What-s-the-relationship-between-Dirichlet-s-function-with-s-function-That-is-n-0-1-n-2n-1-s-with-n-1-1-n-s-

Question Number 140554 by qaz last updated on 09/May/21

{W h a t}^{'} s t h e r e l a t i o n s h i p b e t w e e n D i r i c h l e t β (s) f u n c t i o n w i t h

ζ (s) f u n c t i o n ? T h a t i s \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{s}} w i t h \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{s}} .

Commented by qaz last updated on 09/May/21

H o w t o e v a l u a t e β {(s)}^{'} s s p e c i a l v a l u e ?

Commented by qaz last updated on 09/May/21

I g o t \int_{0}^{π / 2} {(l n \tan x)}^{2} d x

= \int_{0}^{\infty} \frac{{(l n y)}^{2}}{1 + y^{2}} d y

= \int_{0}^{1} \frac{{(l n y)}^{2}}{1 + y^{2}} d y + \int_{1}^{\infty} \frac{{(l n y)}^{2}}{1 + y^{2}} d y

= 2 \int_{0}^{1} \frac{{(l n y)}^{2}}{1 + y^{2}} d y

= 2 \int_{0}^{\infty} \frac{z^{2} e^{- z}}{1 + e^{- 2 z}} d z

= 2 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{\infty} z^{2} e^{- (2 n + 1) z} d z

= 4 \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} = ? ? ?

Commented by Dwaipayan Shikari last updated on 09/May/21

\frac{π}{2} t a n (\frac{π}{2} x) = \frac{1}{1 - x} - \frac{1}{1 + x} + \frac{1}{3 - x} - \frac{1}{3 + x} + \dots

D i f f e r e n t i a t i n g b o t h s i d e s r e s p e c t t o x

\frac{π^{2}}{4} {s e c}^{2} (\frac{π x}{2}) = \frac{1}{{(1 - x)}^{2}} + \frac{1}{{(1 + x)}^{2}} + \frac{1}{{(3 - x)}^{2}} + \frac{1}{{(3 + x)}^{2}} + \dots

A g a i n

\frac{π^{3}}{8} {s e c}^{2} (\frac{π}{2} x) t a n (\frac{π}{2} x) = \frac{1}{{(1 - x)}^{3}} - \frac{1}{{(1 + x)}^{3}} + \frac{1}{{(3 - x)}^{3}} - \frac{1}{{(3 + x)}^{3}} + . .

\Rightarrow \frac{π^{3}}{8} (2) = \frac{1}{{(1 - \frac{1}{2})}^{3}} - \frac{1}{{(1 + \frac{1}{2})}^{3}} + \frac{1}{{(3 - \frac{1}{2})}^{3}} - \frac{1}{{(3 + \frac{1}{2})}^{3}} + \dots

\Rightarrow \frac{π^{3}}{32} = \frac{1}{1^{3}} - \frac{1}{3^{3}} + \frac{1}{5^{3}} - \frac{1}{7^{3}} + \dots

Commented by qaz last updated on 09/May/21

t h a n k s a l o t .

Commented by Ar Brandon last updated on 16/May/21

f (α) = \int_{0}^{\frac{π}{2}} \tan^{α} θ d θ = \frac{Γ (\frac{α + 1}{2}) Γ (\frac{1 - α}{2})}{2}, f (0) = \frac{π}{2}

lnf (α) = \ln Γ (\frac{α + 1}{2}) + \ln Γ (\frac{1 - α}{2}) + \ln (\frac{1}{2})

\frac{f^{'} (α)}{f (α)} = \frac{1}{2} (ψ (\frac{α + 1}{2}) - ψ (\frac{1 - α}{2}))

f^{'} (0) = \frac{π}{4} (ψ (\frac{1}{2}) - ψ (\frac{1}{2})) = 0

f^{″} (α) = \frac{f (α)}{4} (ψ^{'} (\frac{α + 1}{2}) + ψ^{'} (\frac{1 - α}{2})) + \frac{f^{'} (α)}{2} (ψ (\frac{α + 1}{2}) + ψ (\frac{1 - α}{2}))

f^{″} (0) = \frac{π}{8} (\frac{π^{2}}{\sin^{2} (\frac{π}{2})}) + 0 = \frac{π^{3}}{8} = \int_{0}^{\frac{π}{2}} \ln^{2} (\tan θ) d θ