......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


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Question Number 137177 by mnjuly1970 last updated on 30/Mar/21
$......advanced .... calculus.... Φ=Σ_(k=1) ^∞ ((ψ′(k))/k) =Σ_(n=1) ^∞ (a/n^b ) a , b =?? (adapted from brilliant) ................ ψ(k)=^(??) −γ+∫_0 ^( 1) (((1−t^(k−1) )/(1−t)))dt ∴ ψ′(k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/(1−t))dt Σ_(k=1) ^∞ ((ψ′(k))/k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/((1−t)k))dt =∫_0 ^( 1) ((ln(t))/(t(1−t)))(−(t^k /k))dt=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗 𝛗=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗_1 +𝛗_2 where ... ={∫_0 ^( 1) ((ln(t).ln(1−t))/(1−t))dt=𝛗_1 }+{∫_0 ^( 1) ((ln(1−t).ln(t))/t)dt=𝛗_2 } 𝛗_1 =[−(1/2)ln(t)ln^2 (1−t)]_0 ^1 +(1/2)∫_0 ^( 1) ((ln^2 (1−t))/t)dt ∴ 𝛗_1 = (1/2)(2ζ(3))=ζ(3)=^(easy) 𝛗_2 note : ∫_0 ^( 1) ((ln^2 (1−t))/t)dt=2ζ(3) (derived earlier) 𝛗=𝛗_1 +𝛗_2 =2ζ(3)=Σ_(n=1) ^∞ (2/n^3 ) .... 𝚽= Σ_(n=1) ^∞ (a/n^b ) ......⇒a=2 , b=3$
$. . . . . . a d v a n c e d . . . . c a l c u l u s . . . .$ $Φ = \underset{k = 1}{\sum^{\infty}} \frac{ψ^{'} (k)}{k} = \underset{n = 1}{\sum^{\infty}} \frac{a}{n^{b}}$ $a, b = ? ? (a d a p t e d f r o m b r i l l i a n t)$ $. . . . . . . . . . . . . . . .$ $ψ (k) \overset{? ?}{=} - γ + \int_{0}^{1} (\frac{1 - t^{k - 1}}{1 - t}) d t$ $∴ ψ^{'} (k) = \int_{0}^{1} \frac{- t^{k - 1} l n (t)}{1 - t} d t$ $\underset{k = 1}{\sum^{\infty}} \frac{ψ^{'} (k)}{k} = \int_{0}^{1} \frac{- t^{k - 1} l n (t)}{(1 - t) k} d t$ $= \int_{0}^{1} \frac{l n (t)}{t (1 - t)} (- \frac{t^{k}}{k}) d t = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t (1 - t)} d t = ϕ$ $ϕ = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t (1 - t)} d t = ϕ_{1} + ϕ_{2} w h e r e . . .$ $= {\int_{0}^{1} \frac{l n (t) . l n (1 - t)}{1 - t} d t = ϕ_{1}} + {\int_{0}^{1} \frac{l n (1 - t) . l n (t)}{t} d t = ϕ_{2}}$ $ϕ_{1} = {[- \frac{1}{2} l n (t) {l n}^{2} (1 - t)]}_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t$ $∴ ϕ_{1} = \frac{1}{2} (2 ζ (3)) = ζ (3) \overset{e a s y}{=} ϕ_{2}$ $n o t e : \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t = 2 ζ (3) (d e r i v e d e a r l i e r)$ $ϕ = ϕ_{1} + ϕ_{2} = 2 ζ (3) = \underset{n = 1}{\sum^{\infty}} \frac{2}{n^{3}} . . . .$ $Φ = \underset{n = 1}{\sum^{\infty}} \frac{a}{n^{b}} . . . . . . \Rightarrow a = 2, b = 3$
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