nice-mathematics-ln-2-n-1-2n-1-1-n-1-m-n-

Question Number 149339 by mnjuly1970 last updated on 04/Aug/21

\dots n i c e \dots \dots m a t h e m a t i c s \dots

l n (2) - \underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n + 1) - 1}{n + 1} = ?

\dots . m . n \dots .

Answered by Kamel last updated on 04/Aug/21

W e h a v e : Ψ (1 + x) = - γ - \underset{n = 1}{\sum^{+ \infty}} {(- x)}^{n} ζ (n + 1)

= - γ - \underset{n = 1}{\sum^{+ \infty}} x^{2 n} ζ (2 n + 1) + \underset{n = 0}{\sum^{+ \infty}} x^{2 n + 1} ζ (2 n + 2)

= - γ - \underset{n = 1}{\sum^{+ \infty}} x^{2 n} ζ (2 n + 1) + \frac{1}{x} \underset{n = 1}{\sum^{+ \infty}} x^{2 n} ζ (2 n) \dots (1)

A n d : \underset{n = 0}{\sum^{+ \infty}} ζ (2 n) x^{2 n} = - \frac{π x}{2} c o t (π x) = - \frac{1}{2} + \underset{n = 1}{\sum^{+ \infty}} ζ (2 n) x^{2 n} \dots (2)

(1) \land (2) \Rightarrow \underset{n = 1}{\sum^{+ \infty}} x^{2 n + 1} ζ (2 n + 1) = - x Ψ (1 + x) - γ x + \frac{1}{2} - \frac{π x}{2} c o t (π x)

∴ \underset{n = 1}{\sum^{+ \infty}} \frac{ζ (2 n + 1) - 1}{n + 1} = - 2 \int_{0}^{1} x Ψ (1 + x) d x - γ + 2 - \int_{0}^{1} (π x c o t (π x) + \frac{2 x}{1 - x^{2}}) d x

= 2 \int_{0}^{1} L n (Γ (1 + x)) d x - γ + 2 - \underset{x \to 1^{-}}{l i m} (x L n (s i n (π x)) - L n (1 - x^{2})) + \int_{0}^{1} L n (s i n (π x)) d x

= L n (2) - L n (π) + \int_{0}^{1} L n (\frac{π}{s i n (π x)}) d x - γ + \int_{0}^{1} L n (s i n (π x)) d x

= L n (2) - γ

S o : \underset{n = 1}{\sum^{+ \infty}} \frac{ζ (2 n + 1) - 1}{n + 1} = L n (2) - γ

T h e n L n (2) - \underset{n = 1}{\sum^{+ \infty}} \frac{ζ (2 n + 1) - 1}{n + 1} = γ

W h e r e : ζ d e n o t e z e t a - R e i m a n n f u n c t i o n

γ i s t h e E u l e r - M a s c h e r o n i c o n s t a n t .

Ψ i s t h e d i g a m m a f u n c t i o n a n d Γ t h e g a m m a f u n c t i o n .

B E N A I C H A K A M E L

Commented by mnjuly1970 last updated on 04/Aug/21

v e r y n i c e m a s t e r, M r . K a m e l . .

Commented by Kamel last updated on 04/Aug/21

T h a n k y o u .

Commented by Tawa11 last updated on 04/Aug/21

Weldone sir .

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