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

$$$$$$provethat$$$$$$$${\psi }^{″}\left(\frac{1}{4}\right)=-2{\pi }^3-56\zeta \left(3\right)$$$$$$

Commented by aleks041103 last updated on 29/Dec/21

$${\psi }^{″}\left(\frac{1}{4}\right)={\psi }_2\left(\frac{1}{4}\right)={(-1)}^{2+1}2!\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(k+\frac{1}{4})}^{2+1}}=$$$$=-2\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(k+\frac{1}{4})}^3}=-128\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(4k+1)}^3}$$

Commented by amin96 last updated on 29/Dec/21

$$bravo$$

Answered by mindispower last updated on 30/Dec/21

$${\mathrm{\Psi }}^{″}\left(\frac{1}{4}\right)=-128\sum _{n⩾0}\frac{1}{{(4n+1)}^3}$$$${\mathrm{\Psi }}^{″}\left(x\right)-{\mathrm{\Psi }}^{″}(1-x)=-2{\pi }^3(1+{cot}^2\left(\pi x\right))cot\left(\pi x\right)$$$${\mathrm{\Psi }}^{″}\left(\frac{1}{4}\right)-{\mathrm{\Psi }}^{″}\left(\frac{3}{4}\right)=-4{\pi }^3$$$$\zeta \left(3\right)=\sum _{n⩾0}\frac{1}{{(4n+1)}^3}+{\left(\frac{1}{4n+3}\right)}^3+\frac{1}{64{(n+1)}^3}+\frac{1}{8{(2n+1)}^3}$$$$=-\frac{{\mathrm{\Psi }}^{″}\left(\frac{1}{4}\right)+{\mathrm{\Psi }}^{″}\left(\frac{3}{4}\right)}{64}+\frac{1}{8}(\frac{1}{8}\zeta \left(3\right)+\frac{7}{8}\zeta \left(3\right))$$$${\mathrm{\Psi }}^{″}\left(\frac{3}{4}\right)={\mathrm{\Psi }}^{″}\left(\frac{1}{4}\right)+4{\pi }^3$$$$\Rightarrow -\frac{2{\mathrm{\Psi }}^{″}\left(\frac{1}{4}\right)+4{\pi }^3}{64}+\frac{\zeta \left(3\right)}{8}=\zeta \left(3\right)$$$$\Rightarrow {\mathrm{\Psi }}^{″}\left(\frac{1}{4}\right)=-56\zeta \left(3\right)-2\pi$$$$$$

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