Prove-that-1-1-8-2-5-8-32G-4pi-2-trigamma-function-G-catalan-constant-

Question Number 157323 by MathSh last updated on 22/Oct/21

Prove that :

Ψ_{1} (\frac{1}{8}) + Ψ_{2} (\frac{5}{8}) = 32 G + 4 π^{2}

Ψ - trigamma function

G - catalan constant

Answered by mindispower last updated on 22/Oct/21

Ψ (\frac{p}{q}) = \frac{π^{2}}{2 {s i n}^{2} (\frac{p π}{q})} + 2 q \underset{k = 1}{\sum^{[\frac{q - 1}{2}]}} s i n (\frac{2 π k p}{q}) {C l}_{2} (\frac{2 π k}{q})

{C l}_{2} . C l a u s f u n c t i o n

Ψ_{1} (\frac{1}{8}) = \frac{π^{2}}{2 {s i n}^{2} (\frac{π}{8})} + 16 \underset{k = 1}{\sum^{3}} s i n (\frac{π k}{4}) {C l}_{2} (\frac{π k}{4})

= \frac{2 π^{2}}{2 - \sqrt{2}} + 16 (s i n (\frac{π}{4}) {C l}_{2} (\frac{π}{4}) + {C l}_{2} (\frac{π}{2}) + s i n (\frac{π}{4}) {c l}_{2} (\frac{3 π}{4}))

Ψ_{1} (\frac{5}{8}) = \frac{2 π^{2}}{\sqrt{2} + 2} + 16 (s i n (\frac{10 π}{8}) {C l}_{2} (\frac{π}{4}) + s i n (\frac{20 π}{8}) {C l}_{2} (\frac{π}{2}) + s i n (\frac{30 π}{8}) {C l}_{2} (\frac{3 π}{4}))

= \frac{2 π^{2}}{2 + \sqrt{2}} + 16 (- s i n (\frac{π}{4}) {C l}_{2} (\frac{π}{4}) + {C l}_{2} (\frac{π}{2}) - s i n (\frac{π}{4}) {C l}_{2} (\frac{3 π}{4}))

ψ_{1} (\frac{1}{8}) + ψ_{1} (\frac{5}{8})

= \frac{2 π^{2}}{2 + \sqrt{2}} + \frac{2 π^{2}}{2 - \sqrt{2}} + 32 {C l}_{2} (\frac{π}{2})

= 4 π^{2} + 32 {C l}_{2} (\frac{π}{2})

{C l}_{2} (x) = \sum_{n ⩾ 1} \frac{s i n (n x)}{n^{2}}, {C l}_{2} (\frac{π}{2}) = \sum_{n ⩾ 1} \frac{s i n (n \frac{π}{2})}{n^{2}}

s i n (n π) = 0, \forall n \in Z \Rightarrow

{C l}_{2} (\frac{π}{2}) = \sum_{n ⩾ 0} \frac{s i n (n π + \frac{π}{2})}{{(2 n + 1)}^{2}} = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{(2 n + 1)} = β (2) = G

\Leftrightarrow ψ (\frac{1}{8}) + ψ (\frac{5}{8}) = 4 π^{2} + 32 G

Commented by MathSh last updated on 22/Oct/21

Perfect dear Ser, thank you very much

Answered by qaz last updated on 22/Oct/21

ψ^{'} (\frac{1}{8}) + ψ^{'} (\frac{5}{8})

= \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{8})}^{2}} + \frac{1}{{(n + \frac{5}{8})}^{2}}

= 64 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(8 n + 1)}^{2}} + \frac{1}{{(8 n + 5)}^{2}}

= - 64 \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} (x^{8 n} + x^{8 n + 4}) lnxdx

= - 64 \int_{0}^{1} (1 + x^{4}) lnx \underset{n = 0}{\sum^{\infty}} x^{8 n} dx

= - 64 \int_{0}^{1} \frac{(1 + x^{4}) lnx}{1 - x^{8}} dx

= - 32 \int_{0}^{1} (\frac{1}{1 - x^{2}} + \frac{1}{1 + x^{2}}) lnxdx

= - 32 \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} (x^{2 n} + {(- 1)}^{n} x^{2 n}) lnxdx

= 32 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} + \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}

= 32 (1 - 2^{- 2}) ζ (2) + 32 G

= 4 π^{2} + 32 G

Commented by MathSh last updated on 22/Oct/21

Perfect dear Ser, thank you so much

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