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Question Number 140554 by qaz last updated on 09/May/21

$${What}^{\prime }stherelationshipbetweenDirichlet\beta \left(s\right)functionwith$$$$\zeta \left(s\right)function?Thatis\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^s}with\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}.$$

Commented by qaz last updated on 09/May/21

$$Howtoevaluate\beta {\left(s\right)}^{\prime }sspecialvalue?$$

Commented by qaz last updated on 09/May/21

$$Igot{\int }_0^{\pi /2}{\left(ln\tan x\right)}^{2}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{{\left(lny\right)}^2}{1+y^2}dy$$$$={\int }_0^1\frac{{\left(lny\right)}^2}{1+y^2}dy+{\int }_1^{\mathrm{\infty }}\frac{{\left(lny\right)}^2}{1+y^2}dy$$$$=2{\int }_0^1\frac{{\left(lny\right)}^2}{1+y^2}dy$$$$=2{\int }_0^{\mathrm{\infty }}\frac{z^2{e}^{-z}}{1+e^{-2z}}dz$$$$=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^{\mathrm{\infty }}{z}^2{e}^{-(2n+1)z}dz$$$$=4\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^3}=???$$

Commented by Dwaipayan Shikari last updated on 09/May/21

$$\frac{\pi }{2}tan\left(\frac{\pi }{2}x\right)=\frac{1}{1-x}-\frac{1}{1+x}+\frac{1}{3-x}-\frac{1}{3+x}+...$$$$Differentiatingbothsidesrespecttox$$$$\frac{{\pi }^2}{4}{sec}^2\left(\frac{\pi x}{2}\right)=\frac{1}{{(1-x)}^2}+\frac{1}{{(1+x)}^2}+\frac{1}{{(3-x)}^2}+\frac{1}{{(3+x)}^2}+...$$$$Again$$$$\frac{{\pi }^3}{8}{sec}^2\left(\frac{\pi }{2}x\right)tan\left(\frac{\pi }{2}x\right)=\frac{1}{{(1-x)}^3}-\frac{1}{{(1+x)}^3}+\frac{1}{{(3-x)}^3}-\frac{1}{{(3+x)}^3}+..$$$$\Rightarrow \frac{{\pi }^3}{8}\left(2\right)=\frac{1}{{(1-\frac{1}{2})}^3}-\frac{1}{{(1+\frac{1}{2})}^3}+\frac{1}{{(3-\frac{1}{2})}^3}-\frac{1}{{(3+\frac{1}{2})}^3}+...$$$$\Rightarrow \frac{{\pi }^3}{32}=\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+...$$$$$$

Commented by qaz last updated on 09/May/21

$$thanksalot.$$

Commented by Ar Brandon last updated on 16/May/21

$$\mathrm{f}\left(\alpha \right)={\int }_0^{\frac{\pi }{2}}{\tan }^{\alpha }\theta \mathrm{d}\theta =\frac{\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)\mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)}{2},\mathrm{f}\left(0\right)=\frac{\pi }{2}$$$$\mathrm{lnf}\left(\alpha \right)=\ln \mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)+\ln \mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)+\ln \left(\frac{1}{2}\right)$$$$\frac{\mathrm{f}{}^{\prime }\left(\alpha \right)}{\mathrm{f}\left(\alpha \right)}=\frac{1}{2}(\psi \left(\frac{\alpha +1}{2}\right)-\psi \left(\frac{1-\alpha }{2}\right))$$$$\mathrm{f}{}^{\prime }\left(0\right)=\frac{\pi }{4}(\psi \left(\frac{1}{2}\right)-\psi \left(\frac{1}{2}\right))=0$$$$\mathrm{f}{}^{″}\left(\alpha \right)=\frac{\mathrm{f}\left(\alpha \right)}{4}({\psi }^{\prime }\left(\frac{\alpha +1}{2}\right)+{\psi }^{\prime }\left(\frac{1-\alpha }{2}\right))+\frac{\mathrm{f}{}^{\prime }\left(\alpha \right)}{2}(\psi \left(\frac{\alpha +1}{2}\right)+\psi \left(\frac{1-\alpha }{2}\right))$$$$\mathrm{f}{}^{″}\left(0\right)=\frac{\pi }{8}\left(\frac{{\pi }^2}{{\sin }^2\left(\frac{\pi }{2}\right)}\right)+0=\frac{{\pi }^3}{8}={\int }_0^{\frac{\pi }{2}}{\ln }^2\left(\tan \theta \right)\mathrm{d}\theta$$

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