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Question Number 123454 by mnjuly1970 last updated on 25/Nov/20

$$...nicecalculus...$$$$calculate:::$$$$\mathrm{\Omega }\stackrel{???}{=}{\int }_0^{\mathrm{\infty }}\sqrt{x}\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(cos\left(\frac{x}{2^n}\right)\right)dx$$

Answered by Olaf last updated on 25/Nov/20

$$\sin 2\theta =2\sin \theta \cos \theta$$$$\cos \theta =\frac{\sin 2\theta }{2\sin \theta }$$$$\mathrm{Let}\theta =\frac{x}{2^n}$$$$\cos \left(\frac{x}{2^n}\right)=\frac{\sin \left(\frac{x}{2^{n-1}}\right)}{2\sin \left(\frac{x}{2^n}\right)}$$$$\underset{n=1}{\prod ^p}\cos \left(\frac{x}{2^n}\right)=\underset{n=1}{\prod ^p}\frac{\sin \left(\frac{x}{2^{n-1}}\right)}{2\sin \left(\frac{x}{2^n}\right)}=\frac{\sin x}{2^p\sin \left(\frac{x}{2^p}\right)}$$$$2^p\sin \left(\frac{x}{2^p}\right)\underset{\mathrm{\infty }}{\sim }{2}^p\times \frac{x}{2^p}=x$$$$\Rightarrow \underset{n=1}{\prod ^{\mathrm{\infty }}}\cos \left(\frac{x}{2^n}\right)=\frac{\sin x}{x}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\sqrt{x}\underset{n=1}{\prod ^{\mathrm{\infty }}}\cos \left(\frac{x}{2^n}\right)dx$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\sqrt{x}.\frac{\sin x}{x}.dx$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sin x}{\sqrt{x}}dx$$$$\mathrm{Let}t=\sqrt{x},dt=\frac{dx}{2\sqrt{x}}$$$$\mathrm{\Omega }=2{\int }_0^{\mathrm{\infty }}\sin \left(t^2\right)dt$$$$\mathrm{With}{\int }_0^{\mathrm{\infty }}\sin \left(t^2\right)dt=\frac{1}{2}\sqrt{\frac{\pi }{2}}$$$$\left(\mathrm{Fresnel}\mathrm{integral}\right)$$$$\Rightarrow \mathrm{\Omega }=\sqrt{\frac{\pi }{2}}$$

Commented by mnjuly1970 last updated on 26/Nov/20

$$thankyoumrolaf.excellent$$

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