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Question-152464

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Question Number 152464 by mathdanisur last updated on 28/Aug/21
Answered by mindispower last updated on 29/Aug/21
((sin(z))/z)=Π_(n≥1) (1−(z^2 /(n^2 π^2 ))) ((sh(t))/t)=Π_(n≥1) (1+(t^2 /(n^2 π^2 ))) ln(((sh(t))/t))=Σ_(n≥1) ln(1+(t^2 /(n^2 π^2 ))) Σ_(n≥1) ∫_0 ^1 ((ln^(2m) (t)Σ_(k≥0) (−1)^k ((t^2 /(n^2 π^2 )))^(k+1) (1/(k+1)))/t) =Σ_(n≥1) Σ_(k≥1) (1/k).(1/((nπ)^(2k) ))∫_0 ^1 (−1)^(k−1) t^(2k−1) ln^(2m) (t)dt =Σ_(k≥1) (((−1)^(k−1) )/k)Σ_(n≥1) (1/((nπ)^(2k) ))∫_0 ^∞ e^(−(2k)x) x^(2m) dx =Σ_(k≥1) (((−1)^(k−1) )/(k+1))Σ_(n≥1) ((1/(nπ)))^(2k) ((Γ(2m+1))/((2k)^(2m+1) )) =Σ_(k≥1) (((−1)^(k−1) )/(k+1)).(((2m)!)/((2k)^(2m+1) .π^(2k) ))ζ(2k)...A ζ(2k)=B_(2k) .(2^(2k−1) /((2k)!))π^(2k) A⇔Σ_(k≥1) (((−1)^(k−1) )/k).(((2m)!)/((2k)^(2m+1) π^(2k) )).2^(2k−1) π^(2k) .(1/((2k)!))B_(2k) =Σ_(k≥1) (((−1)^(k−1) )/k).((2^(2(k−m−1)) (2m)!)/(k^(2m+1) (2k)!))B_(2k) =Σ_(k≥1) (((−1)^(k−1) 2^(2(k−m−1)) (2m)!)/(k^(2(m+1)) (2k)!))B_(2k)
$$\frac{{sin}\left({z}\right)}{{z}}=\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$$$\frac{{sh}\left({t}\right)}{{t}}=\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}+\frac{{t}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$$${ln}\left(\frac{{sh}\left({t}\right)}{{t}}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}{ln}\left(\mathrm{1}+\frac{{t}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}{m}} \left({t}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \left(\frac{{t}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{{k}+\mathrm{1}}}{{t}} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}}.\frac{\mathrm{1}}{\left({n}\pi\right)^{\mathrm{2}{k}} }\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {t}^{\mathrm{2}{k}−\mathrm{1}} {ln}^{\mathrm{2}{m}} \left({t}\right){dt} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}\pi\right)^{\mathrm{2}{k}} }\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{2}{k}\right){x}} {x}^{\mathrm{2}{m}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}+\mathrm{1}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}\pi}\right)^{\mathrm{2}{k}} \frac{\Gamma\left(\mathrm{2}{m}+\mathrm{1}\right)}{\left(\mathrm{2}{k}\right)^{\mathrm{2}{m}+\mathrm{1}} } \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}+\mathrm{1}}.\frac{\left(\mathrm{2}{m}\right)!}{\left(\mathrm{2}{k}\right)^{\mathrm{2}{m}+\mathrm{1}} .\pi^{\mathrm{2}{k}} }\zeta\left(\mathrm{2}{k}\right)…{A} \\ $$$$\zeta\left(\mathrm{2}{k}\right)={B}_{\mathrm{2}{k}} .\frac{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} }{\left(\mathrm{2}{k}\right)!}\pi^{\mathrm{2}{k}} \\ $$$${A}\Leftrightarrow\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}.\frac{\left(\mathrm{2}{m}\right)!}{\left(\mathrm{2}{k}\right)^{\mathrm{2}{m}+\mathrm{1}} \pi^{\mathrm{2}{k}} }.\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} \pi^{\mathrm{2}{k}} .\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)!}{B}_{\mathrm{2}{k}} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}.\frac{\mathrm{2}^{\mathrm{2}\left({k}−{m}−\mathrm{1}\right)} \left(\mathrm{2}{m}\right)!}{{k}^{\mathrm{2}{m}+\mathrm{1}} \left(\mathrm{2}{k}\right)!}{B}_{\mathrm{2}{k}} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \mathrm{2}^{\mathrm{2}\left({k}−\boldsymbol{{m}}−\mathrm{1}\right)} \left(\mathrm{2}\boldsymbol{{m}}\right)!}{{k}^{\mathrm{2}\left({m}+\mathrm{1}\right)} \left(\mathrm{2}{k}\right)!}{B}_{\mathrm{2}{k}} \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 29/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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