯1) Ξ£_(kβ‰₯1) (((βˆ’1)^(kβˆ’1) x^(kβˆ’1) )/(k.n^k )) :=βˆ’Ξ³+Ξ£_(nβ‰₯1) (1/n)βˆ’Ξ£_(kβ‰₯1) ((((βˆ’1)^(kβˆ’1) x^(kβˆ’1) )/k)Ξ£_(nβ‰₯1) (1/n^k )) :=βˆ’Ξ³+Ξ£_(nβ‰₯1) (1/n)βˆ’Ξ£_(kβ‰₯1) (((βˆ’1)^(kβˆ’1) x^(kβˆ’1) )/k)ΞΆ(k) =βˆ’Ξ³ +ΞΆ(1)βˆ’ΞΆ(1)+((ΞΆ(2)x)/2)βˆ’((ΞΆ(3)x^2 )/3)+... =βˆ’Ξ³+Ξ£_(kβ‰₯2) (((βˆ’1)^k x^(kβˆ’1) ΞΆ(k))/k) ((ln(Ξ“(x+1)))/x)=βˆ’Ξ³+(1/x)Ξ£_(kβ‰₯2) (((βˆ’1)^k x^k ΞΆ(k))/k) ln(Ξ“(x+1))=βˆ’Ξ³x+Ξ£_(kβ‰₯2) (((βˆ’1)^k x^k ΞΆ(k))/k) x:=βˆ’(1/4) β‡’ ln(Ξ“(1βˆ’(1/4)))=(Ξ³/4)+Ξ£_(kβ‰₯2) (((ΞΆ(k))/(k.4^k )) ) ln(Ξ“((3/4)))βˆ’(Ξ³/4)=Ξ£_(kβ‰₯2) (((ΞΆ (k ))/(k . 4^( k) )) )
$$\:\:\:\:\:\:\Gamma\left({x}+\mathrm{1}\right):=\:{e}^{βˆ’\gamma{x}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{{e}^{\frac{{x}}{{n}}} }{\mathrm{1}+\frac{{x}}{{n}}} \\ $$$$\:\:\:\:\:\:\:\:\:{ln}\left(\Gamma\left({x}+\mathrm{1}\right)\right):=βˆ’\gamma{x}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{{x}}{{n}}βˆ’{ln}\left(\mathrm{1}+\frac{{x}}{{n}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{ln}\left(\Gamma\left({x}+\mathrm{1}\right)\right)}{{x}}:=βˆ’\gamma\:+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}}βˆ’\frac{\mathrm{1}}{{x}}\underset{{k}=\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}βˆ’\mathrm{1}} {x}^{{k}} }{{k}.{n}^{{k}} }\right) \\ $$$$\:\:\::=βˆ’\gamma+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}βˆ’\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}βˆ’\mathrm{1}} {x}^{{k}βˆ’\mathrm{1}} }{{k}.{n}^{{k}} } \\ $$$$\:\::=βˆ’\gamma+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}βˆ’\underset{{k}\geqslant\mathrm{1}} {\sum}\left(\frac{\left(βˆ’\mathrm{1}\right)^{{k}βˆ’\mathrm{1}} {x}^{{k}βˆ’\mathrm{1}} }{{k}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{{k}} }\right) \\ $$$$\:\::=βˆ’\gamma+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}βˆ’\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}βˆ’\mathrm{1}} {x}^{{k}βˆ’\mathrm{1}} }{{k}}\zeta\left({k}\right) \\ $$$$\:=βˆ’\gamma\:+\zeta\left(\mathrm{1}\right)βˆ’\zeta\left(\mathrm{1}\right)+\frac{\zeta\left(\mathrm{2}\right){x}}{\mathrm{2}}βˆ’\frac{\zeta\left(\mathrm{3}\right){x}^{\mathrm{2}} }{\mathrm{3}}+… \\ $$$$\:=βˆ’\gamma+\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} {x}^{{k}βˆ’\mathrm{1}} \zeta\left({k}\right)}{{k}} \\ $$$$\:\frac{{ln}\left(\Gamma\left({x}+\mathrm{1}\right)\right)}{{x}}=βˆ’\gamma+\frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} {x}^{{k}} \zeta\left({k}\right)}{{k}} \\ $$$$\:\:\:{ln}\left(\Gamma\left({x}+\mathrm{1}\right)\right)=βˆ’\gamma{x}+\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{k}} {x}^{{k}} \zeta\left({k}\right)}{{k}} \\ $$$$\:\:\:{x}:=βˆ’\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:{ln}\left(\Gamma\left(\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{4}}\right)\right)=\frac{\gamma}{\mathrm{4}}+\underset{{k}\geqslant\mathrm{2}} {\sum}\:\left(\frac{\zeta\left({k}\right)}{{k}.\mathrm{4}^{{k}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)βˆ’\frac{\gamma}{\mathrm{4}}=\underset{{k}\geqslant\mathrm{2}} {\sum}\:\left(\frac{\zeta\:\left({k}\:\right)}{{k}\:.\:\mathrm{4}^{\:{k}} }\:\right) \\ $$
Commented by mindispower last updated on 15/May/21
sir are you student ?
$${sir}\:{are}\:{you}\:{student}\:? \\ $$
Answered by Dwaipayan Shikari last updated on 15/May/21
ψ(z+1)=βˆ’Ξ³+Ξ£_(n=2) ^∞ ΞΆ(n)z^(nβˆ’1) [log(Ξ“(z+1))]_0 ^(βˆ’(1/4)) =βˆ’βˆ«_0 ^((βˆ’1)/4) Ξ³+Ξ£_(n=2) ^∞ ∫_0 ^(βˆ’(1/4)) (βˆ’1)^n ΞΆ(n)z^(nβˆ’1) dz log(Ξ“((3/4)))=(Ξ³/4)+Ξ£((ΞΆ(n))/(n4^n )) Ξ£((ΞΆ(n))/(n4^n ))=log(Ξ“((3/4)))βˆ’(Ξ³/4)
$$\psi\left({z}+\mathrm{1}\right)=βˆ’\gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\zeta\left({n}\right){z}^{{n}βˆ’\mathrm{1}} \\ $$$$\left[{log}\left(\Gamma\left({z}+\mathrm{1}\right)\right)\right]_{\mathrm{0}} ^{βˆ’\frac{\mathrm{1}}{\mathrm{4}}} =βˆ’\int_{\mathrm{0}} ^{\frac{βˆ’\mathrm{1}}{\mathrm{4}}} \gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{βˆ’\frac{\mathrm{1}}{\mathrm{4}}} \left(βˆ’\mathrm{1}\right)^{{n}} \zeta\left({n}\right){z}^{{n}βˆ’\mathrm{1}} {dz} \\ $$$${log}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)=\frac{\gamma}{\mathrm{4}}+\Sigma\frac{\zeta\left({n}\right)}{{n}\mathrm{4}^{{n}} } \\ $$$$\Sigma\frac{\zeta\left({n}\right)}{{n}\mathrm{4}^{{n}} }={log}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)βˆ’\frac{\gamma}{\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 15/May/21
very very nice thank you mr payan... taylor expansion of digamma..
$$\:{very}\:{very}\:{nice} \\ $$$${thank}\:{you}\:{mr}\:{payan}… \\ $$$$\:{taylor}\:{expansion}\:{of}\:{digamma}.. \\ $$

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