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Question Number 156086 by MathSh last updated on 07/Oct/21

$${\psi }^{\left(1\right)}\left(\frac{1}{6}\right)-{\psi }^{\left(1\right)}\left(\frac{5}{6}\right)=10{\psi }^{\left(1\right)}\left(\frac{1}{3}\right)-\frac{20}{3}{\pi }^2$$$$$$

Commented by puissant last updated on 09/Oct/21

$${\psi }^{\left(1\right)}\left(\frac{1}{6}\right)-{\psi }^{\left(1\right)}\left(\frac{5}{6}\right)={\pi }^2(1+{cotan}^2\left(\frac{\pi }{6}\right))$$$$={\pi }^2+{\pi }^2{cotan}^2\left(\frac{\pi }{6}\right)$$$${tan}^2\left(\frac{\pi }{6}\right)=\frac{{sin}^2\frac{\pi }{6}}{{cos}^2\frac{\pi }{6}}=\frac{1}{4}\times \frac{4}{3}=\frac{1}{3}$$$$\Rightarrow {\pi }^2+{\pi }^2{cotan}^2\frac{\pi }{6}={\pi }^2(1+\frac{1}{3})=\frac{4{\pi }^2}{3}...$$

Commented by tabata last updated on 08/Oct/21

$$sirwhatsthemean\psi andwhaystheformolla?$$

Commented by MathSh last updated on 08/Oct/21

$$\mathrm{Thankyou}\mathrm{ser},\mathrm{first}\mathrm{formula}\mathrm{is}\mathrm{wrong}$$

Commented by puissant last updated on 12/Oct/21

$$Mrtabatawehave:$$$${\psi }^{\left(1\right)}\left(x\right)-{\psi }^{\left(1\right)}(1-x)={\pi }^2(1+{cotan}^2\left(\pi x\right))$$

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