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Question Number 168348 by mnjuly1970 last updated on 08/Apr/22

Answered by mathman1234 last updated on 09/Apr/22

$$\mathrm{plz}\mathrm{help}$$$$\mathrm{if}0<\mathrm{a}<\mathrm{b}\mathrm{and}\mathrm{x}>0\mathrm{proove}\mathrm{that}$$$$\frac{2}{\pi }(1-\frac{\mathrm{a}}{\mathrm{b}})⩽\sup \mid \frac{\sin \left(\mathrm{ax}\right)}{\mathrm{ax}}-\frac{\sin \left(\mathrm{ax}\right)}{\mathrm{ax}}\mid ⩽4(1-\frac{\mathrm{a}}{\mathrm{b}})$$

Commented by mr W last updated on 09/Apr/22

$$please{don}^{\prime }tputyournewquestionin$$$$existingthreadsofother{people}^{\prime }s$$$$questions!pleaseopenanewpost$$$$foryourownquestion!$$

Answered by mindispower last updated on 11/Apr/22

$$\mathrm{\Psi }(1+x)=-\gamma -\sum _{n⩾}\zeta (n+1){(-x)}^n$$$$\mathrm{\Psi }(1-x)=-\gamma -\sum _{n⩾1}\zeta (n+1)x^n$$$$\mathrm{\Psi }(1-x)-\mathrm{\Psi }(1+x)=\sum _{n⩾1}\zeta (n+1)(-{\left(x\right)}^n+{(-x)}^n)$$$$=-2\sum _{n⩾0}\zeta (2n+2)x^{2n+1}=\mathrm{\Psi }(1-x)-\mathrm{\Psi }(1+x)$$$$\Rightarrow \sum _{n⩾0}\frac{\zeta (2n+2)}{2n+3}{x}^{2n+3}=-\frac{1}{2}{\int }_0^{x}t(\mathrm{\Psi }(1-t)-\mathrm{\Psi }(1+t))dt$$$$=-\frac{1}{2}{\int }_0^{x}t(\pi cot\left(\pi t\right)-\frac{1}{t})dt$$$$=-\frac{1}{2}{\int }_0^x\pi tcot\left(\pi t\right)+\frac{x}{2}$$$$x=\frac{1}{2}\iff \frac{1}{4}-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}ucot\left(u\right)\frac{du}{\pi }$$$$=\frac{1}{4}+\frac{1}{2\pi }{\int }_0^{\frac{\pi }{2}}ln\left(sin\left(u\right)\right)du\frac{1}{4}-\frac{1}{2\pi }.\frac{\pi }{2}ln\left(2\right)$$$$=\frac{1}{4}-\frac{1}{4}ln\left(2\right)=\sum _{n⩾1}\frac{\zeta \left(2n\right)}{2n+1}.\frac{1}{2^{2n+1}}$$$$$$$$\Rightarrow \sum _{n⩾1}\frac{\zeta \left(2n\right)}{2^{2n-1}(2n+1)}=1-ln\left(2\right)=ln\left(\frac{e}{2}\right)$$$$$$

Commented by mnjuly1970 last updated on 13/Apr/22

$$gratefulsirpower...$$

Commented by mindispower last updated on 13/Apr/22

$$withePleasursirHaveniceDay$$$$RamadanMobarak$$

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