Question Number 156086 by MathSh last updated on 07/Oct/21
$$\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)\:-\:\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)\:=\:\mathrm{10}\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:-\:\frac{\mathrm{20}}{\mathrm{3}}\pi^{\mathrm{2}} \\ $$$$ \\ $$
Commented by puissant last updated on 09/Oct/21
$$\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)−\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)=\pi^{\mathrm{2}} \left(\mathrm{1}+{cotan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\pi^{\mathrm{2}} +\pi^{\mathrm{2}} {cotan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right) \\ $$$${tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)=\frac{{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}}{{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}}=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{4}}{\:\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:\pi^{\mathrm{2}} +\pi^{\mathrm{2}} {cotan}^{\mathrm{2}} \frac{\pi}{\mathrm{6}}=\pi^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{3}}… \\ $$
Commented by tabata last updated on 08/Oct/21
$${sir}\:{whats}\:{the}\:{mean}\:\psi\:{and}\:{whays}\:{the}\:{formolla}\:? \\ $$
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
$${Mr}\:{tabata}\:{we}\:{have}\:: \\ $$$$\:\psi^{\left(\mathrm{1}\right)} \left({x}\right)−\psi^{\left(\mathrm{1}\right)} \left(\mathrm{1}−{x}\right)=\pi^{\mathrm{2}} \left(\mathrm{1}+{cotan}^{\mathrm{2}} \left(\pi{x}\right)\right) \\ $$