......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
|