nice-calculus-prove-that-n-1-2n-n-1-2n-1-1-2-m-n-

Question Number 134662 by mnjuly1970 last updated on 06/Mar/21

\dots . n i c e c a l c u l u s \dots

p r o v e t h a t ::

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

\dots m . n \dots

Answered by Dwaipayan Shikari last updated on 06/Mar/21

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

= \int_{0}^{1} 2 \underset{n = 1}{\sum^{\infty}} \underset{k = 1}{\sum^{\infty}} \frac{x^{2 n}}{k^{2 n}} - \underset{n = 1}{\sum^{\infty}} \underset{k = 1}{\sum^{\infty}} \frac{x^{n}}{k^{2 n}}

= \int_{0}^{1} \underset{k = 1}{\sum^{\infty}} \frac{2 x^{2}}{k^{2} - x^{2}} - \frac{x}{k^{2} - x} d x

= \int_{0}^{1} x \underset{k = 1}{\sum^{\infty}} \frac{1}{k - x} - \frac{1}{k + x} - \frac{1}{2} \sqrt{x} \underset{k = 1}{\sum^{\infty}} \frac{1}{k - \sqrt{x}} - \frac{1}{k + \sqrt{x}} d x

= \int_{0}^{1} x ψ (1 + x) - x ψ (1 - x) - \frac{1}{2} \sqrt{x} ψ (1 + \sqrt{x}) + \sqrt{x} ψ (1 - \sqrt{x}) d x

= 1 - \int_{0}^{1} x π c o t (π x) d x - \frac{1}{2} + \frac{1}{2} \int_{0}^{1} \sqrt{x} π c o t (π \sqrt{x}) d x

= \frac{1}{2} - \int_{0}^{1} x π c o t (π x) d x + \int_{0}^{1} t π c o t (π t) d t = \frac{1}{2}

Commented by mnjuly1970 last updated on 06/Mar/21

t a y e b b a l l a h \dots s i r p a y a n \dots

y o u r s o l u t i o n i s v e r y n i c e

a n d b e t t e r t h a n t h a t m e . .

m e r c e y \dots

Commented by mnjuly1970 last updated on 06/Mar/21

\frac{1}{2} \int_{0}^{1} \sqrt{x} π c o t (π (\sqrt{x})) d x \overset{W H Y ? ?}{=} \int_{0}^{1} t π c o t (π t) d t

p l e a s e e x p l a i n \dots .

Answered by mnjuly1970 last updated on 07/Mar/21

Answered by mnjuly1970 last updated on 07/Mar/21