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

Prove that: Ψ_1 ((1/8)) + Ψ_2 ((5/8)) = 32G + 4π^2 Ψ-trigamma function G-catalan constant

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\Psi_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right)\:+\:\Psi_{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)\:=\:\mathrm{32G}\:+\:\mathrm{4}\pi^{\mathrm{2}} \\ $$$$\Psi-\mathrm{trigamma}\:\mathrm{function} \\ $$$$\mathrm{G}-\mathrm{catalan}\:\mathrm{constant} \\ $$

Answered by mindispower last updated on 22/Oct/21

Ψ((p/q))=(π^2 /(2sin^2 (((pπ)/q))))+2qΣ_(k=1) ^([((q−1)/2)]) sin(((2πkp)/q))Cl_2 (((2πk)/q)) Cl_2 . Claus function Ψ_1 ((1/8))=(π^2 /(2sin^2 ((π/8))))+16Σ_(k=1) ^3 sin(((πk)/4))Cl_2 (((πk)/4)) =((2π^2 )/(2−(√2))) +16(sin((π/4))Cl_2 ((π/4))+Cl_2 ((π/2))+sin((π/4))cl_2 (((3π)/4))) Ψ_1 ((5/8))=((2π^2 )/( (√2)+2))+16(sin(((10π)/8))Cl_2 ((π/4))+sin(((20π)/8))Cl_2 ((π/2))+sin(((30π)/8))Cl_2 (((3π)/4))) =((2π^2 )/(2+(√2)))+16(−sin((π/4))Cl_2 ((π/4))+Cl_2 ((π/2))−sin((π/4))Cl_2 (((3π)/4))) ψ_1 ((1/8))+ψ_1 ((5/8)) =((2π^2 )/(2+(√2)))+((2π^2 )/(2−(√2)))+32Cl_2 ((π/2)) =4π^2 +32Cl_2 ((π/2)) Cl_2 (x)=Σ_(n≥1) ((sin(nx))/n^2 ),Cl_2 ((π/2))=Σ_(n≥1) ((sin(n(π/2)))/n^2 ) sin(nπ)=0,∀n∈Z⇒ Cl_2 ((π/2))=Σ_(n≥0) ((sin(nπ+(π/2)))/((2n+1)^2 ))=Σ_(n≥0) (((−1)^n )/((2n+1)))=β(2)=G ⇔ψ((1/8))+ψ((5/8))=4π^2 +32G

$$\Psi\left(\frac{{p}}{{q}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{p}\pi}{{q}}\right)}+\mathrm{2}{q}\underset{{k}=\mathrm{1}} {\overset{\left[\frac{{q}−\mathrm{1}}{\mathrm{2}}\right]} {\sum}}{sin}\left(\frac{\mathrm{2}\pi{kp}}{{q}}\right){Cl}_{\mathrm{2}} \left(\frac{\mathrm{2}\pi{k}}{{q}}\right) \\ $$$${Cl}_{\mathrm{2}} .\:{Claus}\:{function} \\ $$$$\Psi_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}+\mathrm{16}\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{sin}\left(\frac{\pi{k}}{\mathrm{4}}\right){Cl}_{\mathrm{2}} \left(\frac{\pi{k}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}−\sqrt{\mathrm{2}}}\:+\mathrm{16}\left({sin}\left(\frac{\pi}{\mathrm{4}}\right){Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)+{Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)+{sin}\left(\frac{\pi}{\mathrm{4}}\right){cl}_{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right) \\ $$$$\Psi_{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}+\mathrm{2}}+\mathrm{16}\left({sin}\left(\frac{\mathrm{10}\pi}{\mathrm{8}}\right){Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)+{sin}\left(\frac{\mathrm{20}\pi}{\mathrm{8}}\right){Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)+{sin}\left(\frac{\mathrm{30}\pi}{\mathrm{8}}\right){Cl}_{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right) \\ $$$$=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}+\sqrt{\mathrm{2}}}+\mathrm{16}\left(−{sin}\left(\frac{\pi}{\mathrm{4}}\right){Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)+{Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)−{sin}\left(\frac{\pi}{\mathrm{4}}\right){Cl}_{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right) \\ $$$$\psi_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right)+\psi_{\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}+\sqrt{\mathrm{2}}}+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}−\sqrt{\mathrm{2}}}+\mathrm{32}{Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{4}\pi^{\mathrm{2}} +\mathrm{32}{Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right) \\ $$$${Cl}_{\mathrm{2}} \left({x}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{2}} },{Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({n}\frac{\pi}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} } \\ $$$${sin}\left({n}\pi\right)=\mathrm{0},\forall{n}\in\mathbb{Z}\Rightarrow \\ $$$${Cl}_{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{sin}\left({n}\pi+\frac{\pi}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}=\beta\left(\mathrm{2}\right)={G} \\ $$$$\Leftrightarrow\psi\left(\frac{\mathrm{1}}{\mathrm{8}}\right)+\psi\left(\frac{\mathrm{5}}{\mathrm{8}}\right)=\mathrm{4}\pi^{\mathrm{2}} +\mathrm{32}{G} \\ $$$$ \\ $$

Commented by MathSh last updated on 22/Oct/21

Perfect dear Ser, thank you very much

$$\mathrm{Perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

Answered by qaz last updated on 22/Oct/21

ψ′((1/8))+ψ′((5/8)) =Σ_(n=0) ^∞ (1/((n+(1/8))^2 ))+(1/((n+(5/8))^2 )) =64Σ_(n=0) ^∞ (1/((8n+1)^2 ))+(1/((8n+5)^2 )) =−64Σ_(n=0) ^∞ ∫_0 ^1 (x^(8n) +x^(8n+4) )lnxdx =−64∫_0 ^1 (1+x^4 )lnxΣ_(n=0) ^∞ x^(8n) dx =−64∫_0 ^1 (((1+x^4 )lnx)/(1−x^8 ))dx =−32∫_0 ^1 ((1/(1−x^2 ))+(1/(1+x^2 )))lnxdx =−32Σ_(n=0) ^∞ ∫_0 ^1 (x^(2n) +(−1)^n x^(2n) )lnxdx =32Σ_(n=0) ^∞ (1/((2n+1)^2 ))+(((−1)^n )/((2n+1)^2 )) =32(1−2^(−2) )ζ(2)+32G =4π^2 +32G

$$\psi'\left(\frac{\mathrm{1}}{\mathrm{8}}\right)+\psi'\left(\frac{\mathrm{5}}{\mathrm{8}}\right) \\ $$$$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{64}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{8n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{8n}+\mathrm{5}\right)^{\mathrm{2}} } \\ $$$$=−\mathrm{64}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{8n}} +\mathrm{x}^{\mathrm{8n}+\mathrm{4}} \right)\mathrm{lnxdx} \\ $$$$=−\mathrm{64}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)\mathrm{lnx}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{x}^{\mathrm{8n}} \mathrm{dx} \\ $$$$=−\mathrm{64}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)\mathrm{lnx}}{\mathrm{1}−\mathrm{x}^{\mathrm{8}} }\mathrm{dx} \\ $$$$=−\mathrm{32}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\mathrm{lnxdx} \\ $$$$=−\mathrm{32}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{2n}} +\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{2n}} \right)\mathrm{lnxdx} \\ $$$$=\mathrm{32}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{32}\left(\mathrm{1}−\mathrm{2}^{−\mathrm{2}} \right)\zeta\left(\mathrm{2}\right)+\mathrm{32G} \\ $$$$=\mathrm{4}\pi^{\mathrm{2}} +\mathrm{32G} \\ $$

Commented by MathSh last updated on 22/Oct/21

Perfect dear Ser, thank you so much

$$\mathrm{Perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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