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Question Number 100368 by bemath last updated on 26/Jun/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cos \left(x^n\right)dx=?$$$$wheren=2k,k\in \mathbb{N},k\ne 0$$

Answered by mathmax by abdo last updated on 26/Jun/20

$${\mathrm{A}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\cos \left({\mathrm{x}}^{2\mathrm{n}}\right)\mathrm{dx}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=2{\int }_0^{\mathrm{\infty }}\cos \left({\mathrm{x}}^{2\mathrm{n}}\right)\mathrm{dx}=2\mathrm{Re}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ix}}^{2\mathrm{n}}}\mathrm{dx}\right)\mathrm{but}$$$$\mathrm{changement}{\mathrm{ix}}^{2\mathrm{n}}=\mathrm{t}\mathrm{give}{\mathrm{x}}^{2\mathrm{n}}=\frac{\mathrm{t}}{\mathrm{i}}=-\mathrm{it}={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\mathrm{t}\Rightarrow \mathrm{x}={\left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\mathrm{t}\right)}^{\frac{1}{2\mathrm{n}}}$$$$={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4\mathrm{n}}}{\mathrm{t}}^{\frac{1}{2\mathrm{n}}}\Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ix}}^{2\mathrm{n}}}\mathrm{dx}={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4\mathrm{n}}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\frac{1}{2\mathrm{n}}{\mathrm{t}}^{\frac{1}{2\mathrm{n}}-1}\mathrm{dt}$$$$=\frac{{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4\mathrm{n}}}}{2\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}{\mathrm{t}}^{\frac{1}{2\mathrm{n}}-1}\mathrm{dt}=\frac{\mathrm{\Gamma }\left(\frac{1}{2\mathrm{n}}\right)}{2\mathrm{n}}\times {\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4\mathrm{n}}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=2\times \frac{\mathrm{\Gamma }\left(\frac{1}{2\mathrm{n}}\right)}{2\mathrm{n}}\times \cos \left(\frac{\pi }{4\mathrm{n}}\right)\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\frac{1}{\mathrm{n}}\mathrm{\Gamma }\left(\frac{1}{2\mathrm{n}}\right)\cos \left(\frac{\pi }{4\mathrm{n}}\right)\mathrm{rest}\mathrm{to}\mathrm{find}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}....\mathrm{be}\mathrm{continued}...$$$$$$

Commented by mathmax by abdo last updated on 26/Jun/20

$$\mathrm{thnk}\mathrm{you}\mathrm{sir}.$$

Commented by Ar Brandon last updated on 26/Jun/20

wow, cool !

Commented by maths mind last updated on 26/Jun/20

$$nicesiricomplet\mathrm{\Gamma }\left(\frac{1}{2n}\right)=\frac{\pi }{\mathrm{\Gamma }(1-\frac{1}{2n})sin\left(\frac{\pi }{2n}\right)}$$$$A_n=\frac{1}{n}\frac{.\pi }{\mathrm{\Gamma }(1-\frac{1}{2n})sin\left(\frac{\pi }{2n}\right)}cos\left(\frac{\pi }{4n}\right),\frac{1}{n}=t$$$$\Rightarrow A_n=\frac{\pi t}{\mathrm{\Gamma }(1-\frac{t}{2})sin\left(\frac{\pi t}{2}\right)}cos\left(\frac{\pi }{4}t\right)$$$$\underset{t\to 0}{\lim }\frac{\pi t}{sin\left(\frac{\pi t}{2}\right)}=\frac{2}{\pi },weget\frac{2}{\pi \mathrm{\Gamma }\left(1\right)}cos\left(0\right)=\frac{2}{\pi }$$$$$$

Commented by Ar Brandon last updated on 27/Jun/20

Are you talking to yourself, Sir ?��

Commented by mathmax by abdo last updated on 27/Jun/20

$$\mathrm{i}\mathrm{talk}\mathrm{to}\mathrm{sir}\mathrm{mind}\mathrm{what}\mathrm{happen}\mathrm{to}\mathrm{you}...$$

Commented by Ar Brandon last updated on 27/Jun/20

Oh sorry ! �� I just feel you both are alike. ����

Commented by maths mind last updated on 27/Jun/20

$$withepleasursir$$

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