Question Number 140554 by qaz last updated on 09/May/21
$${What}'{s}\:{the}\:{relationship}\:{between}\:{Dirichlet}\:\beta\left({s}\right)\:{function}\:{with} \\ $$$$\zeta\left({s}\right)\:{function}\:?\:{That}\:{is}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{s}} }\:\:{with}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }. \\ $$
Commented by qaz last updated on 09/May/21
$${How}\:{to}\:{evaluate}\:\beta\left({s}\right)'{s}\:{special}\:{value}? \\ $$
Commented by qaz last updated on 09/May/21
$${I}\:{got}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({ln}\mathrm{tan}\:{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy}+\int_{\mathrm{1}} ^{\infty} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{z}^{\mathrm{2}} {e}^{−{z}} }{\mathrm{1}+{e}^{−\mathrm{2}{z}} }{dz} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {z}^{\mathrm{2}} {e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){z}} {dz} \\ $$$$=\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=??? \\ $$
Commented by Dwaipayan Shikari last updated on 09/May/21
$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+… \\ $$$${Differentiating}\:{both}\:{sides}\:{respect}\:{to}\:{x} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \left(\frac{\pi{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{2}} }+… \\ $$$${Again} \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right){tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{3}} }+.. \\ $$$$\Rightarrow\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }+… \\ $$$$\Rightarrow\frac{\pi^{\mathrm{3}} }{\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }+… \\ $$$$ \\ $$
Commented by qaz last updated on 09/May/21
$${thanks}\:{a}\:{lot}. \\ $$
Commented by Ar Brandon last updated on 16/May/21
$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{\alpha} \theta\mathrm{d}\theta=\frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)}{\mathrm{2}},\:\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{lnf}\left(\alpha\right)=\mathrm{ln}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{ln}\Gamma\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)+\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=\mathrm{0} \\ $$$$\mathrm{f}\:''\left(\alpha\right)=\frac{\mathrm{f}\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)+\psi'\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{2}}\left(\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)+\psi\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:''\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)}\right)+\mathrm{0}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{tan}\theta\right)\mathrm{d}\theta \\ $$