Question Number 137177 by mnjuly1970 last updated on 30/Mar/21
![......advanced .... calculus.... Φ=Σ_(k=1) ^∞ ((ψ′(k))/k) =Σ_(n=1) ^∞ (a/n^b ) a , b =?? (adapted from brilliant) ................ ψ(k)=^(??) −γ+∫_0 ^( 1) (((1−t^(k−1) )/(1−t)))dt ∴ ψ′(k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/(1−t))dt Σ_(k=1) ^∞ ((ψ′(k))/k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/((1−t)k))dt =∫_0 ^( 1) ((ln(t))/(t(1−t)))(−(t^k /k))dt=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗 𝛗=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗_1 +𝛗_2 where ... ={∫_0 ^( 1) ((ln(t).ln(1−t))/(1−t))dt=𝛗_1 }+{∫_0 ^( 1) ((ln(1−t).ln(t))/t)dt=𝛗_2 } 𝛗_1 =[−(1/2)ln(t)ln^2 (1−t)]_0 ^1 +(1/2)∫_0 ^( 1) ((ln^2 (1−t))/t)dt ∴ 𝛗_1 = (1/2)(2ζ(3))=ζ(3)=^(easy) 𝛗_2 note : ∫_0 ^( 1) ((ln^2 (1−t))/t)dt=2ζ(3) (derived earlier) 𝛗=𝛗_1 +𝛗_2 =2ζ(3)=Σ_(n=1) ^∞ (2/n^3 ) .... 𝚽= Σ_(n=1) ^∞ (a/n^b ) ......⇒a=2 , b=3](Q137177.png)
$$\:\:\:\:\:\:\:\:\:\:\:......{advanced}\:\:\:\:\:....\:\:\:\:\:{calculus}.... \\ $$$$\:\:\:\:\:\:\:\Phi=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi'\left({k}\right)}{{k}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}}{{n}^{{b}} } \\ $$$$\:\:\:\:\:\:{a}\:,\:{b}\:=??\:\left({adapted}\:{from}\:{brilliant}\right) \\ $$$$\:\:\:\:\:\:\:................ \\ $$$$\:\:\:\:\:\:\:\psi\left({k}\right)\overset{??} {=}−\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}−{t}^{{k}−\mathrm{1}} }{\mathrm{1}−{t}}\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\psi'\left({k}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{t}^{{k}−\mathrm{1}} {ln}\left({t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi'\left({k}\right)}{{k}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{t}^{{k}−\mathrm{1}} {ln}\left({t}\right)}{\left(\mathrm{1}−{t}\right){k}}{dt} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right)}{{t}\left(\mathrm{1}−{t}\right)}\left(−\frac{{t}^{{k}} }{{k}}\right){dt}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right)}{{t}\left(\mathrm{1}−{t}\right)}{dt}=\boldsymbol{\phi} \\ $$$$\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right)}{{t}\left(\mathrm{1}−{t}\right)}{dt}=\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} \:\:{where}\:... \\ $$$$\:\:\:\:\:\:\:=\left\{\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right).{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}−{t}}{dt}=\boldsymbol{\phi}_{\mathrm{1}} \right\}+\left\{\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right).{ln}\left({t}\right)}{{t}}{dt}=\boldsymbol{\phi}_{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}_{\mathrm{1}} =\left[−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$$\:\:\:\:\:\therefore\:\:\boldsymbol{\phi}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\zeta\left(\mathrm{3}\right)\right)=\zeta\left(\mathrm{3}\right)\overset{{easy}} {=}\boldsymbol{\phi}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{note}\::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}=\mathrm{2}\zeta\left(\mathrm{3}\right)\:\left({derived}\:{earlier}\right) \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} =\mathrm{2}\zeta\left(\mathrm{3}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Phi}=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}}{{n}^{{b}} }\:\:......\Rightarrow{a}=\mathrm{2}\:\:,\:{b}=\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$