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Question Number 117380 by mnjuly1970 last updated on 11/Oct/20

$$...provethat...$$$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{1}{2\sqrt{x}}sin({\pi }^{2}x+\frac{1}{x})dx=\frac{1}{\sqrt{8\pi }}$$$$$$$$m.n.1970$$

Commented by mindispower last updated on 11/Oct/20

$$niceone$$

Answered by mnjuly1970 last updated on 13/Oct/20

$$solution::$$$$\mathcal{R}ecall::{\int }_0^{\mathrm{\infty }}sin\left(z^2\right)dz\stackrel{..fresnelintegral..}{=}\sqrt{\frac{\pi }{8}}$$$$$$$$\mathrm{\Omega }\stackrel{t=\sqrt{x}}{=}{\int }_0^{\mathrm{\infty }}sin({\pi }^2{t}^2+\frac{1}{t^2})dt$$$$\pi \mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\pi sin({\pi }^2{t}^2+\frac{1}{t^2})dt\left(i\right)$$$$\pi \mathrm{\Omega }\stackrel{t=\frac{1}{\pi u}}{=}\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\pi sin(\frac{1}{u^2}+{\pi }^2{u}^2)\frac{du}{u^2}$$$$\pi \mathrm{\Omega }={\int }_0^{\mathrm{\infty }}sin({\pi }^2{u}^2+\frac{1}{u^2})\frac{du}{u^2}\left(ii\right)$$$$\left(i\right)+\left(ii\right)::$$$$2\pi \mathrm{\Omega }={\int }_0^{\mathrm{\infty }}(\pi +\frac{1}{x^2})sin[{(\pi x-\frac{1}{x})}^2+2\pi ]du$$$$2\pi \mathrm{\Omega }\stackrel{\pi x-\frac{1}{x}=y}{=}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}sin\left(y^2\right)dy$$$$2\pi \mathrm{\Omega }\stackrel{\mathcal{R}ecall}{=}2\sqrt{\frac{\pi }{8}}\Rightarrow \mathrm{\Omega }=\sqrt{\frac{1}{8\pi }}✓$$$$...\mathrm{♣}\mathcal{M}.\mathcal{N}.july.1970\mathrm{♣}...$$$$\mathrm{♠}peacebeuponyou\mathrm{♠}$$$$$$

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