advanced-calculus-n-1-n-n-I-havefound-pi-4-36-

Question Number 137592 by mnjuly1970 last updated on 04/Apr/21

\dots \dots a d v a n c e d \dots . . c a l c u l u s \dots .

Ω = \underset{n = 1}{\sum^{\infty}} \frac{ψ^{″} (n)}{n} = ? ? ?

I h a v e f o u n d :: Ω = - \frac{π^{4}}{36} \dots!

Answered by Dwaipayan Shikari last updated on 04/Apr/21

\underset{n = 1}{\sum^{\infty}} \frac{ψ^{″} (n)}{n}

= - \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} . \frac{{l o g}^{2} (x) x^{n - 1}}{1 - x}

= - \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{{l o g}^{2} (x) \frac{x^{n - 1}}{n}}{1 - x} d x

= \int_{0}^{1} \frac{{l o g}^{2} (x) l o g (1 - x)}{x (1 - x)} d x

= \int_{0}^{1} \frac{{l o g}^{2} (x) l o g (1 - x)}{x} + \int_{0}^{1} \frac{{l o g}^{2} (x) l o g (1 - x)}{1 - x} d x

= [{l o g}^{2} (x) \underset{n = 1}{\sum^{\infty}} \frac{- x^{n}}{n^{2}}] + 2 \int_{0}^{1} l o g (x) \underset{n = 1}{\sum^{\infty}} \frac{x^{n - 1}}{n^{2}} d x + \int_{0}^{1} \frac{{l o g}^{2} (1 - x) l o g (x)}{x} d x

= 2 l o g (x) \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n^{3}} - 2 \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{x^{n - 1}}{n^{3}} d x + ℵ

= - \frac{π^{4}}{45} + ℵ = - \frac{π^{4}}{36}

ℵ = - \frac{π^{4}}{180} (M a y b e d o n e p r e v i o u s l y)

Commented by mnjuly1970 last updated on 04/Apr/21

t h a n k y o u s o m u c h m r p a y a n \dots

y e s y o u a r e r i g h t

ϕ = \int_{0}^{1} \frac{l n (x) . {l n}^{2} (1 - x)}{x} d x = \frac{- π^{4}}{180}

⇓ ⇓ ⇓

Answered by mnjuly1970 last updated on 04/Apr/21

Answered by mnjuly1970 last updated on 04/Apr/21