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

What′s the relationship between Dirichlet β(s) function with ζ(s) function ? That is Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^s )) with Σ_(n=1) ^∞ (1/n^s ).

$${What}'{s}\:{the}\:{relationship}\:{between}\:{Dirichlet}\:\beta\left({s}\right)\:{function}\:{with} \\ $$$$\zeta\left({s}\right)\:{function}\:?\:{That}\:{is}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{s}} }\:\:{with}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }. \\ $$

Commented by qaz last updated on 09/May/21

How to evaluate β(s)′s special value?

$${How}\:{to}\:{evaluate}\:\beta\left({s}\right)'{s}\:{special}\:{value}? \\ $$

Commented by qaz last updated on 09/May/21

I got ∫_0 ^(π/2) (lntan x)^2 dx =∫_0 ^∞ (((lny)^2 )/(1+y^2 ))dy =∫_0 ^1 (((lny)^2 )/(1+y^2 ))dy+∫_1 ^∞ (((lny)^2 )/(1+y^2 ))dy =2∫_0 ^1 (((lny)^2 )/(1+y^2 ))dy =2∫_0 ^∞ ((z^2 e^(−z) )/(1+e^(−2z) ))dz =2Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ z^2 e^(−(2n+1)z) dz =4Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 ))=???

$${I}\:{got}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({ln}\mathrm{tan}\:{x}\right)^{\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy}+\int_{\mathrm{1}} ^{\infty} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({lny}\right)^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{z}^{\mathrm{2}} {e}^{−{z}} }{\mathrm{1}+{e}^{−\mathrm{2}{z}} }{dz} \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {z}^{\mathrm{2}} {e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){z}} {dz} \\ $$$$=\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=??? \\ $$

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

(π/2)tan((π/2)x)=(1/(1−x))−(1/(1+x))+(1/(3−x))−(1/(3+x))+... Differentiating both sides respect to x (π^2 /4)sec^2 (((πx)/2))=(1/((1−x)^2 ))+(1/((1+x)^2 ))+(1/((3−x)^2 ))+(1/((3+x)^2 ))+... Again (π^3 /8)sec^2 ((π/2)x)tan((π/2)x)=(1/((1−x)^3 ))−(1/((1+x)^3 ))+(1/((3−x)^3 ))−(1/((3+x)^3 ))+.. ⇒(π^3 /8)(2)=(1/((1−(1/2))^3 ))−(1/((1+(1/2))^3 ))+(1/((3−(1/2))^3 ))−(1/((3+(1/2))^3 ))+... ⇒(π^3 /(32))=(1/1^3 )−(1/3^3 )+(1/5^3 )−(1/7^3 )+...

$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+... \\ $$$${Differentiating}\:{both}\:{sides}\:{respect}\:{to}\:{x} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \left(\frac{\pi{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{2}} }+... \\ $$$${Again} \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right){tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{3}} }+.. \\ $$$$\Rightarrow\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }+... \\ $$$$\Rightarrow\frac{\pi^{\mathrm{3}} }{\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }+... \\ $$$$ \\ $$

Commented by qaz last updated on 09/May/21

thanks a lot.

$${thanks}\:{a}\:{lot}. \\ $$

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

f(α)=∫_0 ^(π/2) tan^α θdθ=((Γ(((α+1)/2))Γ(((1−α)/2)))/2), f(0)=(π/2) lnf(α)=lnΓ(((α+1)/2))+lnΓ(((1−α)/2))+ln((1/2)) ((f ′(α))/(f(α)))=(1/2)(ψ(((α+1)/2))−ψ(((1−α)/2))) f ′(0)=(π/4)(ψ((1/2))−ψ((1/2)))=0 f ′′(α)=((f(α))/4)(ψ′(((α+1)/2))+ψ′(((1−α)/2)))+((f ′(α))/2)(ψ(((α+1)/2))+ψ(((1−α)/2))) f ′′(0)=(π/8)((π^2 /(sin^2 ((π/2)))))+0=(π^3 /8)=∫_0 ^(π/2) ln^2 (tanθ)dθ

$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{\alpha} \theta\mathrm{d}\theta=\frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)}{\mathrm{2}},\:\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{lnf}\left(\alpha\right)=\mathrm{ln}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{ln}\Gamma\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)+\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=\mathrm{0} \\ $$$$\mathrm{f}\:''\left(\alpha\right)=\frac{\mathrm{f}\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)+\psi'\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{2}}\left(\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)+\psi\left(\frac{\mathrm{1}−\alpha}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:''\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)}\right)+\mathrm{0}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{tan}\theta\right)\mathrm{d}\theta \\ $$

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