Question Number 114777 by mathdave last updated on 21/Sep/20
Answered by maths mind last updated on 21/Sep/20
![(√x)=t⇒dx=2tdt =∫_0 ^(π/2) 2tcl_2 (t)dt =∫_0 ^(π/2) 2tΣ_(k≥1) ((sin(kt))/k^2 )dt =2Σ_(k≥1) ∫_0 ^(π/2) ((sin(kt))/k^2 )tdt ∫_0 ^(π/2) ((sin(kt))/k^2 )tdt=(1/k^2 ){[((−cos(kt))/k)t]_0 ^(π/2) +∫_0 ^(π/2) ((cos(kt))/k)dt} =(π/2)[((−cos(((kπ)/2)))/k^3 )]+(1/k^4 )(sin(k(π/2))) sin(((kπ)/2))=0 if k=2m sin(((kπ)/2))=(−1)^m if k=2m+1 cos(((kπ)/2)) =1 si k=4m,cos(((kπ)/2))=−1 ifk=4m+2 cos(((kπ)/2))=0 if k=4m+1 or 4m+3 so we get Σ_(k≥1) (π/2).[((−cos(((kπ)/2)))/k^3 )]=Σ_(m≥0) (π/2).[((−cos((4m+2)(π/2)))/((4m+2)^3 ))]Σ_(m≥1) −(π/2)((cos(4m.(π/2)))/((4m)^3 )) +Σ_(m≥0) ((sin((π/2)(2m+1)))/((2m+1)^4 )) =Σ_(m≥0) (π/2).(1/(8(2m+1)^3 ))+−(π/2).Σ_(m≥1) (1/(64m^3 ))+Σ_(m≥0) (((−1)^m )/((2m+1)^4 )) =(π/(16))Σ_(m≥0) (1/((2m+1)^3 ))+Σ_(m≥0) (((−1)^m )/((2m+1)^4 )) β(s)=Σ_(n≥0) (((−1)^n )/((2n+1)^s )) Dirichlet Betta function Σ(((−1)^m )/((2m+1)^4 ))=β(4),Σ(1/((2m+1)^3 ))=(ζ(3)−(1/8)ζ(3))=(7/8)ζ(3) we get =(π/(16)).(7/( 8))ζ(3)−(π/(128))ζ(3)+β(4)=((6π)/(128))ζ(3)+β(4)= ∫_0 ^(π/2) cl_2 ((√t))dt=2∫_0 ^1 tcl_2 (t)dt=2.[((6π)/(128))+β(4)] =((12π)/(128))ζ(3)+2β(4)=((3π)/(32))ζ(3)+2β(4)](https://www.tinkutara.com/question/Q114791.png)
$$\sqrt{{x}}={t}\Rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{tcl}_{\mathrm{2}} \left({t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{t}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({kt}\right)}{{k}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({kt}\right)}{{k}^{\mathrm{2}} }{tdt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({kt}\right)}{{k}^{\mathrm{2}} }{tdt}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left\{\left[\frac{−{cos}\left({kt}\right)}{{k}}{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({kt}\right)}{{k}}{dt}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\left[\frac{−{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{{k}^{\mathrm{3}} }\right]+\frac{\mathrm{1}}{{k}^{\mathrm{4}} }\left({sin}\left({k}\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$${sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)=\mathrm{0}\:\:{if}\:{k}=\mathrm{2}{m} \\ $$$${sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)=\left(−\mathrm{1}\right)^{{m}} \:\:{if}\:{k}=\mathrm{2}{m}+\mathrm{1} \\ $$$${cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)\:\:=\mathrm{1}\:{si}\:{k}=\mathrm{4}{m},{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)=−\mathrm{1}\:{ifk}=\mathrm{4}{m}+\mathrm{2} \\ $$$${cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)=\mathrm{0}\:{if}\:{k}=\mathrm{4}{m}+\mathrm{1}\:{or}\:\mathrm{4}{m}+\mathrm{3} \\ $$$${so}\:{we}\:{get}\:\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\pi}{\mathrm{2}}.\left[\frac{−{cos}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{{k}^{\mathrm{3}} }\right]=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\pi}{\mathrm{2}}.\left[\frac{−{cos}\left(\left(\mathrm{4}{m}+\mathrm{2}\right)\frac{\pi}{\mathrm{2}}\right)}{\left(\mathrm{4}{m}+\mathrm{2}\right)^{\mathrm{3}} }\right]\underset{{m}\geqslant\mathrm{1}} {\sum}−\frac{\pi}{\mathrm{2}}\frac{{cos}\left(\mathrm{4}{m}.\frac{\pi}{\mathrm{2}}\right)}{\left(\mathrm{4}{m}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$+\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{m}+\mathrm{1}\right)\right)}{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{3}} }+−\frac{\pi}{\mathrm{2}}.\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{64}{m}^{\mathrm{3}} }+\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{\pi}{\mathrm{16}}\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{3}} }+\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\beta\left({s}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{s}} }\:\:{Dirichlet}\:{Betta}\:{function} \\ $$$$\Sigma\frac{\left(−\mathrm{1}\right)^{{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}} }=\beta\left(\mathrm{4}\right),\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{3}} }=\left(\zeta\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right)=\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$$${we}\:{get} \\ $$$$=\frac{\pi}{\mathrm{16}}.\frac{\mathrm{7}}{\:\mathrm{8}}\zeta\left(\mathrm{3}\right)−\frac{\pi}{\mathrm{128}}\zeta\left(\mathrm{3}\right)+\beta\left(\mathrm{4}\right)=\frac{\mathrm{6}\pi}{\mathrm{128}}\zeta\left(\mathrm{3}\right)+\beta\left(\mathrm{4}\right)= \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cl}_{\mathrm{2}} \left(\sqrt{{t}}\right){dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {tcl}_{\mathrm{2}} \left({t}\right){dt}=\mathrm{2}.\left[\frac{\mathrm{6}\pi}{\mathrm{128}}+\beta\left(\mathrm{4}\right)\right] \\ $$$$=\frac{\mathrm{12}\pi}{\mathrm{128}}\zeta\left(\mathrm{3}\right)+\mathrm{2}\beta\left(\mathrm{4}\right)=\frac{\mathrm{3}\pi}{\mathrm{32}}\zeta\left(\mathrm{3}\right)+\mathrm{2}\beta\left(\mathrm{4}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$