prove-that-1-4-2pi-3-56-3-

Question Number 162377 by mnjuly1970 last updated on 29/Dec/21

p r o v e t h a t

ψ^{″} (\frac{1}{4}) = - 2 π^{3} - 56 ζ (3)

Commented by aleks041103 last updated on 29/Dec/21

ψ^{″} (\frac{1}{4}) = ψ_{2} (\frac{1}{4}) = {(- 1)}^{2 + 1} 2! \underset{k = 0}{\sum^{\infty}} \frac{1}{{(k + \frac{1}{4})}^{2 + 1}} =

= - 2 \underset{k = 0}{\sum^{\infty}} \frac{1}{{(k + \frac{1}{4})}^{3}} = - 128 \underset{k = 0}{\sum^{\infty}} \frac{1}{{(4 k + 1)}^{3}}

Commented by amin96 last updated on 29/Dec/21

b r a v o

Answered by mindispower last updated on 30/Dec/21

Ψ^{″} (\frac{1}{4}) = - 128 \sum_{n ⩾ 0} \frac{1}{{(4 n + 1)}^{3}}

Ψ^{″} (x) - Ψ^{″} (1 - x) = - 2 π^{3} (1 + {c o t}^{2} (π x)) c o t (π x)

Ψ^{″} (\frac{1}{4}) - Ψ^{″} (\frac{3}{4}) = - 4 π^{3}

ζ (3) = \sum_{n ⩾ 0} \frac{1}{{(4 n + 1)}^{3}} + {(\frac{1}{4 n + 3})}^{3} + \frac{1}{64 {(n + 1)}^{3}} + \frac{1}{8 {(2 n + 1)}^{3}}

= - \frac{Ψ^{″} (\frac{1}{4}) + Ψ^{″} (\frac{3}{4})}{64} + \frac{1}{8} (\frac{1}{8} ζ (3) + \frac{7}{8} ζ (3))

Ψ^{″} (\frac{3}{4}) = Ψ^{″} (\frac{1}{4}) + 4 π^{3}

\Rightarrow - \frac{2 Ψ^{″} (\frac{1}{4}) + 4 π^{3}}{64} + \frac{ζ (3)}{8} = ζ (3)

\Rightarrow Ψ^{″} (\frac{1}{4}) = - 56 ζ (3) - 2 π