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Question Number 195320 by Erico last updated on 30/Jul/23

$${\mathrm{I}}_{\mathrm{n}}={\int }_0^{+\mathrm{\infty }}{t}^{-2t}{sin}^{2n}tdt$$$$\mathrm{Prove}\mathrm{that}{\mathrm{I}}_{\mathrm{n}}=\frac{1}{1-e^{-2\pi }}{\underset{0}{\int }}^{\pi }{e}^{-2t}{sin}^{2n}tdt$$$$\mathrm{and}{\mathrm{I}}_{\mathrm{n}}\underset{\mathrm{\infty }}{\backsim }\frac{1}{2sh\left(\pi \right)}\sqrt{\frac{\pi }{n}}$$

Answered by witcher3 last updated on 31/Jul/23

$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{1-{\mathrm{e}}^{-2\pi }}{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_{\mathrm{k}\pi }^{(\mathrm{k}+1)\pi }{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{t}-\mathrm{k}\pi =\mathrm{x}\Rightarrow \mathrm{dt}=\mathrm{dx}$$$$=\sum _{\mathrm{k}⩾0}{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{x}-2\mathrm{k}\pi }{\sin }^{2\mathrm{n}}(\mathrm{x}+\mathrm{k}\pi )\mathrm{dx}$$$$=\sum _{\mathrm{k}⩾0}{\mathrm{e}}^{-2\mathrm{k}\pi }{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{x}}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$$$=\frac{1}{1-{\mathrm{e}}^{-2\pi }}{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{x}}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}...$$$${\mathrm{J}}_{\mathrm{n}}={\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{x}}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{By}={\left[\frac{{\mathrm{e}}^{-2\mathrm{x}}}{-2}{\sin }^{2\mathrm{n}}\left(\mathrm{x}\right)\right]}_0^{\pi }+\frac{1}{2}{\int }_0^{\pi }2\mathrm{ncos}\left(\mathrm{t}\right){\sin }^{2\mathrm{n}-1}\left(\mathrm{t}\right){\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt}$$$$=\mathrm{n}{\int }_0^{\pi }\cos \left(\mathrm{t}\right){\sin }^{2\mathrm{n}-1}\left(\mathrm{t}\right){\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt}$$$$=\mathrm{n}\left[\frac{{\mathrm{e}}^{-2\mathrm{t}}}{-2}\cos \left(\mathrm{t}\right){\sin }^{2\mathrm{n}-1}\left(\mathrm{t}\right)\right]+\frac{\mathrm{n}}{2}{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{t}}[-{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)+(2\mathrm{n}-1){\cos }^2\left(\mathrm{t}\right){\sin }^{2\mathrm{n}-2}]\mathrm{dt}$$$$=\frac{\mathrm{n}}{2}{\int }_0^{\pi }[(-2\mathrm{n}){\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)+(2\mathrm{n}-1){\sin }^{2\mathrm{n}-2}]{\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt}$$$$=-{\mathrm{n}}^2{\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{t}}{\sin }^{2\mathrm{n}}\left(\mathrm{t}\right)\mathrm{dt}+\frac{\mathrm{n}(2\mathrm{n}-1)}{2}{\int }_0^{\pi }{\sin }^{2\mathrm{n}-2}\left(\mathrm{t}\right){\mathrm{e}}^{-2\mathrm{t}}\mathrm{dt}$$$$=-{\mathrm{n}}^2{\mathrm{J}}_{\mathrm{n}}+\frac{\mathrm{n}(2\mathrm{n}-1)}{2}{\mathrm{J}}_{\mathrm{n}-1}$$$$\iff (1+{\mathrm{n}}^2){\mathrm{J}}_{\mathrm{n}}=\frac{\mathrm{n}(2\mathrm{n}-1)}{2}{\mathrm{J}}_{\mathrm{n}-1}\iff \frac{{\mathrm{J}}_{\mathrm{n}}}{{\mathrm{J}}_{\mathrm{n}-1}}=\frac{\mathrm{n}(2\mathrm{n}-1)}{2(1+{\mathrm{n}}^2)}$$$$\Rightarrow \underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}\frac{{\mathrm{J}}_{\mathrm{k}}}{{\mathrm{J}}_{\mathrm{k}-1}}=\frac{{\mathrm{J}}_{\mathrm{n}}}{{\mathrm{J}}_0}=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}\frac{\mathrm{k}(2\mathrm{k}-1)}{2(\mathrm{k}+\mathrm{i})(\mathrm{k}-\mathrm{i})}$$$$=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}\frac{2\mathrm{k}(2\mathrm{k}-1)}{4(\mathrm{k}+\mathrm{i})(k-i)}=\frac{\left(2n\right)!}{4^{\mathrm{n}}}.\frac{\mathrm{\Gamma }(1-\mathrm{i})\mathrm{\Gamma }(1+\mathrm{i})}{\mathrm{\Gamma }(\mathrm{n}+1-\mathrm{i})\mathrm{\Gamma }(\mathrm{n}+1+\mathrm{i})}$$$$\mathrm{\Gamma }(1+\mathrm{i})\mathrm{\Gamma }(1-\mathrm{i})=\mathrm{i}\mathrm{\Gamma }(1-\mathrm{i})\mathrm{\Gamma }\left(\mathrm{i}\right)=\frac{\mathrm{i}\pi }{\sin \left(\mathrm{i}\pi \right)}$$$$=\frac{\pi }{\mathrm{sh}\left(\pi \right)}$$$$\mathrm{\Gamma }(\mathrm{n}+1\underset{-}{+}\mathrm{i})\sim \sqrt{2\pi \mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}{\mathrm{n}}^{\mathrm{n}\underset{-}{+}\mathrm{i}}$$$$\mathrm{\Gamma }(2\mathrm{n}+1)={\mathrm{n}}^{2\mathrm{n}}.2^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}2\sqrt{2\pi \mathrm{n}}$$$$\frac{{\mathrm{J}}_{\mathrm{n}}}{{\mathrm{J}}_0}\sim \frac{2\sqrt{2\pi \mathrm{n}}.4^{\mathrm{n}}.{\mathrm{n}}^{2\mathrm{n}}{\mathrm{e}}^{-2\mathrm{n}}}{2\pi \mathrm{n}.{\mathrm{e}}^{-2\mathrm{n}}{\mathrm{n}}^{2\mathrm{n}}.4^{\mathrm{n}}}.\frac{\pi }{\mathrm{sh}\left(\pi \right)}=\frac{2\pi }{\mathrm{sh}\left(\pi \right)\sqrt{2\pi \mathrm{n}}}$$$${\mathrm{J}}_0={\int }_0^{\pi }{\mathrm{e}}^{-2\mathrm{x}}=\frac{1-{\mathrm{e}}^{-2\pi }}{2}$$$${\mathrm{J}}_{\mathrm{n}}\sim \frac{1-{\mathrm{e}}^{-2\pi }}{2\mathrm{sh}\left(\pi \right)}.\frac{2\pi }{\sqrt{2\pi \mathrm{n}}}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{1-{\mathrm{e}}^{-2\pi }}{\mathrm{J}}_{\mathrm{n}}\sim \frac{1}{\mathrm{sh}\left(\pi \right)}.\sqrt{\frac{\pi }{2\mathrm{n}}}$$$$$$$$$$$$$$$$$$

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